高中数学建模论文 精品

全国中学生数理化学科能力展示活动优秀论文精选
一、初试获奖比例：一等奖5%；二等奖10%；三等奖15%。
二、复试
复试资格：初试获三等奖及以上者入围复试。
复试形式：数学：提交数学建模论文；物理、化学：提交实验报告或建模论文。建模论文或实验报告单独评奖。
三、论文提交方式：以电子邮件发送至指定邮箱，学生自行提交，发送邮件主题须注明：学生姓名、学校 、年级、
学科、联系方式、指导老师。未注明相关信息视为无效提交，不计成绩。
四、论文提交指定邮箱：bjshulihua@ ，截止时间：2013年4月10日。
五、总决赛入围资格：初试成绩60%+论文成绩40%进行综合排序选拔。
4月底 公布论文获奖名单，5月中旬评选总决赛入围名单及颁发论文获奖证书，总决赛报名时间：6月份；比赛时
间：7月下旬（具体时间地点待定）。
六、建模论文写作说明：
（一）、建模论文的标准组成部分
建模论文作为一种研究性学习有意义的尝试，可以 锻炼学生发现问题、解决问题的能力.一般来说，建模论文的标
准组成部分由论文的标题、摘要、正文、 结论、参考文献等部分组成.现就每个部分做个简要的说明.
1. 题目
题目给评 委第一印象，一定要避免指代不清，表达不明的现象.建议将论文所涉及的模型或所用的计算方式写入题
目.如“用概率方法计算商场打折与返券的实惠效应”.
2. 摘要
摘要是论文中 重要的组成部分.摘要应该使用简练的语言叙述论文的核心观点和主要思想.如果你有一些创新的地
方， 一定要在摘要中说明.进一步，必须把一些数值的结果放在摘要里面，例如：“我们的最终计算得出，对于消费者
来说，打折比返券的实惠率提高了23%.”摘要应该最后书写.在论文的其他部分还没有完成之前,你 不应该书写摘要.
因为摘要是论文的主旨和核心内容的集中体现，只有将论文全部完成且把论文的体系罗 列清楚后，才可写摘要.
摘要一般分三个部分.即：用什么模型，解决什么问题；通过怎样的思路来解决问题；最后结果怎么样.
3. 正文

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全国中学生数理化学科能力展示活动优秀论文精选
正文是论文的核心 ，也是最重要的组成部分.在论文的写作中，正文是从“提出问题—分析问题—选择模型—建立
模型—得 出结论”的方式来逐渐进行的.其中，提出问题、分析问题清晰简短.而选择模型和建立模型应该是目标明确、< br>数据详实、公式合理、计算精确.在正文写作中，应尽量不要用单纯的文字表述，尽量多地结合图表和数据 ，尽量多
地使用科学语言，这会使得论文的层次上升.
4. 结论
论文的 结论集中表现了这篇论文的成果，可以说，只有论文的结论经得起推敲，论文才可以获得比较高的评价.结
论的书写应该注意用词准确，与正文所描述或论证的现象或数据保持绝对的统一.并且一定要对结论进行自我点 评，
最好是能将结论推广到社会实践中去检验.
5. 参考资料
在论文中，如果使用了其他人的资料.必须在论文后标明引用文章的作者、应用来源等信息.
（二）、建模论文的写作步骤
1. 确定题目
选择一个你感兴趣的生活中 的问题作为研究对象，并根据研究对象设置论文题目.最好是找一位或几位老师帮助安
排研究课题.在确 定好课题后，应该写一个写作计划给指导老师看看，并征求他们对该计划的建议.
2. 开展科研课题
去图书馆、互联网上查阅与课题相关的资料，观察有关的事件，收集与课题相关的信息.同时如 果有条件的话，可
以去拜访相关领域的专家和学者.然后将前期所收集到的资料与自己所学的相关知识组 织在一起，进行论文的结构论
证.完成这些工作后，你应该要制定一个课题时间安排表，这样能保证书写 论文的循序渐进.记住在开始写论文后一定
要不断地和老师、家长进行沟通，让老师和家长斧正论文中出 现的明显错误，并能提出一些更好的研究建议.在论文
写作结束以后，一定要得出结论.记住，在论文的 结果出来后，有可能得出的结果与假设并不相符，这个并不重要，
不要强行改变结果来迎合假设.只要你 在论述过程中严格地按照科学方法进行，你的论文还是相当有价值的.最后，需
要很好地写一份摘要.摘 要的字数应该是论文字数的十分之一左右.
3. 完成论文写作
完整的论文在完成 以上步骤之后就可以新鲜出炉了，完成论文后，一定要再看一遍自己的论文有无错别字、计算
错误、图形 的移位或偏差等.最后，在论文的结尾处应该写上感谢的话，感谢帮助你完成这篇论文的所有人。
七、全国中学生数理化学科能力展示活动相关信息，家长可在“理科学科能力评价网”查询。
网址： 北京组委会电话：62202861 62205108

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全国中学生数理化学科能力展示活动优秀论文精选
八、往届优秀建模论文及实验报告范文可到点击百度链接网盘下载。
高中组：http:relink?shareid=214534&uk=1244926876
初中组：http:relink?shareid=214535&uk=1244926876
九 、公益讲座PPT：http:relink?shareid=238984&uk=1244926876

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全国中学生数理化学科能力展示活动优秀论文精选

简评北京主要路口交通状况及改进措施
北京首师大附中高一年级：马森

摘要：现如今北京车辆越来越多，主要路口天天暴堵，人民深受其苦。现测定红绿灯时间，观察周
边环境 分析长安街附近两路口的交通状况，致堵原因，提出我设想的改进方案。
关键词：红绿灯，公交站，堵车。
正文：
前言：
问题背景：红绿灯及公 交车的出现，给交通以极大的便利。红绿灯让繁华都市的交通更加有秩序，
根据1968年联合国?道路 交通及道路标志信号协定?，绿灯是通行信号，面对绿灯的车辆可通行、左
转、右转，除非另一种标志禁 止某种转向，左右转弯车辆都必须让合法地正在路口内行驶的车辆和
过人行横道的行人优先通行。红灯是 禁行信号，面对红灯的车辆必须在交叉路口的停车线后停车。
黄灯是警告信号，面对黄灯的车辆不能越过 停车线，但车辆已十分接近停车线而不能安全停车时可
进入交叉路口。我国?道路交通安全法?也规定， 黄灯亮时，若车前轮已过停车线则可继续通行，若
车前轮未过停车线则必须将车停在停车线后。城市公共 交通最早出现在英国，1829年伦敦出现了第
一辆公共马车，至今已有180年历史，期间公交事业经 历了发展、兴旺、衰退和目前的复兴阶段。
但对现在人口和私家车数量都日益增长的北京市来说，一个红 绿灯的设计或是公交车线路、车站的
位置都可能影响整条路的交通，本文探讨的就是这一问题。
研究价值：研究长安街路口的交通状况和拥堵原因，提出合理化建议，有助于改进交通，又因为长
安街 沿线是北京的中心部位，更具有代表价值。
概括论文主要内容、研究结论：
长安街西单路口 红绿灯设计较为合理，但西单路口往北的第一个路口红绿灯的设计欠合理，另外中
间有几个公交车站，严 重阻碍交通。
使用方法：
恰表记下红绿灯的时间。
骑自行车绕路口一圈及路口周围100米，观察致堵原因。
询问交通协管员及路人。
主体：
长安街西单路口：
晚
（1）. 路口平面图：
高

峰

拥

西单图书大厦
堵
西单文化广场

中国银行
易拥堵



中国光大银行
商 场



（2）. 红绿灯时间（中午12：00）
长 安 街

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全国中学生数理化学科能力展示活动优秀论文精选

红
绿
东西向(主)
91s
90s
南北向(支)
100s
75s
车流量(一个绿灯内)
东西向 214辆 (单向)
南北向 156辆 (单向)
（3）. 分析
西单是北京的繁华区，因此这个路口是长安街沿线上的重要路口，在长安街交通上占有举足
轻重的地位。 这个路口设计得较为合理，东西方向是长安街主干线，因此红灯稍短些，绿灯稍长些，
在早晚高峰时间段 便于车辆通行。对于南北方向，由南往北方向较为合理，路口南边道路很宽，且
车站不会阻碍交通，红灯 时车辆较为分散，不会形成大规模拥堵现象，而路口北边由于有大面积的
商业区，车辆行人都很多，且道 路狭窄，易造成堵车（具体原因谈到下个路口会讲到），绿灯时间稍
短有利于路口北车辆的疏散，这样由 南向北不会太堵。而由北向南欠合理，一旦晚高峰到来，路北
道路宽，车少，红灯将大量车辆积压在狭窄 的北边车道，导致晚5：00至6：30路口北车道上堵车
将近500米。这个路口的另一个问题是由东 向西开的车辆右转时和直行的南北向车辆打架。这个路
口无论东西向还是南北向车流辆都非常大,有时右 转车辆强行转弯,或是发生剐蹭,或导致大量南北
向车辆和东西向车辆被堵,严重时还有可能波及西单图 书大厦前的小路口。
综述：这是个重点路口，交通信号灯时间及车站位置都较合理，
有四个 平面立交，交通状况良好，只是晚高峰由北往南的车辆在路北易发生拥堵，以及由东向西的
右转处易拥堵
路人评价：大多数路人认为路口设计合理。
西单路口往北第一个路口：
路口平面图





堵



商场区
堵
车
西单商场

站
红绿灯时间（12：30）

红
绿
南→北
56s
68s
北→南
57s
68s
东→西
100s
35s
西→东
95s
30s
分析
这个路口的设计 不甚合理，情况也比较复杂，先从商场区旁边说起，南北方向就是西单路口往北的
这条路，上文已经说过 南北方向的路非常窄，由南往北两条车道，由北往南两条车道，这条路从西
单路口往北第一个平面立交入 口到该路口在早高峰的时候暴堵，经常是连动都动不了，原因之一就
是西单商场旁边的这个车站，这个车 站设计的很不是位置，为什么说呢？由于建设北京城的需要，
西单商场占用了很大一块地方，导致这条道 路被急剧压缩，这条路没有辅路，所以就导致车站被直
接设在主路上，而这又不是一路车进站，有五、六 路，所以结果就是本来就窄的两条车道的其中一
条长时间被几辆大车牢牢占据，别的车根本不能走，而且 不少司机还偷懒，大车司机进站前走左边

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全国中学生数理化学科能力展示活动优秀论文精选
车道，要进站时才并到右边 车道，可是狭窄的道路根本并不过去，这就导致后面的车经常动弹不得，
直接反映在道路上就是长达50 0米的车队，早高峰时还进场堵到西单路口，我想这也是西单路口南
北向绿灯短的原因吧。
另 一个堵车原因就要归罪于路口的信号灯设置了，东西向的设置很人性化，由东向西的绿灯要长5
秒，这是 因为从东边左转的车辆经常很多，先开5秒绿灯让左转车先过避免堵塞，事实上这是在道
路狭窄的情况下 很好用的办法，但是东西向的绿灯是在是太短了，这使晚高峰时由东往西的道路上
经常也要排二，三百米 的长队，既然东西向的绿灯可以长5秒，我不明白为什么南北向的不行？事
实上不这样做的结果是容易加 剧道路拥堵，早高峰时由南向北的两条车道一条被左转车占据，像由
北向南的直行车根本不肯让左转车， 这就使左边车道上排着一大串左转车，而且还是有车辆快到路
口才并线，这就使车站到路口的300米又 变成了停车场，直接后果就是整条路都变成了停车场，给
赶早高峰的市民带来极大不便。
综述 ：本路口信号灯的设置极其有问题，早晚高峰都是一通狂堵，再加上一个不恰当的车站，使“西
单”在人 民心中就意味着“堵车”。
合理化建议：
西单路口：由北向南的绿灯应延长5～10秒，以便疏通晚高峰时北边过多的车辆。
西单路口往北的第一个路口：由南向北绿灯延长5～10秒，
或取消左转弯，这样可以使早晚高峰时的左转弯车先过，避免左转车和直行车卡住。东西向绿灯再
延长1 0秒。以便更好地疏通车辆。
对造成拥堵的车站的建议：这个车站应该是为了西单商场
而设 计的，不过我还是认为应撤掉这个车站（改站），或者改到路口北边也可以，毕竟不应为商场而
毁掉长安 街的交通。
反思与评价：
应用前景：
西单路口向北的第一个路口由北向南绿灯延 长10秒应该很好办，东西向延长绿灯时间也不难，
但由于人们的需要和长久的习惯，撤掉车站可能有一 定的难度，随之而来的就是另一路口的红绿灯
变化也有一定的难度。
问题：
由于时 间仓促，我只在中午，在我家附近的长安街实验了一次，并没有在早晚高峰进行实验，
所得数据可能不是 很有代表性，另外要彻底改善北京的交通，还应在其他拥堵路口，在不同时间进
行多次实验。
参考资料：
《道路交通那个及道路标志信号协定》
《道路交通安全法》











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全国中学生数理化学科能力展示活动优秀论文精选

投篮中的数学问题
北京四中高一年级：毛天奇 指导老师：杨虎
摘要：
我是一名篮球运动爱好者。本文试图使用我们所学过的数学和物理知识，包 括方程、函数、矢
量（向量）及 运动和力等，基于对定点远投空心篮运动过程的分析，建立初步的数学 模形，并对出
手高度、投篮力的方向和角度进行分析，得出了提高投篮命中率的有关结果，可供训练时参 考。
关键词：力 方向 矢量 函数 分解 篮球
前言：
远投3分球在现代 篮球比赛中十分重要。我是一个篮球投手，因此对投篮运动的分析很感兴趣，
在学习了有关数理知识后， 试图用建立数学模型的方法，分析在某一距离投篮，手对球的作用力大
小方向如何，如何让人能够控制好 出手力度，才能使篮球的命中率最大，从而使我们的训练和比赛
有科学的依据，而不只是靠运气。 对于定点远投，篮球出手高度、速度和角度是影响命中率的三大因素。本文在分析投蓝的运动学过
程 的基础上，运用相关数学知识建立了定点远投空心篮的数学模型，讨论了各关键因素之间的关系，
并计算 有关的数据。
主体
基本数据：篮球场长度为28米，宽15米，篮筐高3.05米，g可取 10Nkg。篮球质量一般约为
0.6kg。
我们可以用平面代替立体，暂且不考虑三维的情况，即默认无侧面刮来的风。
那么可以建立2维直角坐标系对问题进行简化研究，如下图：

y
B
F
C
A
x

图1

o

3.那么，我们可以进行有关于A点所施的力F的计算了。之前，为了计算简便，我们将受力 时间t
设为0.1s（可以知道这一时间很短，为简化问题，设它为0.1s），下面我们将在这一将球 ，其轨迹

7

全国中学生数理化学科能力展示活动优秀论文精选 < br>和球筐中心抽象到同一平面内的坐标系中计算F和人与篮筐距离L及人跳起投篮时球的高度h之间
的关系。
解： 将F正交分解：设F与水平方向交角为b。水平分力F1，竖直分力F2。F=ma
（为过程简略部分省略单位）
V（水平）= F10.6kg×0.1s= F16 T=L V（水平）V（竖直）=（F2-0.6kg×g）0.6kg×0.1s= F26-1
（1）
2
又有x=vt-12gt
2
3.05-h=V（竖直）× T-5T (2)
连立（1）（2）由T=T的关系可得：（F1=F×cosb F2=F×sinb）
222
tanb×L×F×cosb-6L×F×cosb-180L
22
F×cosb
=3.05-h

2222222312
-6L×cosb+ （36L×cosb -2196L cosb+720 cosb×h×L+720tanb×cosb×L ）
F=
222
6.1 cosb-2 cosb×h-2tanb×L×cosb b（min）=arctan
（3.05-1.8）L= arctan1.25L（取不到）
b∈（arctan524，∏2） （实际可能在该域波动，但要有一定角度。） （不妨设L=6 h=1.8，）
将得到的函数输入几何画板中，得到的在定义域中的图像很明显单调递增。（由于横坐 标问题，用了
弧度制，对值域影响较大，它的图像看起来就像一条直线，其实不是。）
8
6
4
2
-55
-2

图2
让我们继续扩展一些：如果有来自你左方或右方的风，不用害怕投不进，在前面的基础上在与刚刚
的平面垂直的平面中改变力的方向即可。简单来说，就是往和风向相反的方向用力投篮，这也就是
最基本的对于速度合成与分解的矢量三角形。
红：风的速度。蓝：球的速度。黑：构成矢量三角形的辅助向量。绿：对球力的方向，给球的速度。

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全国中学生数理化学科能力展示活动优秀论文精选
图3



5．如果你愿意用这公式，在确定好你爱的出手角 度后，不难得出一个确定的力，但是，对于力的大
小机体恐怕没有更好的概念，控制不好这个力，但是将 N这个单位直观化不就好了？易知一个鸡蛋
重量为0.5N，设置新的力的单位记为JD，1JD读作1 鸡蛋力1JD=0.5N。
得出力的值后，换算为鸡蛋力，力的大小就直观了，找几个鸡蛋举举看，感 受一下，对于投篮力度
的掌握大有益处。

讨论
下面我们对于上 面推出的关系式作解释。作为投手，首先要确定自己的投篮的弧度，不忽略风，
若逆风投篮，由于弧度越 大，相对球在空中时间越长，这样风对于球的影响也就越大，故应减小投
篮角度。反之亦可作类似论述。 关系式是理想状况下的，忽略了风，因此，角度由你的喜好而定，
通过方程可求出两个值。都可以吗？显 然不是，观察第一个图，有两点经过了篮筐的高度，我们要
的是右边的，由前面的式子可得，两个值选择 大的那个值即可
这也是本关系式缺陷所在，要取舍一下。我们得到了一个函数，得到的结论告诉我们， 在理想状况
下，你投篮角度越大，力度也要增大。这一研究得出的结论如果运用好，无疑可以将这门体育 运动
变得更加理性化，让投手们对自己的投篮更有信心，而不是去求神拜佛。这一理论，将对于篮球教< br>学大有裨益。但是它不是万能的，除了上面提到的小缺陷外，这个结论忽略了实际中不可或缺的影
响因素，因此会有偏差，保证必进是不现实的，它只能增大进球概率。
注：本文没有引用任何专业的文献资料。工具：《几何画板》












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圆形广场的地下灯排布问题
北京十一学校高一年级：肖菁
摘要：本文为一个广场设 计了安装地下灯的方案。本文通过构建理想模型，利用圆的半径、圆心角、
弓形面积及三角函数、反三角 函数进行研究，综合美观、节约、光照率等问题，得到一个较合适的
方案。
关键词：地下灯排布、光照率、三角函数、圆
壹：问题的提出：
某将要修建的商业 街欲在其中包含一个半径约为11米圆形广场，该广场的地下将安装数盏地下灯
（最好不超过50盏，不 少于30盏），且希望这些地下灯排列有序，尽量不浪费光源（即它们的光照
面尽量不重叠，重叠率（= 重叠面积总面积）不要大于2%），且不要有连续大片无光照区域（光照
率（=总光照面积总面积）不要 小于80%）。本文针对此，设计了一种方案。
贰：建立模型：
把该广场看作一个大圆，一个地下灯的光照面为一个小圆。
叁：制定方案
在广场正 中安装一盏灯，再按光照范围围绕它安装三圈灯。
设大圆半径为R，取r=R7为小圆半径。即在大圆内 有一个与
之同圆心的小圆，再围绕它放置三圈小圆，如图1。

肆：研究计算：
如图2设最外圈一个小
圆对应以O为圆心，6r为
半径的圆的圆心角为a。
三角形AO
1
O中，
tan
(a2)=r6r=16，

2[1][2]
∵
tan
a=2
tan
(a2) [ 1-
tan
(a2)],
∴
tan
a=1235,a≈18.92°，
36018.92=19余0.52，∴最外圈安19盏灯较好。

同理，则如图3设从外数第二圈一个小圆对应以O为圆心，4r为半径的圆的圆心角为b。
三角形BO
2
O中，
tan
(b2)=r4r=14，
∵
tan
b=2
tan
(b2) [ 1-
tan
2
(b2)],
∴
tan
b=815,b≈28.07°，
36028.07=12余23.16，∴从外数第二圈安13盏灯较好。




同理，则如图4设从内数第二圈一个小圆对
应以O为圆心，2r为半径的圆的圆心角为c。
三角形CO
3
O中，
tan
(c2)=r2 r=12，

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全国中学生数理化学科能力展示活动优秀论文精选

2
∵
tan
c=2
tan
(c2) [ 1-
tan
(c2)],
∴
tan
c=43,c≈53.13°，
36053.13=6余41.22，∴从内数第二圈安7盏灯较好。
且从外数第二圈的小圆有互相重叠的部分，且为美观，使重叠部分
均匀分布，则局部如图5。
设一个重叠部分面积为S
1
。如图6。
sin
b
1
=O
2
B
1
O
2
O，
∴O
2
B
1
=O
2
O*
sin
b
1
，
其中b
1
=360(13*2)，O
2
O=4r，
∴O
2
B
1
≈0.96r，
∴
cos
d= O
2
B
1
O
2
D≈0.96，
d≈16.26°，
2
∴S
1
4=d∏r360- O
2
B
1
* B
1
D2
222
=d∏r360- 0.5* O
2
B
1
*
tan
d≈0.0075r，
2
∴13S
1
=0.39r，
同理从内数第二圈的小圆有互相重叠的部分，且为美观，使重叠部分均匀分布，则局部如图7。
设一个重叠部分面积为S
2
。如图8。
sin
c1=O
3
C
1
O
3
O，
∴O
3
C
1
=O
3
O*
sin
c1，
其中c
1
=360(7*2)，O
3
O=2r，
∴O
3
C
1
≈0.87r，
∴
cos
e= O
3
C
1
O
3
E≈0.87，
e≈29.54°，
2
∴S
2
4=e∏r360- O
3
C
1
*C
1
E2
222
=e∏r360- 0.5*O
3
C
1
*
tan
e≈0.0433r，
2
∴7S
2
=1.21r，

又19+13+7+1=40（盏），
2222
光照面积=40*∏r-0.39 r-1.21r=(40∏-1.6)r，
22
光照率=(40∏-1.6)r49∏r≈80.6%，
222
重叠面积=0.39 r+1.21r=1.6 r，
22
重叠率=1.6 r49∏r≈1.0%，
综上所述，在半径为R的大圆内铺半 径为R4的小圆，共需40盏灯。其中，光照率约为80.6%，重
22
叠率约为1.0%。其 中，R=11m，∏r≈7.76m。
伍：讨论分析：
本文通过构建理想模型，解决现实生 活中的问题，考虑了多方面的因素。但由于对一些数据进行
近似处理，因而存在一定的误差，但此研究对 类似的排布问题有着一定的参考和启发价值。
陆：结论&应用：
本文给该广场设计的方案是 选用光照面积为7.76平方米的地下灯40盏。排布方式如下图1，
外圈19盏，从外数第二圈13盏 ，从内数第二圈7盏，中间
一盏，其中间两圈中每圈相邻两灯的光照面有重叠。此方案
的光照率 约为80.6%，重叠率约为1.0%。
此结论可应用于此广场的地下灯排布。同时，此研究思路也

11

全国中学生数理化学科能力展示活动优秀论文精选
可为其他的地下灯、地上路灯、凉亭、树木的排布（种植）提供参考借鉴。
另外，从节约考虑 ，使用此方案时可不同时开所有灯，可只开其中某圈，或其他一部分灯，从而达
到既节省又美观的效果。
柒：参考文献：
[1]人民教育出版社 课程教材研究所 中学数学教材实验研究组 编著，普通高中课程标准实验教科
书 数学 必修四，2007年4月
[2]乔家瑞 国运之 裘大彭 编著，超级高中数理化生公式定理，2007年9月
附：
使用计算器计算反三角函 数的方法：1.输入三角函数数值，2.按下第二功能键，再按下
sin
(或
cos< br>或
tan
)键，3.按输入（或等于）键，即可得角度（注：1、2两步可颠倒，具体视 自己的计算器而定）。
























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关于合适教室形状的探究
北京第九中学高二年级：马原
摘要：本文在保证听课质量 、保护学生视力健康、满足目前班级平均人数上课需要前提下，通过数
学模型和几何分析方法，探讨了各 种形状教室的最小面积要求、视听和采光效果，为教室形状的选
择提供基本的科学依据。同时，在探究过 程中，也给出了桌椅的最佳排列，窗户的最佳面积分配方
案。
关键词：教室形状， 教室最佳面积， 声场分布， 采光系数。
1.引言：
记得刚升入高一步 入北京四中的校园，我就被那新颖的正六边形教室所吸引，在这里上课的一年
里，也感受到了这种建筑布 局带来的舒适。后来我了解到，国外的一些学校也采用了正六边形教室。
走访国内的多所学校，我发现大 部分仍旧采用的是传统的矩形教室，有一部分使用了正方形教室。
那么，教室不同形状的原因是什么？而 哪一种更合适呢？
[1][2] [3]
通过查阅资料，阅读了教育部发布的中小学校 建筑设计规范、房屋建筑学中对讲演厅音响
[1]
效果要求，以及建筑采光设计标准。这些资料 对教室设计提出一些规定与评价标准。依照这些规
定和标准，我运用数学模型，作图计算分析各种形状教 室的特点。本文的文章结构概述如下：
教室面积是教室设计中重要的经济因素，本文第2节首先 考虑在满足使用要求的前提下，应尽量
提高空间利用率。将各项要求用不等式表达，通过求解不等式，得 到在满足要求前提下，各种形状
教室所需的最小面积，相比而言，矩形教室所需面积最小。但是，正方形 教室和正六边形教室能够
缩小最后一排学生与黑板的距离，具有更良好的视觉条件。
音 质对听课质量的影响很大，本文第3节进行了音质分析。根据波的传播与反射原理，利用几何
画板，给出 教室内声波的分布。由分布图可见，在矩形教室里，教室前方的声波密度比后方的大；
在正方形教室里， 前方与后方的声波密度分布的差异要小些；正六边形教室的声波密度分布最均匀，
音质最好。
教室的自然采光效果影响学生视觉。本文第4节根据采光设计标准,分别计算各形状教室单侧采
光和双侧 采光的采光系数最低值Cmin,。Cmin越大，说明采光效果越好。计算结果表明，在窗户总
面积相 同前提下，无论是单侧采光还是双侧采光，矩形教室的采光效果最好，而正方形教室和正六
边形教室的采 光效果较为相近。
在最后一节，我总结了分析的结果，对教室形状选择有一定的帮助。
2、教室面积
教室面积是教室设计中重要的经济因素，在满足使用要求的前提下，应尽量提高 空间利用率。
我通过设计合理的桌椅排列、教室布局，计算出各形状教室所需的最小面积，比较它们对面 积的要
求，同时也作为后面声压、采光分析的教室面积和尺寸。
[1]
为了满足学生的活动范围、视线要求，按有关规范有如下规定：
一、课桌椅的 排距：小学不宜小于850mm，中学不宜小于900mm；纵向走道宽度均不应小于550mm。
课桌 端部与墙面（或突出墙面的内壁柱及设备管道）的净距离均不应小于120mm。
二、前排边座的学生与黑板远端形成的水平视角不应小于30°。
三、教室第一排课桌前沿与黑板 的水平距离不宜小于2000mm；教室最后一排课桌后沿与黑板的
水平距离：小学不宜大于8000m m，中学不宜大于8500mm。教室后部应设置不小于600mm的横向走道。
假设：1、不考虑结构与施工的影响，教室均为形状规则的普通教室。
2、每间教室容纳50名学生、50套桌椅。
3、课桌规格为600mm×400mm，桌椅排拒为900mm，纵向走道宽度为550mm。 4、不采用3张桌子拼在一起的形式（因为中间的同学活动不便），为了减小面积，用尽量多地采用

13

全国中学生数理化学科能力展示活动优秀论文精选
2张桌子合并的形式。
5、为美观，每排、每列不少于2套桌椅，桌椅排列应规则齐整。
2.1矩形教室设计：
(以下计算过程不保留，所得边长精确至dm，面积保留1位小数)
为减小面积，应全部使用2张桌子合并的形式。
设教室开间为a mm，进深为b mm，每排x张合并桌子，共y排（a,b,x,y
?N*
）
?
a?2?6 00x+550(x-1)+2?120
?
b?2000+900y+600
?
?
2000+900y?8500

?
?
2x(y-1) ?50?2xy
?
2y(x-1)<50
?
?
解得:
2
当x=4，y=7时，有最小面积，开间6700mm，进深8900mm。S≈60.0m
经 过在几何画板上排列作图,图1(a,b)，满足视角要求，符合条件。于是得到了矩形教室的最小面
积 和最佳排列
6700mm
30
6700mm
30
8900mm8900mm
（a）
(b)
图1 矩形教室最小面积、尺寸及桌椅最佳排列
2.2正方形教室设计
由于正方形教室的进深增大，所以可分两种情况讨论。
(1)全部使用2张桌子合并的形式。
教室开间、进深为a mm，每排x张合并桌子，共y排（a,x,y
?N*
）
?
a?2?600 x+550(x-1)+2?120
?
a?2000+900y+600
?
?
2000+900y?8500

?
?
2x(y-1) ?50?2xy
?
2y(x-1)<50
?
?
当x=y=5时， a≥8440
(2)在教室边侧有一单列，其余为2张桌子合并的形式。
设教室开间、进深为a mm，每排x张合并桌子，共y排（a,x,y
?N*
）

14

全国中学生数理化学科能力展示活动优秀论文精选
?
a?2?600x+550x+600+2?120
?
a?2000+900y+6 00
?
?
2000+900y?8500

?
?
2xy ?50?2xy+y
?
2x(y-1)+y-1<50
?
?
解得：
x=4,y=6时，a有最小值8000
由于2个单列容纳的桌椅数与1个合并的列相同， 但占用的面积会更多，同样多于2列的单列会被
合并的列或合并的列和1单列替代，所以不用再讨论。
2
综上，当x=4,y=6 时，有最小面积，此时a=8000mm S=64.0m排列作图，图2(a,b)满足视角要求，符合条件。于是得到了矩形教室的最小面积和最佳排列。
2.3正六边形教室设计
由于正
8000mm
8000mm
六边形
30
30
教室不
像矩形
和正方
形那样
8000mm
8000mm
规整，所
以每排
的桌椅
数都不
尽相同。
我采用
（b）
（a）
的方法
图2 正方形教室最小面积、尺寸及桌椅最佳排列
是先根
据桌椅的排数假设正六边形的边长，然后计 算能容纳的最多桌椅数，若桌椅数少于50套，那么适当
增加边长，进行调整，使教室满足使用要求。（ 为减小面积，应尽量多地使用2张桌子合并的形式。）
设共y排, 边长为a,第n排有x
n
张合并课桌,
因为 2000+900y≤8500（a,y
?N*
）
所以y≤7
30
y=7，此时a=5200 mm
设第n排有x
n
张合并课桌,
w
图3，对角线m前方与后方的桌椅 分别于要满足不同的
条件，因为前方桌椅每排数量限制因素是边缘的前桌脚
m
与墙面保 持一定距离，而后方桌椅的限制因素是边缘的
后椅与墙面保持一定距离， w=a×
0
cos30=4503.3，则2000+900×2?
图3 正六边形教室
第三排恰通过对角线m, 需同时满足两个条件。
00
600+550)(x-1)+1200+240-[900(n-1)+2000]×tan3 0×2≤5200
0
3≤n≤7时，(2×600+550)(x-1)+1200+240 -[900(7-n)+600]×tan30×2≤5200
x
1
≤4 x
2
≤5 x
3
≤5 x
4
≤5 x
5
≤4 x
6
≤4 x
7
≤3

15

全国中学生数理化学科能力展示活动优秀论文精选
?
x
n=1
7
n
≤30,此时可容纳60套桌椅，远大于所需的50张。
(2) y=6，此时a=4700 mm
w=a×cos30=4070.3,则2000+900×2?
第三排恰通过对角线m, 需同时满足两个
条件。
0
00
3≤n≤7时，(2×600+550)(x-1)+1200+240-[ 900(6-n)+600]×tan30×2≤4700
x
1
≤4 x
2
≤4 x
3
≤5 x
4
≤4 x
5
≤3 x
6
≤3
5000mm
?
x< br>n=1
6
n
≤23<25,不满足条件。
30
若有4700≤y’ ≤5200，使教室满足使用要求，
由于第一排和最后一排 分别受到视角和墙的
限制，所以应适当加大2、3、4、5排的课桌
数。为加大面积利用率，应 使教室的中心通过
对角线m，可估算，第三排应通过对角线。设
其被对角线平分。
x
2
=x
4
=5时，y’ ≥5000.0
x
3
=6时，y’ ≥5354.8
x
5
=4时，y’ ≥4644.0
图4 正六边形教室最小面积、尺寸及最佳
经作图验证和适当调整，发现当y ’=5000.0mm
2
时恰满足条件，如图4,此时S≈65.0m
当y≤5时，每排需容纳的桌椅数增多，教室边长会进一步增加，面积会进一步扩大
于是上述布局可得到正六边形教室的最小面积和最佳排列。

综上，在满足 基本使用要求的前提下，矩形教室所需的面积最小，正方形教室次之，正六
边形教室所需面积最大，即要 求最高。建筑用地的大小会限制教室形状的选择。但是，正方形
教室和正六边形教室能够缩小最后一排学 生与黑板的距离，具有更良好的视觉条件。
3、 音质比较
[2]
音质对听课质量 的影响很大。按照房屋建筑学中相关内容，讲演厅等对音质的主要要求是：语言
的清晰度和声场分布均匀 。为此常利用墙面和顶棚做声的反射面，利用声音的反射来加强厅内声压
不足的部位、以达到声场分布的 均匀。由于声能的反射遵从与光学反射相同的准则，所以可以通过
作图分析各形状教室声场分布是否均匀 。
假设：1、各教室的墙面材料相同，即对声能的反射和吸收能力相同。
2、不考虑教室顶棚对声能的反射，而只考虑平面内声能的反射。
3、不考虑桌椅及教室其它构件对声音的作用，只考虑墙面的作用。
4、由于声能在反射的过程中会 有所损耗，所以只考虑声能一次反射后的声场分布。最小面积
下3种形状教室的声场分布如图5（a,b ,c）：

16

全国中学生数理化学科能力展示活动优秀论文精选

JI
分析：
H
G8000mm
B
A
3种 形
状的
教室
中，反
8900mm
射的
声能
均不在室
C
F
内聚
(c)
(b)
(a)
焦，声
场比
图5 声场分布
较均
匀。比较后，发现矩形教室中，教 室前方的声能比后方明显集中、空白较少，前方声场的密度比后
方大；正方形教室的前后差异没有矩形教 室明显，更好些；正六边形教室的声场分布最均匀，前后
差异小，音质最好。

4、采光效果
研究依据：
[4]
根据建筑采光设计标准,采光标准的数量评价指标以采光系数
C
表示, 侧面采光取采光系数最低
值作为标准值。学校教室采光系数最低值的要求为2.0%
侧面采光:
C
min= C
d
′?
K
τ
′?
K
ρ
′?
K
w

?
K
c

式中
C
d
′—侧窗窗洞口的采光系数，可按本标准第5.0.5 条的规定取值；
K
τ
′—侧面采光的总透射比；
K
ρ
′—侧面采光的室内反射光增量系数，可按本标准附录D 表D-5 的规定取值；
K
w
—侧面采光的室外建筑物挡光折减系数，可按本标准附录D 表D-6 的规定
取值
K
c
—侧面采光的窗宽修正系数，应取建筑长度方 向一面墙上的窗宽总和与建筑长度之比。
注:1.在Ⅰ、Ⅱ、Ⅲ类光气候区(不包含北回归线以南的地 区)，应考虑晴天方向系数(
K
f
)，可按本标
准附录D 表D-3 的规定取值。
计算点的确定：单侧采光时，计算点应定在距侧窗对面内墙面1m
双侧采光时，用B
1
=Ac
1
(Ac×LAd)确定

研究思路：
[4]
根据建筑采光设计标准，学校教室的窗地面积比取标准15。分别 计算各形状教室单侧采光和双侧
采光的采光系数最低值Cmin, Cmin越大，采光效果越好。对于 双侧采光，由于只知道窗户总面积，
不知道两侧窗户的面积，所以建立一侧窗户面积与Cmin的函数关 系，取Cmin最大的情况，这种情
况下的窗户面积分配即为最佳分配。

设教室总 面积为A，教室开间为L,进深为b，正六边形教室边长为a,窗户总面积为A
c
,双侧采光时 南
向窗户面积为A
c1
, 北向窗户面积为A
c2
,窗高为h
c
,窗户总宽度为b
c
, 双侧采光时南向窗户窗总宽度为
b
c1
, 北向窗户窗总宽度为b
c2
。计算点与窗户距离为B, 双侧采光时与南北向窗户距离分别为B
1
,B
2
。
6700mm5000mm

17

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假设：1、教室均为单跨.
2、为增强采光效果，更符合现在大多数学校的教室朝向，教室采用南北朝向，单侧采光窗南向，双侧采光窗南向和北向。
3、窗户采用普通玻璃，查表得：t=0.8；铝窗、单层窗 ，t
c
=0.75;窗户垂直清洁，t
w
=0.9。
所以
K
τ
′= t

? t
c
? t
w
=0.54
4、教室内的平均反射比ρ取0.5（可参见标准）
5、教室外无遮挡，即
K
w
=
1.00
6、取纬度40度时的晴天方向系数，南向时，
K
f
=1.55,北向时，
K
f
=1.00
7、窗高统一为2m(走访的多所学校的窗户高度在2m左右。)
相关数据：
在建筑采
表1
光设计标
[4]
准中，
K
′与
ρ
Bhc有上
表1的关
系：
当ρ=0.5
时，为便
于使
用，我
图6(a).单侧采光时Kp'与Bhc的函数关系
图6 (b).双侧采光时Kp'与Bhc的函数关系
用
f(x) = -0.1464x
2
+ 1.4336x + 0.4
f'(x) = -0.0893x2 + 0.8907x + 0.24
R
2
= 0.9995
Excel
R2 = 0.9992
4.5
3
4对数据
3.5
2.5
进行逆
Kp
3
Kp
2多项式 (Kp)
2.5
合，得
多项式 (Kp)
1.5
2到了
K
1.5
1
1
ρ
’与
0.5
0. 5
Bhc
Bhc
Bhc的
0
0
0123456
02 46
函数关
系，如
图6(a,b)：

C
d’
与Bh
c
有图7的关系：







图7
当
L
=b，时，C
d’
(%)与Bh
c
的函数关系如
图8

K
p
'
K
p
'

18

全国中学生数理化学科能力展示活动优秀论文精选

2.3.1 矩形教室采光效果(以下结果均保留2位有效数字)
2
A
c
=15A=15×60=12m
2
b
c
=A
c
h
c
=6m
2.3.1.1 单侧采光
根据《建筑采光设计标准》, 单侧采光时，计算点应定在距侧窗对面内墙面1m
所以B=b-1=5.7m

Bh
c
=5.72=2.85
2
K
p
’=f(2.85)=-0.1464×2.85+1.4336×2.85 +0.4=3.30
Lb=8.906.70=1.33, 在侧面采光计算图N表中近似取为L=1b
432
C
d
’=g(2.85) =0.1196×2.85-1.6982×2.85+8.9857×2.85-21.509×2.85+2 0.772100=1.04%
K
c
=b
c
L=68.9=0.67
C
min
= C
d
′?
K
τ
′?
K
ρ
′?
K
w

?
K
c

?
K
f

=1.04%×0.54×3.30×1×0.67×1.55=1.9%
2.3.1.2 双侧采光
南向窗宽为b
c1
(01
<6),则北向窗宽为b
c2
b
c2
=6-b
c1
=A
c1
( Ac×LA)=bc
1
×bbc= bc
1
×6.76
B
2
=A
c2
( Ac×LA)=bc
2
×bbc=(6-bc
1
)×6.76
令x
1
= B
1
h
c
= b
c1
×6.712, x
2
= B
2
h
c
=(6-b
c1
)×6.712
K
c1
=b
c1
L= bc
1
8.9
K
c2
=b
c2
L=(6-bc
1
)8.9
C
min
=C
min1
+C
min2

= [g(x
1
)100]×0.54×f’(x
1
)×1×(bc
1< br>8.9)×1.55+[g(x
2
)100]×0.54×f’(x
2
)×1×
[(6-bc
1
)8.9]×1
在几何画板中作出C
m in
与b
c1
的函数关系图象并取出定义域中的最大值，如图9(a,b)
(a)
图9 矩形教室双侧采光曲线
(b)
当bc
1
在[2.13,3.58]内，C
min
可取到最大值2.0%,此时晴天和阴天时，均满足《 标准》中采光系数最
低值要求。
2.3.2 正方形教室采光效果
2
A
c
=15A=15×64=12.8m
2
b
c
=Achc=6.4m

19

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2.3.2.1 单侧采光
与矩形教室单侧采光同理，
有 C
min
= C
d
′?
K
τ
′?
K
ρ
′?
K
w

?
K
c

?
K
f


=0.7%×0.54×3.62×1×0.8×1.55=1.70%
2.3.2.2 双侧采光
令x
1
= B
1
h
c
= b
c1
×0.625, x
2
= B
2
h
c
=(6.4-b
c1
)×0.625
C
min
=C
min1
+C
min2

= [g(x
1
)100]×0.54×f’(x
1
)×1×(bc
1< br>8)×1.55+[g(x
2
)100]×0.54×f’(x
2
)× 1× [(6-bc
1
)8]
×1
在几何画板中作出C
min与b
c1
的函数关系图象并取出定义域中的最大值，如图10(a,b)

26
24
f?x? = -0.0893?x
2
+0.8907?x+0.24
g?x? = ???0.11 96?x
4
-1.6982?x
3
?+8.9857?x
2
?-21.509?x?+20.772
h?x? = x?0.625
q?x? = ?6.4-x??0.625
22
20
18
r?x? =
?
?
g?h?x??
100
?
?0.54?f?h?x???1?x
8
??
?
?1.55+
g?q?x??
100
?
1 6
?0.54?f?q?x???1??6.4-x?
14
8
12
?
?1
1m
C:(1.717,0.019)
D:(3.318,0.019)
C
51015
10
m
8
F
G
6
4
2
D
25-25-20-15-10-5
1.5
20
230
2.533.544.5
-2
当
b
c1
在
图10正方形教室双侧采光曲线
[1.

72,3.32]内，C
min< br>可取到最大值1.9%,此时晴天时，满足《标准》中采光系数最低值要求。阴天时，
不满足采光 系数最低值要求。
2.3.3 正六边形教室采光效果
2
A
c
=15A=15×65=13m
2
b
c
=A
c
h
c
=6.5m
图11 单侧采光简图
2.3.3.1 单侧采光
由于正六边形的窗户存在一定角度，为计算方便， 设两面墙上的窗户面积
相同，b
c
’=3.75m.
根据标准, 单侧采光 时，计算点应定在距侧窗对面内墙面1m,所以计算点取对角线m上距距侧窗对
面内墙面均为1m的位置 F上。如图11.由于两窗的采光效果完全相同，所以可先计算一个窗户的采
光系数最低值，乘2即可。

C
min
= 2×C
d
′?
K
τ
′?
K
ρ
′?
K
w

?
K
c

?
K
f

=2×0.53%×0.54×3.74×1×0.5×1.55=1.7%
2.3.2.2 双侧采光
南向窗户总宽为bc
1
(01
<6.5),则北向 窗户总宽为bc
2
，设两侧的窗户面积分别相等。由于计算点
并不固定，所以双侧采光 可以按两侧总面积计算，而不用分别带入每个窗户的面积，最终结果不变。
b
c2
=6.5-b
c1
令x
1
= B
1
h
c
= b
c1
×0.665, x
2
= B
2
h
c
=(6.5-b
c1
)×0.665
(a)
(b)

20

全国中学生数理化学科能力展示活动优秀论文精选
C
min
=C
min1
+C
min2

= [g(x
1
)100]×0.54×f’(x
1
)×1×(b
c1< br>7.5)×1.55+[g(x
2
)100]×0.54×f’(x
2
)×1×
[(6-b
c1
)7.5]×1
在几何画板中作出C
m in
与b
c1
的函数关系图象并取出定义域中的最大值，如图10(a,b)
30
f?x? = -0.0893?x
2
+0.8907?x+0.24< br>28
26
24
g?x? = ???0.1196?x
4
-1 .6982?x
3
?+8.9857?x
2
?-21.509?x?+20. 772
22
h?x? = 0.665?x
20
18
16
q?x? = ?6.5-x??0.665
14
r?x? =
?
?
-30
g?h?x??
?0.54?1?f?h?x???x
100
7.5
??
?1.55+
?
?
-5
12
g?q?x??
?0.54?1?f?q?x????6.5-x?
100
7.5
10
86
4
?
?
?1
C:(1.654,0.019)
D:( 2.722,0.019)
CD
1.822.22.42.62.833.23.43.63. 844.24.44.6
2
-35-25-20-15-105101520
1.4< br>25
1.6
-2
-4
(a)
-6
(b)
图11正六边形教室双侧采光曲线
当b
c1
在[1.65,2.72]内， C
min
可取到最大值1.9%,此时晴天时，满足《标准》中采光系数最低值要求。
阴天时，不满足采光系数最低值要求。

综上，无论是单侧采光还是双侧采光，矩形教室的采 光效果最好，而正方形教室和正六边形教室的
采光效果较为相近，若要达到采光标准，还需采取一些措施 ，如：增大创洞口面积、采用透射性能
更好的玻璃等。其实，在实际生活中，为了更好地采暖和其他因素 ，常增大南向窗户的面积，所以
各教室的采光系数最低值可能达不到计算值，需要再采取措施。上面的分 析与计算还只是对采光效
果初步的探究，具体操作还需练习生活和多种因素。

3.结论分析
通过上述探讨，我发现3种形状的教室其实各有千秋，正方形教室和正六边形教 室对面积的要
求相对较高，对于面积较小的校园，矩形教室更为适用。正六边形教室拥有最好的视听效果 ，正方
形教室次之，对听课质量和视力健康都很有帮助。在采光方面，矩形教室较好，正方形教室和正六
边形教室最好采用双侧采光或采取一些其他措施，保证自然光线的充足。实际设计时，还要考虑其
他相关因素。

参考文献：
[1]
[2]
[3]
[4]
中小学校建筑设计规范(GBJ99-86)[S].
刘建荣、龙世潜主编.房屋建筑学[M]. 中央广播电视大学出版社，1984年9月.
郑忱主编.房屋建筑学[M]. 中央广播电视大学出版社，1994年2月第一版..
建筑采光设计标准 (GBT 50033-2001)[S].


21

全国中学生数理化学科能力展示活动优秀论文精选
自行车存放问题
湖南省 娄底市 双峰县第一中学 451 作者：朱天成 指导老师：贺永铁


摘要 本文以双峰县一中存车区为例，试图说明在具有 确定面积和形状的区域内如何摆放最大数
量的自行车问题。
学生使用的交通工具普遍是自行车 ，因此学校必须有适当的自行车存放区域。但由于场地有限，会
出现摆放拥挤的情况。
关键词 确定面积和形状的区域摆放最大数量的自行车 方程思想

现双峰县一中高中存车区（长6288cm；宽936cm）俯视平面图如图
1所示（图附后 ），阴影部分为自行车停放区域（简称“存车带”——分别
用Ⅰ、Ⅱ、Ⅲ、Ⅳ标明），均宽170cm, ；空白部分为通道，其中标出“A左”、
“B右”等符号的带状通道（简称“车道”）均宽85cm，图 中存车带箭头所
示方向为车辆停置时车头指向E
1
、E
2
、为出、入 口。带有箭头的虚线表示学
生进校后进入存车区的路线。
据调查，现一中高中学生骑车上 学的人数约为1400人，进校时间集中在
6:10—6:20约400人。此时存车区内拥挤，疏通缓 慢。故需对车辆的摆放
做适当的调整，以增大疏通速率。
分析堵塞原因：（1）车道狭窄，推车通过车道较慢。
（2）存车带分布不当，引起学生过于集中。
一、 分析问题
据统计:

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全国中学生数理化学科能力展示活动优秀论文精选
1)6:1 0~6:20到校学生以37人分钟的平均速度进入存车区.因在Ⅰ、Ⅱ存车
带存车都将进入A道,所以 单位时间通过A道的人数(设为2p)是B.C两道的
2倍,即P
A
=2P
B
=2P
C
2)车道的宽度d不小于75cm.当d=75cm时,人推车恰好通过, 且此时速度最小,
等于50cms.
3)车道每增宽1cm，通行速度增加2cms.
4)若干道在110cm的基础上继续增宽,通行速率将不再增加.
据以上四点计算当车道宽度各为多少时,相同人数全部通过车道用时最
短.
设:当A道宽为（75+x）cm时，用时最省。
因为：车道宽必须大于等于75cm，且A.B.C车道总宽为（85×3）cm，
0≤x ≤30
此时，2p人通过A道的时间
T
A
(x)=2p×
6288
（0≤x≤30）
50?2x
其中2x为A道增加的宽度所对应的速率增加值。
同理，p人通过B 或C道所用的时间T
B
(x)=T
C
（x）=p×
(0≤x≤30)
因为4p人同时通过A、B、C道，
所以4p人全部通过的时间是T
A
(x) T
B
(x) T
C
(x)中较大的一个。
即max｛T
A
(x)、T
B
（x）、T
C
（x）｝=6188×p×T(x).
6288

50?2?
30
2
?x
21
其中，T（x）=max｛50?2x
30?x
｝.
50?2?
2

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全国中学生数理化学科能力展示活动优秀论文精选
所以最短时间即为：当x∈ [0,30]时T（x）的最小值。在[0,30]上考察函数T
（x），
T
a
(x)=
2
是减函数， T
b
=
50 ?2x
1
30?x
50?2?
2
是增函数，
∴当x
0
满足T
a
（x）=T
b
（x）时，T（x）达到最小值T（x< br>0
）.
解方程
2
=
50?2x
1
50?2 ?
30?x
2
得x
0
=27.5﹙㎝﹚
∴当 A道宽为75+27.5=102.5(㎝),B、C道各宽75+1.25=76.25(㎝)时,通过相同人数所用时间最短.
二、分析问题(2)
1)若进入A左(或B左)，则存车后需原 路返回.据统计,在未出A左(或B左)
之前约六人随后进入A左(或B左)相遇,每与一人相遇,需停 留t
0
(t
0
为搁放车
所需时间)，且在第七人未进入之前可出A左 (或B左)。
2）若进入A中（或B中），则由E1进，E2出，不需停留。
现右存车带长 固定为860cm（不考虑右存车带的变动）,则左,中存车带
共长4556cm。
试计算，当左，中存车带长度之比为何时，通过相同的人数所用时间最
少。
据统计：
1）6:10—6:20通A左和A中的人数约为155人；通过B左和B中的人
数约为78人。
2）在6:10之前约有400人已到校，占1400人的29%，所以可粗略的认
为从 各道口向内延伸的0.4倍车道长已有车辆存放（如图1交叉的阴影部

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全国中学生数理化学科能力展示活动优秀论文精选
分）。即6:10—6:20到校的学生至少需推车通过的0.4倍的车道长。
3）每人放车时间约15s。
（注：C道临存车区外部，不存在（2）中问题，不考虑）
设：左道长xcm,
则中道长（4556-x）cm.
由（2）的结论得出：A道 通行速率为50+2×27.5=105≈100㎝s;B道通
行速率为52.5㎝∕s≈50cm∕s ;不推车速率为100cm∕s.
又进入某车道存车的人数与该道长度成正比，
∴进A道存 车所用时间T
A
左
=（进入时间+返回时间+停留时间+存车时
间）×人数
=（
同理
0.4x0.4x15x
??6
×+15）××78，
1
4556 ?x4556?x
?15
）T
A
中
=（通行时间+存车时间）×人数 =（×155，
1004556
0.4?(4556?x)0.6?(4556?x)455 6?x
??15
]× 同理T
B
中
=[×78.
5010 04556
0.4x0.4x15x
??
6×
?15）
××155
1
T
B
左
=（
∴T
左
=T
A< br>左
+T
B
左
，T
中
=T
A
左
=T
B
中
.
与问题（1）同理，学生全部通过的时间为max﹛T
左
（x），T
右
（x）﹜
设T（x）=max﹛T
左< br>（x），T
右
（x）﹜，当x
0
满足T
左
（x）=T
（x）时，T（x）达到最小值T（x
0
）.
解方程
0.4x0 .4x15
??6??15
）
1001002
x0.4x0.4x15x45 56?x4556?x
?155?(??6??15)??78?(?15)??155
＋×< br>45561564556
右
（

25

全国中学生数理化学科能力展示活动优秀论文精选
[
0.4?(45 56?x)0.6?(4556?x)4556?x
??15
]×
?78
.
501004556
整理得
0.466x
2
-41548.904x+70763573.31=0
解得x=1737（㎝）
∴左车道长（即左存车道长）为1737㎝，中车道长为4556- 1737=2819
㎝时，通过相同的人数用时最少。
三、调整
1） 将A道加长至102.5㎝，BC道相应变窄。
2） 将左部的存车带加长至1737㎝，中部的存车带缩短至2819㎝。
调整后的存车区俯视平面图如图2所示。
注：1）如图1、图2，存车区外向内52㎝的区域无车篷， 故其最外层
为车道。因此，按存车带箭头所示方向最佳。
2） E
1
、 E
2
所正对的存车带1，其面积较小且不影响周围车域的车辆存
放，故在上述计算中忽 略不计。
3）
4）
由E
2
进入中车道的人数很少，忽略不计。
同时，在摆放自行车时分开班级，前教学楼的同学将自行车放在图
示Ⅰ、Ⅱ、Ⅲ、Ⅳ左边的位置 ，后教学楼的同学应放在右边的位置上。
这样更节约时间。



参考文献：《高中数学教材知识资料包》 北京教育出版社

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全国中学生数理化学科能力展示活动优秀论文精选
2013年高中数学论文 图形计算器应用能力测试活动学生 数学建模2
乒乓球比赛是中国队在世界上占优势的比赛项目，我们 学校也有自己的乒乓球社团。因此我
们决定在乒乓球比赛的出场策略上进行研究，以便帮助学校找到比赛 的最佳策略。
就乒乓球比赛五局三胜方案进行了建模，当A队以
?
i
次序出 场、B队以
?
j
次序出场时，设
这时A队每一局比赛获胜的概率是一个不变的 常数
p
ij
，并且假设各局是否获胜是相互独立的，则5
局比赛就是一个独立 重复试验序列。 设
?
是A队在5局比赛中获胜的局数，显然，
?
服从二项分 布
b(5,p
ij
)
，再求得数学期望 ，要比较A，B两队实力的大小，可 以比较两队在每一局比赛中获胜的
平均概率大小。这是一个博弈问题，设A队以概率
x
1
,x
2
,x
3
采用策略
?
1
,
?
2
,
?
3
，由概率公式可知，
当B队采用纯策略
?
j
时，求A队的得分（最后获胜概率），所以，整个问题就可以表示成一个线性
规划 问题，对于B队，也可以列出类似的线性规划问题，正好是上述A队问题的对偶问题。解这个
对偶的线性 规划问题，可以求得：B队采用策略
?
1
,
?
2
,
?
3
的概率。
?
i
表示A队选手的出场顺序（i=1，2，3）；
；
?
j
表示B队选手的出场顺序（j=1，2，3）
p
ij
表示A队每一局比赛获胜的概率；
?
表示A队在5局比赛中获胜的局数；
q
ij
表示在五局三胜制比赛中A队最后获胜的概率；
Q
表示 当A队以
?
i
次序出场、B队以
?
j
次序出场时，在五局三 胜制比赛中A队最后获胜的概率组
成的矩阵；
P 表示在9种不同的出场次序下A队每局获胜的概率组成的矩阵；
x
1
,x
2
,x
3
表示A队以概率
x< br>1
,x
2
,x
3
采用策略
?
1
,< br>?
2
,
?
3
；
z
表示A队采用混合策略时，不管B队采用什么策略，A队的得分（最后获胜概率）;
问题描述
自2001年10月1日起，国际乒联改用11分制等新规则。中国乒乓球老将王家声认为，规 则改变的
实践效果的检验标准是三个有利于：要有利于运动的推广，有利于形成对抗激烈，场面精彩的比 赛，
有利于它的市场开发和赞助商利益。
11分制的实行，使比赛增加偶然性增加， 让一些二三流选手也有机会战胜一流选手。“但这个
偶然性应有个度”王家声说：“如果这个偶然性大到 世界顶尖高手也纷纷被无名小卒淘汰，三四流
选进决赛，那它就不是好规则了。”乒乓球11分制利弊如 何，是否会象羽毛球7分制一样实行不久就
取消呢？

28

全国中学生数理化学科能力展示活动优秀论文精选
1．试对11分制的5盘3胜与21分制的3盘2胜制作定量的比较分析；
2．试对11分制的7盘4胜和21分制的5盘3胜制作定量的比较分析；
3．综合评价及建议 。

乒乓球比赛11分制规定
1 一局比赛中，先得11分的一方为胜方。10平后，先多得2分的一方为胜方。
2 一场比赛单数局组 成。在获得每2分之后，接发球方即成为发球方，依此类推，直至该局比赛结束，
或者直至双方比分都达 到10分或者实行轮换发球法，这时，发球和接发球次序仍然不变，但每人只
轮发1分球。
3 一局中首先发球的一方，在该场下一局应首先接发球。一局中，在某一方位比赛的一方，在该场下
一局应 换到另一方位。在决胜局中，一方先得5分时，双方应交换方位。
4 当球停放在选手张开和伸平的手掌内时，才可以进行发球。 从球离开运动员手掌的那一刻到球被
击中，球都应该在球台平面的高度之上和在发球选手的端线之后。
5 当球被击中时，发球选手或他的双打队友的身体与衣服的任何部分都不能在球与网之间的范围内。
此项目的是：防止在接发球选手视线以外的隐蔽式发球。
6 发球一方只要没有发过（包括漏发）都算对方得分。



问题分析
（1）、对11分制的5盘3胜与21分制的3盘2胜制作定量的比较分析
A、B两乒乓球队 进行一场五局三胜制的乒乓球赛，两队各派3名选手上场，并各有3种选手的
出场顺序（分别记为
序出场而B队以
?
1
,
?
2
,
?
3 和
?
1
,
?
2
,
?
3
）< br>?
。根据过去的比赛记录，可以预测出如果A队以
i
次
a
ij
局。由此得矩阵
?
j
次序出场，则打满5局A队可胜
R?(a
ij
)
如下：
?
1
?
2
R?
?
3
4
?
?
4
?
1
?
?

?
1
?
2
?
2
?
?
0
?
3
?
?
5
1
3
3

（1） 根据矩阵R 能看出哪一队的实力较强吗？
（2） 如果两队都采取稳妥的方案，比赛会出现什么结果？
（3） 如果你是A队的教练，你会采取何种出场顺序？
（4） 比赛为五战三胜制，但矩阵R 中的元素却是在打满五局的情况下得到的，这样的数据处理
和预测方式有何优缺点？
首先要弄 清楚：矩阵
R
中的元素
a
ij
到底表示什么意思？是不是表示：如果 A队以
?
i
次序出场
而B队以
?
j
次序出场，A队 在5局中可以百分之百保证一定会胜
a
ij
局？显然不是这个意思，比较
合理 的看法，应该认为它只是对A队平均获胜局数的一个估计。

29

全国中学生数理化学科能力展示活动优秀论文精选
在我们做实验或观察事物时 ，往往存在着随机现象，其结果可能是很多结果中的一个，而不能
在试验前或观察前完全进行预言或确定 。在我们无奈数学建模过程中经常遇到随机现象，所谓初等
概率法就是利用初等概率的相关知识来对随机 现象进行研究的方法。
初等概率法是建立在古典概率、几何概率和统计概率基础上的。根据大数定律， 在无穷多次独
立试验下，具有有限散度的随机变量X观测值的算术平均值X在概率上收敛于此随机变量的 数学期
望E(X)；事件A出现的频率
n
A
则在概率意义上收敛于时间A发生 的概率P(A)。在已知事件B法伤
的条件下，事件A发生的条件概率P(A|B)，就是在重复的试验 中，当B发生时事件A发生的概率。
如果我们直观地把B的频率P(B)当做事件B发生的概率，把事件 A、B同时发生的概率P(A|B)看作
是事件AB发生的频率，那么P(A|B)看成是事件AB发生 的频率，那么P(A|B)的初等定义为

P(A|B)=P(AB)P(B),P(B)>0

当A队以
?
i
次序出场、B队以
?
j
次序出场时，设这时A队每一局比赛获胜的概率是一 个不变的
常数
p
ij
，并且假设各局是否获胜是相互独立的（实际上也许并不 是这样，但是题目中给我们的信
息太少，我们只能这样假设）。这样，5局比赛就是一个独立重复试验序 列。
设
?
是A队在5局比赛中获胜的局数，显然，
?
服从 二项分布
b(5,p
ij
)
，概率分布为
kk
P{
?
?k}?C
5
p
ij
(1?p
ij
)
5?k
，
k?0,1,?,5
。

容易求得它的数学期望为
E
?
?5p
ij
。
如果我们认为矩阵
R
中元素
a
ij
给出的数据，不是完全确定的结果，而是估计A队在5局比赛
中平均获胜的局数，则有
a
ij
?E
?
?5p
ij
。
这样，就可以得到
p
ij
的估计值
p
ij
?
对应于矩阵
E
?
a
ij
?
。
55
?
21 4
?
??
R?(a
ij
)?
?
034
?< br>，
?
?
531
?
?

我们可以得到这样一个矩阵


30

全国中学生数理化学科能力展示活动优秀论文精选
?
0.40.20.8
?
?
。
00.60.8
P?(p
ij
)
?
?
??
?
?
10.60 .2
?
?

要比较A，B两队实力的大小，可以比较两队在每一局比赛中获胜 的平均概率大小。矩阵
P?(p
ij
)
中的9个元素，是在9种不同的出场次 序下A队每局获胜的概率。假设这9种不同的出
场次序出现的概率相同，都是
19
，那 么，根据全概率公式，就可以求出A队在每一局比赛中获胜
的平均概率

0.4?0 .2?0.8?0?0.6?0.8?1?0.6?0.24.6
??0.511111
，
99

这个概率超过了
0.5
，也就是说，从每一局比赛来说，A队 的实力比B队略微强一些。
以上是从每一局比赛获胜概率的大小来比较实力，但是，比赛实际 上是五局三胜制，要在五局
三胜制比赛中最后获胜，才是真正获胜。
（2）、1分制的7盘4胜和21分制的5盘3胜制作定量的比较分析
A队最后获胜，可以分成下列几种情况：
（1）A队连胜三局。这种情况的概率为
p
ij
；
（2）在前三局中A队胜二局，最后A队又胜一局。这种情况的概率为
22
C
3
p
ij
(1?p
ij
)p
ij
?3p
i
3
j
(1?p
ij
)
；
3
（3）在前四局中A队胜二局，最后A队又胜一局。这种情况的概率为
22
C
4
p
ij
(1?p
ij
)
2
p
ij
?6p
i
3
j
(1?p
ij
)
2< br> ；

把这三种情况加起来，就得到在五局三胜制比赛中A队最后获胜的概率


q
ij
?
p
i
3
j
?3p
i
3
j
(1?p
ij
)?6p
i
3
j
(1?p
ij
)
2

?p
i
3
j
[1?3(1?p
ij
)?6(1?2p
ij
?p
i
2
j
)]?p
i
3
j
(10?15p
ij
?6p
i
2
j
)
。

?
0.40.20.8
?
??
根据以上公式，从矩阵
P?(p
ij)
?00.60.8
出发，可以计算出这样一个矩阵
??
?
?
10.60.2
?
?


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全国中学生数理化学科能力展示活动优秀论文精选
?
0.317440.057920.94208
?
?
。
Q?(q
ij
)?
?
00.682560.94208
??
??
0.682560.05792
?
1
?

矩阵
Q
中元素
q
ij
表示：当A队以
?
i
次序出场、 B队以
?
j
次序出场时，在五局三胜制比赛中A
队最后获胜的概率（也就是B 队最后失败的概率）。
如果两队都随机排阵，9种出场次序出现的可能性相等，都是
19，根据全概率公式，就可以
算出A队在五局三胜制比赛中最后获胜的平均概率

0.31744?0.05792?0.94208?0?0.68256?0.94208?1?0.6825 6?0.05792

9
?0.520284
。

这个数字大于
0.5
，同样也说明A队的实力比较强。
下面来看，如果两队都采取稳妥的方案，比赛会出现什么结果？
什么是“稳妥的方案”？我们的理解是 ：所谓“稳妥的方案”，就是对自己的每一种出场次序，
都考虑最坏的情况，求出在最坏的情况下，我方 失败的概率是多少，然后在各种出场次序中，选择
一种最坏情况下失败概率最小的出场次序，作为我方的 排阵方案。

?
1
从矩阵
Q?(q
ij
)?< br>?
2
?
3
?
1
?
2
?
3< br>?
0.317440.057920.94208
?
?
0
?< br>
0.682560.94208
??
??
0.682560.057 92
?
1
?

当博弈的双方都只采用一种固定的策略（称为 纯策略）时，两人零和博弈问题要得到一个稳定
的解，矩阵
Q?(q
ij
)< br>中必须有一个鞍点（Saddle Point），即在同一列中取到最大值、又在同一行
中取到最小值的元素。
设A队以概率< br>x
1
,x
2
,x
3
采用策略
?
1< br>,
?
2
,
?
3
。因为
x
1
,x
2
,x
3
是概率，所以必须满足
x
1
?x< br>2
?x
3
?1
，
x
1
?0
，
x
2
?0
，
x
3
?0
。
设
z
是A队采用这种混合策略时，不管B队采用什么策略，A队的得分（最后获胜概率）能够
保证的 最小值。
由全概率公式可知，当B队采用纯策略
?
j
时，A队的得分（最后 获胜概率）为
q
1j
x
1
?q
2j
x
2
?q
3j
x
3
，
j?1,2,3
。
因为
z
是A队的得分（最后获胜概率）能够保证的最小值，所以必须有
q< br>1j
x
1
?q
2j
x
2
?q
3j< br>x
3
?z
，
j?1,2,3
。

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全国中学生数理化学科能力展示活动优秀论文精选
容易看出，只要上 述不等式成立，当B队以某种概率混合采用各种策略时，A队的得分同样也
可以保证大于
z，所以不必另外再列式子了。
A队的目标，是要使得这个能够保证的最小得分达到最大， 所以，整个问题就可以表示成一个
线性规划问题：


目标函数
maxz


?
q
11
x
1
?q
21
x
2
?q
31
x
3
?z
?< br>qx?qx?qx?z
121222323
?
?
约束条件 ?
q
13
x
1
?q
23
x
2
?q
33
x
3
?z
。
?
x
1
?x
2
?x
3
?1
?
?
?
x
1
?0,x
2
?0,x
3
?0

解这个线性 规划问题，可以求得：A队采用策略
?
1
,
?
2
,
?
3
的概率应该分别为
，
x
1
?0.235987，
x
2
?0.303772
，
x
3
?0.46 0241

当A队采用这种混合策略时，A队能够保证的获胜概率为
z?0.535135
。
对于B队，也可以列出类似的线性规划问题，正好是上述A队问题的对偶问题。
解这个对偶的线性规划 问题，可以求得：B队采用策略
?
1
,
?
2
,
?< br>3
的概率应该分别为
，
y
2
?0.192556
，
y
3
?0.428543
，
y
1
?0.378901

当B队采用这种混合策略时，B队能够保证的获胜概率为
z
?
?0.464865
。
混合策略是以一定的概率 混合采用各种策略，但是实际上，在一次具体的比赛中，必须取定其
中的一种策略，到底取那一种呢？这 又是一个值得研究的问题。所以我们学校如果在以后要参加乒
乓球比赛，应该将这种计分的状况考虑在内 ，避免意外的出现。

（3）、乒乓球11分制的利弊的综合评价及建议
由本模型可以看出11分制是可以接受的。因为它使比赛的“偶然性”增加，使比赛更加惊险，
优势选手 与稍弱的选手之间的竞技更具悬念性，二三流选手打败一流选手进入决赛的可能性更大，
更能吸引观众。




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美国数学建模论文
2010 Problems

Problem A
Problem: Bicycle Club
Several cities in the US are starting bike share programs. Riders can pick up and drop off a bicycle at any
rental station. These bicycles are typically used for short trips within the city center, either one-way or
roundtrip. The idea is to help people get around town on a bike instead of a car. Those making longer trips
(such as commuting to work) are likely to use their own bikes. Some of the challenges are how to
determine where to locate the rental stations, how many bikes to have at each station, howwhere to add
new locations as the program grows, how many bikes to move to another location and when (time of day,
day of week). The downtown city maps, the bike rental locations and the number of bikes at each location
for Chicago, Denver and Des Moines are available from the following websites:
http:
http:
http:
You have been asked to develop an efficient bike rental program for these cities.
? List the trafficbike usage and other information that you would need to collect in order to plan the bike
rental program for these cities.
? Develop a mathematical model that the city could use to plan the program, including the location of new
rental stations for the next 5 years.
? Assume that the bike usage in the program will grow by 30% per year.

In your analysis consider the existing bike paths in the city center, attractions such as museums, theaters,
etc in the city center, and the other transportation hubs in the city center. When your analysis is complete,
prepare a short letter to the mayor explaining the benefits and recommendations of your analysis.


Problem B

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Problem: Curbing City Violence
A regional city has had lots of problems with gangs and violence over the years. The mayor, chief of police,
and city council need your help. Data are available for the following: Incidents of violence, Homicides,
Assaults, Regional Population (Census data), Unemployment, Unemployment rate, High School enrollment,
High school drop outs, Graduation rate, Drop out rate, Prison population, Released on parole, Parole
violations, Percent of parole violations, and Juvenile Inmates. Analyze and model these data to give the city
a plan to reduce violence. After you complete your analysis and model, prepare a news release for the
mayor briefly outlining your proposals that recommend a campaign strategy to curb the violence.
Problem B Data


















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2009 Problems

Problem A
Problem: Water, Water Everywhere
Fresh water is the limiting constraint for development in much of the United States. Devise an effective,
feasible, and cost-efficient national water strategy for 2010 to meet the projected needs of the United States
in 2025. In particular, address storage and movement, de-salinization, and conservation as some of the
possible components of your strategy. Consider economic, physical, cultural, and environmental effects.
Provide a position paper for the United States Congress outlining your approach, its costs, and why it is the
best choice for the nation.


Problem B
Problem: Tsunami (
Recent events have reminded us about the devastating effects of distant or underwater earthquakes. Build a
model that compares the devastation of various-sized earthquakes and their resulting Tsunamis on the
following cities: San Francisco, CA; Hilo, HI; New Orleans, LA; Charleston, SC; New York, NY; Boston,
MA; and any city of your choice. Prepare an article for the local newspaper that explains what you
discovered in your model about one of these cities.












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2008 Problems

Problem A
Problem: National Debt and National Crisis
Mathematical modeling involves two equally important steps – building models based on real world
situations and interpreting predictions made by those models back in the real world. This problem places
equal emphasis on both steps.

We are at the start of the 2008 U.S. presidential elections, and one important area of debate is sure to be the
national debt. As high school students, you have a particular interest in this subject since you are the people
who will pay off or at least manage the national debt in the future. The rate at which the national debt
changes depends on the difference between federal income (primarily taxes) and federal expenditures. Your
first task is to build a model that can be used to help understand the national debt and make forecasts based
on different assumptions. As usual, modeling involves a balance between so much complexity that the
model may be intractable and so little complexity that it is unrealistic and useless. Your model needs, at the
very least, to allow you to consider different tax policies and different expenditure policies.

As usual, raw numbers don't carry much information. Those numbers must be placed in context. For
example, total national debt is less meaningful than national debt per capita. In addition, you must be
careful about inflation. Many analysts look at the ratio between national debt and gross domestic product as
a good indicator of the impact of the national debt. Others worry about the cost of servicing the national
debt. This cost is affected by both the size of the national debt and the interest rate the government must
pay to borrow money. You may want to look at the Wikipedia article

http:iNational_debt_by_U.S._presidential_terms
for some figures involving the ratio between national debt and gross domestic product.
TASKS:
1. Build a model that can be used to help understand the national debt and make forecasts based on
different assumptions. You must provide justification for the various elements of your model and
you must also test the sensitivity of your model to various parameters.
2. Use your model to compare at least two alternative plans for the years 2009-2017. Your plans
should be based on different tax and spending policies that are reasonable and politically feasible.

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全国中学生数理化学科能力展示活动优秀论文精选
Use your model to compare the impact on the national debt and then impact on the nation in
general of your policies.
3. Prepare a letter to the new president advising him of your model.



Problem B
Problem: Going Green
The United States can address its national carbon footprint in two ways: by reducing carbon dioxide
emissions or by increasing carbon dioxide consumption (sequestration). Assume that the total U.S. carbon
dioxide emissions are capped at 2007-2008 levels indefinitely. What should the U.S. do to increase carbon
dioxide consumption to achieve national carbon neutrality with minimal economic and cultural impact? Is
it even possible to achieve neutrality? Model your solution to show feasibility, effectiveness, and costs.
Prepare a short summary paper for the U.S. Congress to persuade them to adopt your plan.













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2007 Problems

Problem A
Problem: Smoke Alarms
Fire is one of the leading causes of accidental deaths. It is important for everyone to take every preventative
measure and precaution possible to be ready to deal with a fire emergency.
More than half of all fatal fires occur between 10 p.m. and 6 a.m. when everyone in the home is usually
asleep. Smoke alarms are necessary to alert you to fires when you sleep. Will smoke alarms allow enough
time to evacuate safely?
Build a mathematical model to determine the number and locations of smoke alarms to provide the
maximum time for evacuation. Also include a model to determine the number and location of at-home fire
extinguishers to have available. Build a mathematical model for evacuation of a family from both one and
two story homes.
Prepare an advertisement for your local fire department to pass out to the community that includes the main
results of your mathematical models.

One Story Home


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Two Story Home
Downstairs


Upstairs



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Problem B
Problem: Car Rentals
Some people rent a car when they are going on a long trip. They are convinced they save money. Even if
they do not save money, they feel that the knowledge that
is the rental company's
conditions renting a car is a more appropriate option. Determine mileage limits on one's own car and a
break-even value of





























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2006 Problems

Problem A
Problem: Inflation of the Parachute
To view and print Problem A, you will need to have the Adobe Acrobat Reader installed in your Web
browser. Downloading and installing acrobat is simple, safe, and only takes a few ad
Acrobat Here.
Click the Title Below To View a PDF of Problem A:
Inflation of the Parachute


Problem B
Problem: A South Sea Island Resort
A South Sea island chain has decided to transform one of their islands into a resort. This roughly circular
island, about 5 kilometers across, contains a mountain that covers the entire island. The mountain is
approximately conical, is about 1000 meters high at the center, appears to be sandy, and has little
vegetation on it. It has been proposed to lease some fire-fighting ships and wash the mountain into the
harbor. It is desired to accomplish this as quickly as possible.
Build a mathematical model for washing away the mountain. Use your model to respond to the questions
below.
?
?
?
?
How should the stream of water be directed at the mountain, as a function of time?
How long will it take using a single fire- fighting ship?
Could the use of 2 (or 3, 4, etc.) fire-fighting ships decrease the time by more than a factor of 2 (or
3, 4, etc.)?
Make a recommendation to the resort committee about what do.



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2005 Problems

Problem A
Problem: Modeling Ocean Bottom Topography
Background:

A marine survey ship maps ocean depth by using sonar to reflect a sound pulse off the ocean floor. Figure
A shows the ship’s location at B on the surface of the ocean. The sonar apparatus aboard the ship is capable
of emitting sound pulses in an arc measuring from 2 to 30 degrees. In two dimensions this arc is shown
within Figure A by
solid lines BA and BC.
When a sonar sound pulse hits the bottom of the ocean, the pulse is reflected off the ocean bottom the same
way a billiard ball is reflected off a pool table; that is, the angle of incidence
reflection
equals the angle of
, and the emanating sound pulses are displayed by the dashed lines and the
as illustrated in Figure B. Since the ship is moving when the sound pulse is emitted, it will
pick up a reflected sound pulse at location F in this picture. The actual depth of the water is the length of
BD in Figure A

Figure A

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全国中学生数理化学科能力展示活动优秀论文精选

Figure B
Useful Information:
Oceanography vessels usually travel at a speed of 2ms while Navy vessels travel at 20ms. The sonar
apparatus aboard these ships is capable of emitting sound pulses in an arc measuring from 2 to 30 degrees.
The typical speed at which a sonar sound pulse is emitted is 1500ms.
Devise a model for mapping the topography of the ocean bottom. Write a letter to the science editor of your
local paper summarizing your findings.

Problem B
Problem: Gas Prices, Inventory, National Disasters, and the Mighty Dollar
It appears from the economic reports that the world uses gasoline on a very short supply and demand scale.
The impact of any storm, let alone Hurricane Katrina, affects the costs at the pumps too quickly. Let’s
restrict our study to the continental United States.
Over the past six years, Canada has been the leading foreign supplier of oil to the United States, including
both crude and refined oil products. (Petroleum Supply Monthly, Table S3 - Crude Oil and Petroleum
Product Imports, 1988-Present. See page 5 for Canadian exports to the United States.)
?
?
?
?
Canada was the largest foreign supplier of oil to the United States again in 2004, for the sixth year
running (from 1999, when the country displaced Venezuela, to 2004 inclusive).
In 2002, Canada supplied the United States with 17 percent of its crude and refined oil imports —
more than any other foreign supplier at over 1.9 million barrels per day.
Western Canadian crude oil is imported principally by the U.S. Midwest and the Rocky Mountain
states.
Eastern Canada's offshore oil is imported principally by the U.S. East Coast states, and even by
some Gulf Coast states.
Many refiners are buying enough to serve motorists' current needs, but not enough to rebuild stocks.
are looking to buy the oil when they need it,” according to The Washington Post.

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全国中学生数理化学科能力展示活动优秀论文精选
about the future, they hold Washington Post:Crude Oil Imports to U.S. Slow With
War 33103.)
Build a better model for the oil industry for its use and consumption in the United States that is fair to both
the business and the consumer. You can build your model based on a peak day.
Write a letter to the President’s energy advisor summarizing your findings.


2004 Problems

Problem A
Motel Cleaning Problem
Motels and hotels hire people to clean the rooms after each evening's use. Develop a mathematical model
for the cleaning schedule and use of cleaning resources. Your model should include consideration of such
things as stay-overs, costs, number of rooms, number of rooms per floor, etc. Draft a letter to the manger of
a major motel or hotel complex that recommends your model to help them in the management of their
operation.

Problem B
The Art Gallery Security System
An art gallery is holding a special exhibition of small watercolors. The exhibition will be held in a
rectangular room that is 22 meters long and 20 meters wide with entrance and exit doors each 2 meters
wide as shown below. Two security cameras are fixed in corners of the room, with the resulting video
being watched by an attendant from a remote control room. The security cameras give at any instant a
. They rotate backwards and forwards over the field of vision, taking 20 seconds to
complete one cycle.
For the exhibition, 50 watercolors are to be shown. Each painting occupies approximately 1 meter of wall
space, and must be separated from adjacent paintings by 1 meter of empty wall space and hang 2 meters
away from connecting walls. For security reasons, paintings must be at least 2 meters from the entrances.
The gallery also needs to add additional interior wall space in the form of portable walls. The portable walls
are available in 5-meter sections. Watercolors are to be placed on both sides of these ensure

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全国中学生数理化学科能力展示活动优秀论文精选
adequate room for both patrons who are walking through and those stopped to view, parallel walls must be
at least 5 meters apart throughout the gallery. To facilitate viewing, adjoining walls should not intersect in
an acute angle.
The diagrams below illustrate the configurations of the gallery room for the last two exhibits. The present
exhibitor has expressed some concern over the security of his exhibit and has asked the management to
analyze the security system and rearrange the portable walls to optimize the security of the exhibit.
Define a way to measure (quantify) the security of the exhibit for different wall configurations. Use this
measure to determine which of the two previous exhibitions was the more secure. Finally, determine an
optimum portable wall configuration for the watercolor exhibit based on your measure of security.
Figure 1: Exhibit Configuration: November 3-25, 2003



Figure 2: Exhibit Configuration: March 4-29, 2001

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2003 Problems

Problem A
What is it worth?
In 1945, Noah Sentz died in a car accident and his estate was handled by the local courts. The state law
stated that 13 of all assets and property go to the wife and 23 of all assets go to the children. There were
four children. Over the next four years, three of the four children sold their shares of the assets back to the
mother for a sum of $$1300 each. The original total assets were mainly 75.43 acres of land. This week, the
fourth child has sued the estate for his rightful inheritance from the original probate ruling. The judge has
ruled in favor of the fourth son and has determined that he is rightfully due monetary compensation. The
judge has picked your group as the jury to determine the amount of compensation.
Use the principles of mathematical modeling to build a model that enables you to determine the
compensation. Additionally, prepare a short one-page summary letter to the court that explains your results.
Assume the date is November 10, 2003.
To print a copy of problem A in pdf format click here.


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Problem B
How fair are major league baseball parks to the players?
Consider the following major league baseball parks: Atlanta Braves, Colorado Rockies, New York Yankees,
California Angles, Minnesota Twins, and Florida Marlins.
Each field is in a different location and has different dimensions. Are all these parks
fair or unfair is each park. Determine the optimal baseball
Outfield
Dimensions
Center
field
408
401
415
408
408
404
Right
center
361
390
375
385
367
385
Wall
Height
Left
Field
8
8
8
8
13
8
Center
Field
8
8
8
7
13
8

Right
field
330
330
350
314
327
345

Right
Field Area of Fair Ter
18
8
14
10
23
8
110,000
115,000
117,000
113,000
111,000
115,000
Left Left
Franchise field center
Angels
Braves
Rockies
Yankees
Twins
Marlins

330
335
347
318
343
330
376
380
390
399
385
385


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全国中学生数理化学科能力展示活动优秀论文精选
To print a copy of problem B in pdf format click here.















2002 Problems

Problem A: School Busing
Consider a school where most of the students are from rural areas so they must be bused. The buses might
pick up all the students and go to the elementary school and then continue from that school to pick up more
students for the high school.
A clear alternative would be to have separate buses for each school even though they would need to trace
over the same routes. There are, of course, restrictions on time (no student should be in the bus more than
an hour), drivers, equipment, money and so forth.
How can you set up school bus routes to optimize budget dollars while balancing the time on the bus for
various school groups? Build a mathematical model that could be used by various rural and perhaps urban
school districts. How would you test the model prior to implementation? Prepare a short article to the
school board explaining your model, its assumptions, and its results.

Problem B: The Falling Ladder

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全国中学生数理化学科能力展示活动优秀论文精选
A ladder 5 meters long is leaning against a vertical wall with its foot on a rug on the floor. Initially, the foot
of the ladder is 3 meters from the wall. The rug is pulled out, and the foot of the ladder moves away from
the wall at a constant rate of 1 meter per second. Build a mathematical model or models for the motion of
the ladder. Use your model (or models) to find the velocity at which the top of the ladder hits the floor and
the distance the top of the ladder will be from the wall at the moment that it hits the ground.















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November 2001 Problems
Problem A: Adolescent Pregnancy
You are working temporarily for the Department of Health and Environmental Control. The director is
concerned about the issue of teenage pregnancy in their region. You have decided that your team will
analyze the situation and determine if it is really a problem in this region. You gather the following 2000
data.
ty Age Age 18- 10-14 15-17 18-19 10-14 15-17 18-19
19 births birth-
10-14 births births births-unmarried births- unmarried
Pregnant Pregnant unmarried
Pregnant
350
303
422
201
156
523
263
330
123
467
421
179
571
567
691
356
357
970
434
427
221
950
713
311
17
13
29
18
11
33
9
16
10
24
18
8
281 437 16
206 466 13
307 546 28
184 326 15
109 254 10
442 803 32
201 345 7
256 444 14
113 199 9
446 686 22
343 615 15
145 261 7
164
151
251
137
99
293
113
160
78
279
219
114
193
233
366
180
161
396
168
210
106
331
328
162
Age
15-17
1 29
2 24
3 40
4 21
5 16
6 44
7 17
8 23
9 13
10 41
11 28
12 9
1998
Age Pregnancies Births
10-14 320 231
15-17 4041 3222

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18-19 6387 5164
1999
Age Pregnancies Births
10-14 309 208
15-17 3882 3048
18-19 6714 5391

Build a mathematical model and use it to determine if there is a problem or not. Prepare an article for the
newspaper that highlights your result in a novel mathematical relationship or comparison that will capture
the attention of the youth.
Problem B: Skyscrapers
Skyscrapers vary in height , size (square footage), occupancy rates, and usage. They adorn the skyline of
our major cities. But as we have seen several times in history, the height of the building might preclude
escape during a catastrophe either human or natural (earthquake, tornado, hurricane, etc). Let's consider the
following scenario. A building (a skyscraper) needs to be evacuated. Power has been lost so the elevator
banks are inoperative except for use by firefighters and rescue personnel with special keys.
Build a mathematical model to clear the building within X minutes. Use this mathematical model to state
the height of the building, maximum occupation, and type of evacuation methods used. Solve your model
for X = 15 minutes, 30 minutes, and 60 minutes.















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January 2001 Problems
Problem A: Design of an Airline Terminal
The design of airline terminals varies widely. The sketches below show airline terminals from several
cities. The designs are quite dissimilar. Some involve circular arcs; others are rectangular; some are quite
irregular. Which is optimal for operations? Develop a mathematical model for airport design and operation. Use
your model to argue for the optimality of your specified design. Explain how it would operate.


Boston-Logan International Munich International




CharlotteDouglas International Ronald Reagan Washington National


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Pittsburgh International





Problem B: Forest Service

Your team has been approached by the Forest Service to help allocate resources to fight wildfires.

In particular, the Forest Service is concerned about wildfires in a wilderness area consisting of small trees and
brush in a park shaped like a square with dimensions 80 km on a side. Several years ago, the Forest Service
constructed a network of north-south and east-west firebreaks that form a rectangular grid across the interior of
the entire wilderness area. The firebreaks were built at 5 km intervals.

Wildfires are most likely to occur during the dry season, which extends from July through September in this
particular region. During this season, there is a prevailing westerly wind throughout the day. There are
frequent lightning bursts that cause wildfires.

The Forest Service wants to deploy four fire-fighting units to control fires during the next dry season. Each unit
consists of 10 firefighters, one pickup truck, one dump truck, one water truck (50,000 liters), and one bulldozer
(w truck and trailer). The unit has chainsaws, hand tools, and other fire- fighting equipment. The people can
be quickly moved by helicopter within the wilderness area, but all the equipment must be driven via the existing
firebreaks. One helicopter is on standby at all times throughout

the dry season.

Your task is to determine the best distribution of fire-fighting units within the wilderness area. The Forest
Service is able to set up base camps for those units at sites anywhere within the area. In addition, you are asked
to prepare a damage assessment forecast. This forecast will be used to estimate the amount of wilderness likely
to be burned by fire as well as acting as a mechanism for helping the Service determine when additional
fire- fighting units need to be brought in from elsewhere.



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2000 Problems

Problem A: Bank Robbers
The First National Bank has just been robbed (the position of the bank on the map is marked). The clerk
pressed the silent alarm to the police station. The police immediately sent out police cars to establish road
blocks at the major street junctions leading out of town. Additionally, 2 police cars were dispatched to the
bank.

See the attached map.
The Bank is located at the corner of 8th Ave. and Colorado Blvd. and is marked with the letter B. The main
exits where the two road blocks are set up are at the intersection of Interstate 70 and Colorado Blvd, and
Interstate 70 (past Riverside Drive). These are marked with a RB1 and RB2 symbol.
?
?
?
Assume the robbers left the bank just before the police cars arrived. Develop an efficient algorithm
for the police cars to sweep the area in order to force the bank robbers (who were fleeing by car)
into one of the established road blocks.
Assume that no cars break down during the chase or run out of gas.
Further assume that the robbers do not decide to flee via other transportation means.
Problem B: Elections
It is almost election time and it is time to revisit the electoral vote process. The constitution and its
amendments have provided a subjective method for awarding electoral votes to states. Additionally, a state
popular vote, no matter how close, awards all electoral votes to the winner of that plurality. Create a
mathematical model that is different than the current electoral system. Your model might award fractional
amounts of electoral votes or change the methods by which the number of electoral votes are awarded to
the states. Carefully describe your model and test its application with the data from the 1992 election (in
the attached table). Justify why your model is better than the current model.

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1999 Problem
Major thoroughfares in big cities are usually highly congested. Traffic lights are used to allow cars to cross the
highway or to make turns onto side streets. During commuting hours, when the traffic is much heavier than on
any cross street, it is desirable to keep traffic flowing as smoothly as possible. Consider a two-mile stretch of a
major thoroughfare with cross streets every city block. Build a mathematical model that satisfies both the
commuters on the thoroughfare as well as those on the cross streets trying to enter the thoroughfare as a function
of the traffic lights. Assume there is a light at every intersection along your two-mile stretch.

First, you may assume the city blocks are of constant length. You may then wish to generalize to blocks of
variable length.


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For office use only
T1 ________________
T2 ________________
T3 ________________
T4 ________________
Team Control Number
8038
A

For office use only
F1 ________________
F2 ________________
F3 ________________
F4 ________________

Problem Chosen
Team #8038
February 23, 2010
Summary
Baseball is a popular bat-and-ball game involving both athletics and wisdom. There are
strict restrictions on the material, size and manufacture of the bat. It is vital important to
transfer the maximum energy to the ball in order to give it the fastest batted speed during the
hitting process. Firstly, this paper locates the center-of-percussion (COP) and the viberational
node based on the single pendulum theory and the analysis of bat vibration.

With the help of
the synthesizing optimization approach, a mathematical model is developed to execute the
optimized positioning for the “sweet spot”, and the best hitting spot turns out not to be at the
end of the bat. Secondly, based on the basic model hypothesis, taking the physical and
material attributes of the bat as parameters, the moment of inertia and the highest batted ball
speed (BBS) of the “sweet spot” are evaluated using different parameter values, which
enables a quantified comparison to be made on the performance of different bats. Thus finally
explained why Major League Baseball prohibits “corking” and metal bats.
In problem I, taking the COP and the viberational node as two decisive factors of the
“sweet zone”, models are developed respectively to study the hitting effect from the angle of
energy conversion. Because the different “sweet spots” decided by COP and the viberational
node reflect different form of energy conversion, the “space-distance” concept is introduced
and the “Technique for Order Preferenceby Similarity to Ideal Solution (TOPSIS) is used to
locate the “sweet zone” step by step. And thus, it is proved that the “sweet spot” is not at the
end of the bat from the two angles of specific quantitative relationship of the hitting effects
and the inference of energy conversion.
In problem II, applying new physical parameters of a corked bat into the model
developed in Problem I, the moment of inertia and the BBS of the corked bat and the original
wood bat under the same conditions are calculated. The result shows that the corking bat
reduces the BBS and the collision performance rather than enhancing the “sweet spot” effect.
On the other hand, the corking bat reduces the moment of inertia of the bat, which makes the
bat can be controlled easier. By comparing the two conflicting impacts comprehensively, the

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conclusion is drawn that the corked bat will be advantageous to the same player in the game,
for which Major League Baseball prohibits “corking”.
In problem III, adopting the similar method used in Problem II, that is, applying different
physical parameters into the model developed in Problem I, calculate the moment of inertia
and the BBS of the bats constructed by different material to analyze the impact of the bat
material on the hitting effect. The data simulation of metal bats performance and wood bats
performance shows that the performance of the metal bat is improved for the moment of
inertia is reduced and the BBS is increased. Our model and method successfully explain why
Major League Baseball, for the sake of fair competition, prohibits metal bats.
In the end, an evaluation of the model developed in this paper is given, listing its
advantages s and limitations, and providing suggestions on measuring the performance of a
bat.












Key words:


sweet spot, moment- of-inertia, Center-of-Percussion, Bat-Ball
Coefficient-of-Restitution, Batted-Ball Speed


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Contents
Summary ...................................... .................................................. .................................................5 8
Contents.................................... .................................................. .................................................. ...60
ement of the Problem ................... .................................................. ....................................61
is of the Problem ...................................... .................................................. .......................61
2.1 Analysis of Problem I ........................................ .................................................. ..............61
2.2 Analysis of Problem II .. .................................................. .................................................. .62
2.3 Analysis of Problem III .............. .................................................. .....................................62
Assumptions and Symbols .......................... .................................................. ....................63
3.1 Model Assumptions . .................................................. .................................................. ......63
3.2 Symbols ......................... .................................................. .................................................. 63
ng and Solution ........................... .................................................. ....................................64
4.1 Modeling and Solution to Problem I ............... .................................................. ................64
4.1.1 Model Preparation ... .................................................. .............................................64
4.1.2 Solutions to the two “sweet spot” regions .......................................... ....................66
4.1.3 Optimization Model Based on TOPSIS Method ..................... ...............................68
4.1.4 Verifying the “sweet spot” is not at the end of the bat ...........................................70
4.2 Modeling and Solution to Problem II ...... .................................................. ........................70
4.2.1 Model Preparation ...................................... .................................................. ..........70
4.2.2 Controlling variable method analysis ......................................... ............................71
4.2.3 Analysis of corked bat and wood bat
[5][6]
........ .................................................. .....72
4.2.4 Reason for prohibiting corking
[4]
............................... ..........................................73
4.3 Modeling and Solution to Problem III ..... .................................................. ........................74
4.3.1 Analysis of metal bat and wood bat
[8][9]
............ .................................................. ..74
4.3.2 Reason for prohibiting the metal bat
[4]
.................................. ................................75
ths and Weaknesses of the Model .......................... .................................................. ..........76
ths ............................. .................................................. .............................................76
5.2 Weaknesses ............................... .................................................. .......................................76
nces .................................................. .................................................. ................................76







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ement of the Problem
Explain the “sweet spot” on a baseball bat.
Every hitter knows that there is a spot on the fat part of a baseball bat where maximum
power is transferred to the ball when hit. Why isn’t this spot at the end of the bat? A simple
explanation based on torque might seem to identify the end of the bat as the sweet spot, but
this is known to be empirically incorrect. Develop a model that helps explain this empirical
finding.
Some players believe that “corking” a bat (hollowing out a cylinder in the head of the bat
and filling it with cork or rubber, then replacing a wood cap) enhances the “sweet spot” effect.
Augment your model to confirm or deny this effect. Does this explain why Major League
Baseball prohibits “corking”?
Does the material out of which the bat is constructed matter? That is, does this model
predict different behavior for wood (usually ash) or metal (usually aluminum) bats? Is this
why Major League Baseball prohibits metal bats?
is of the Problem
2.1 Analysis of Problem I
First explain the “sweet spot” on a baseball bat, and then develop a model that helps
explain why this spot isn’t at the end of the bat.
[1]

There are a multitude of definitions of the sweet spot:
1) the location which produces least vibrational sensation (sting) in the batter's hands
2) the location which produces maximum batted ball speed
3) the location where maximum energy is transferred to the ball
4) the location where coefficient of restitution is maximum
5) the center of percussion
For most bats all of these at different locations on the bat, so one is
often forced to define the sweet spot as a region.
If explained based on torque, this “sweet spot” might be at the end of the bat, which is
known to be empirically incorrect. This paper is going to explain this empirical paradox by
exploring the location of the sweet spot from a reasonable angle.
Based on necessary analysis, it can be known that the sweet zone, which is decided by the
center-of-percussion (COP) and the vibrational node, produces the hitting effect abiding by
the law of energy conversion. The two different sweet spots respectively decided by the COP
and the viberational node reflect different energy conversions, which forms a two-factor
influence. This situation can be discussed from the angle of “space-distance” concept, and the
“Technique for Order Preference by Similarity to Ideal Solution (TOPSIS)” could be used.
[2]

The process is as follows: first, let the sweet spots decided by the COP and the viberational

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node be “optional sweet spots”; second, define the regions that these optional sweet spots may
appear as the “sweet zones”, and the length of each sweet zone as distance; then, the sweet
spot could be located by sequencing the sweet zones of the two kinds on the bat. Finally,
compare the maximum hitting effect of this sweet spot with that of the end of the bat.
2.2 Analysis of Problem II
Problem II is to explain whether “corking” a bat enhances the “sweet spot” effect and
why Major League Baseball prohibits “corking”.
[4]

In order to find out what changes will occur after corking the bat, the changes of the bat’s
parameters should be analyzed first:
1) The mass of the corked bat reduces slightly than before;
2) Less mass (lower moment of inertia) means faster swing speed;
3) The mass center of the bat moves towards the handle;
4) The coefficient of restitution of the bat becomes smaller than before;
5) Less mass means a less effective collision；
6) The moment of inertia becomes smaller
.[5][6]

By analyzing the changes of the above parameters of a corked bat, whether the hitting
effect of the sweet spot has been changed could be identified and then the reason for
prohibiting “corking” might be clear.
2.3 Analysis of Problem III
First, explain whether the bat material imposes impacts on the hitting effect; then,
develop a model to predict different behavior for wood or metal bats to find out the reason
why Major League Baseball prohibits metal bats?
[1][4]

The mass (M) and the center of mass (CM) of the bat are different because of the material
out of which the bat is constructed. The changes of the location of COP and moment of inertia
(
I
bat
) could be inferred.
[2][3]

Above physical attributes influence not only the swing speed of the player (the less the
moment of inertia--
I
bat
is, the faster the swing speed is) but also the sweet spot effect of the
ball which can be reflected by the maximum batted ball speed (BBS).
The BBS of different material can be got by analyzing the material parameters that affect
the moment of inertia. Then, it can be proved that the hitting effects of different bat material
are different.

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Assumptions and Symbols
3.1 Model Assumptions
1) The collision discussed in this paper refers to the vertical collision on the “sweet
spot”;
2) The process discussed refers to the whole continuous momentary process starting
from the moment the bat contacts the ball until the moment the ball departs from the
bat;
3) Both the bat and the ball discussed are under common conditions.
3.2 Symbols
Table 3-1
Symbols
k

Instructions
a kinematic factor
the rotational inertia of the object about its pivot point
the mass of the physical pendulum
the location of the center-of-mass relative to the pivot point
the distance between the undetermined COP and the pivot
the gravitational field strength
the moment-of- inertia of the bat as measured about the pivot
point on the handle
the swing period of the bat on its axis round the pivot
the length of the bat
the distance from the pivot point where the ball hits the bat
vibration frequency
the mass of the ball
I
0

M

d

L

g

I
bat

T

S

z

f

m
ball


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ng and Solution
4.1 Modeling and Solution to Problem I
4.1.1 Model Preparation
1) Analysis of the pushing force or pressure exerted on hands
[1]


Fig. 4-1
As showed in Fig. 4-1:
? If an impact force
F
were to strike the bat at the center-of-mass (CM) then
point
P
would experience a translational acceleration - the entire bat would attempt
to accelerate to the left in the same direction as the applied force, without rotating
about the pivot point. If a player was holding the bat in hisher hands, this would
result in an impulsive force felt in the hands.
? If the impact force
F
strikes the bat below the center-of-mass, but above the
center-of- percussion, point
P
would experience both a translational acceleration in
the direction of the force and a rotational acceleration in the opposite direction as
the bat attempts to rotate about its center-of-mass. The translational acceleration to
the left would be greater than the rotational acceleration to the right and a player
would still feel an impulsive force in the hands.
? If the impact force strikes the bat below the center-of-percussion, then
point
P
would still experience oppositely directed translational and rotational
accelerations, but now the rotational acceleration would be greater.

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? If the impact force strikes the bat precisely at the center-of-percussion, then the
translational acceleration and the rotational acceleration in the opposite direction
exactly cancel each other. The bat would rotate about the pivot point but there
would be no net force felt by a player holding the bat in hisher hands.
? Define point
O
as the center-of- percussion（COP）
1）Locating the COP
According to physical knowledge, it can be determined by the following method:
Instead of being distributed throughout the entire object, let the mass of the
physical pendulum
M
be concentrated at a single point located at a distance L from
the pivot point. This point mass swinging from the end of a string is now a
pendulum, and its period would be the same as that of the original
physical pendulum if the distance
L
was
L?
I
bat
（4-1）
Md
This location
L
is known as the
A solid object which oscillates about a fixed pivot point is called a physical
pendulum. When displaced from its equilibrium position the force of gravity will
attempt to return the object to its equilibrium position, while its inertia will cause it
to overshoot. As a result of this interplay between restoring force and inertia the
object will swing back and forth, repeating its cyclic motion in a constant amount
of time. This time, called the period, depends on the mass of the object
M
, the
location of the center-of-mass relative to the pivot point
d
, the rotational inertia of
the object about its pivot point
I
0
and the gravitational field strength
g
according to
T?2
?
2) Analysis of the vibration:
[1]


I
0
(4-2)
Mgd

Fig. 4-2
As showed in Fig. 4-2, mechanical vibration occurs when the bat hits the ball. Hands feel
comfortable only when the holding position lies in the balance point. The batting point is the
vibration source. Define the position of the vibration source as the vibrational node. Now this

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vibrational node is one of the optional “sweet spots”.
4.1.2 Solutions to the two “sweet spot” regions
1) Locating the COP
[1][4]


? Determining the parameters：

a. mass of the bat
M
;
b. length of the bat
S
(the distance between Block 1 and Block 5 in Fig 4-3);
c. distance between the pivot and the center-of-mass
d
( the distance between
Block 2 and Block 3 in Fig. 4-3);
d. swing period of the bat on its axis round the pivot
T
(take an adult male as an
example: the distance between the pivot and the knob of the bat is 16.8cm (the
distance between Block 1 and Block 2 in Fig. 4-3);
e. distance between the undetermined COP and the pivot
L
(the distance between
Block 2 and Block 4 in Fig. 4-3, that is the turning radius)
.

Fig. 4-3

Table 4-1
knob
pivot
the center- of-mass（CM）
the center of percussion (COP)
the end of the bat

[1][4]
? Calculation method of COP
:
distance between the undetermined COP and the pivot:
T
2
g
L?
(
g
is the gravity acceleration） （4-3）
2
4
?
Block 1
Block 2
Block 3
Block 4
Block 5

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moment of inertia:
T
2
MgL
I
0
?
(
L
is the turning radius,
M
is the mass) （4-4）
4
?
2

? Results:
The reaction force on the pivot is less than 10% of the bat-and-ball collision force.
When the ball falls on any point in the “sweet spot” region, the area where the collision
force reduction is less than 10% is
(0.9L,1.1L)
cm, which is called “Sweet Zone 1”.
2) Determining the vibrational node
The contact between bat and ball, we consider it a process of wave
the bat excited by a baseball of rapid flight, all of these modes, (as well as some additional
higher frequency modes) are excited and the bat vibrates .We depend on the frequency
modes ,list the following two modes:
The fundamental bending mode has two nodes, or positions of zero displacement). One
is about 6-12 inches from the barrel end close to the sweet spot of the bat. The other at about
24 inches from the barrel end (6 inches from the handle) at approximately the location of a
right-handed hitter's right hand.

Fig. 4-4 Fundamental bending mode 1 (215 Hz)
The second bending mode has three nodes, about 4.5 inches from the barrel end, a
second near the middle of the bat, and the third at about the location of a right-handed hitter's
left hand.

Fig4-5. Second bending mode 2 (670 Hz)
The figures show the two bending modes of a freely supported baseball bat. The handle

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end of the bat is at the right, and the barrel end is at the left. The numbers on the axis
represent inches (this data is for a 30 inch Little League wood baseball bat). These figures
were obtained from a modal analysis experiment. In this opinion we prefer to follow the
convention used by Rod Cross[2] who defines the sweet zone as the region located between
the nodes of the first and second modes of vibration (between about 4-7 inches from the barrel
end of a 30-inch Little League bat).

Fig. 4-6 The figure of “Sweet Zone 2”
The solving time in accordance with the searching times and backtrack times. It is
objective to consider the two indices together.
4.1.3 Optimization Model Based on TOPSIS Method

Table4-2

swing
period
T

bat
mass
M

bat CM
length position
S

d

coefficient initial swing
of velocispeed
restitution
ty
v
in

v
bat

BBCOR
ball
mass
m
ball

wood
876.086.4 27.7m15.3
bat 0.12s 41.62cm 0.4892 850.5g
15g cm s ms
(ash)

Adopting the parameters in the above table and based on the quantitative regions in sweet
zone 1 and 2 in 4.1.2, the following can be drawn:
[2]

cm)
Sweet zone 1 is
(0.9L,1.1L)
=(50cm,57.8358
*
cm,55.23cm)
Sweet zone 2 is
(L
*
1
,L
2
)
=
(48. 41
As shown in Fig 4-3, define the position of Block 2 which is the pivot as the origin of the
number axis, and
x
as a random point on the number axis.
1) Optimization modeling
[2]

The TOPSIS method is a technique for order preference by similarity to ideal solution

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whose basic idea is to transform the integrated optimal region problem into seeking the
difference among evaluation objects—“distance”. That is, to determine the most ideal position
and the acceptable most unsatisfactory position according to certain principals, and then
calculate the distance between each evaluation object and the most ideal position and the
distance between each evaluation object and the acceptable most unsatisfactory position.
Finally, the “sweet zone” can be drawn by an integrated comparison.
Step 1 : Standardization of the extent value
Standardization is performed via range transformation,
x
*
?
dimensionless quantity，and
x?x
min
,
x
*
is a
x
max
?x
min
x
*
?[0,1]

x
min
?min0{.9L,L
*
1
}x
max
?max1.{1L,L
*
2
}x?(x
mi
,
n
x
ma
)
x
;
Step 2:

Determining the most ideal position
x
*?
and the acceptable most unsatisfied
position
x
*?

*
Assume that the most ideal position is
x
*?
?min{x
1
}
, and the acceptable most
*
unsatisfied position is
x
*?
?max{x
2
}
;
Step 3: Calculating the distance
The Euclidean distance of the positive ideal position is:

The Euclidean distance of the negative ideal position is:
d
?
?
d
?
?
?(x
*
?x
*?
)
?
(x
*
?x*?
)

Step 4:

Seeking the integrated optimal region
The integrated evaluation index of the evaluation object is:
d
?
b?
?
d?d
?
… …………………………（4-5）


2) Optimization positioning
Considering bat material physical attributes of normal wood, when the period is
T?0.12s
and the vibration frequency is
f?520
HZ, the ideal “sweet zone” extent can
be drawn as
?
51.32cm,55.046cm
?
. As this consequence showed, the “sweet spot”
cannot be at the end the bat. This conclusion can also be verified by the model for problem
II.

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4.1.4 Verifying the “sweet spot” is not at the end of the bat
1) Analyzed from the hitting effect
According to Formula 4-11 and Table 4-2, the maximum batted-ball-speed of the
“sweet spot” can be calculated as
BBS
sweet
?27.4ms
, and the maximum
batted-ball-speed of the bat end can be calculated as
BBS
end
?22.64ms
. It is obvious
that the “sweet spot” is not at the end of the bat.

2) Analyzed from the energy
According to the definition of “sweet spot” and the method of locating the “sweet spot”,
energy loss should be minimized in order to transfer the maximum energy to the ball. When
considering the “sweet spot” region from angle of torque, the position for maximum torque
is no doubt at the end of the bat. But this position is also the maximum rebounded point
according to the theory of force interaction. Rebound wastes the energy which originally
could send the ball further.
To sum up the above points: it can be proved that the “sweet spot” is not at the end of
the bat by studying the quantitative relationship of the hitting effect and the inference of the
energy transformation.
4.2 Modeling and Solution to Problem II
4.2.1 Model Preparation

1) Introduction to corked bat
[5][6]
:

Fig 4-7
As shown in Fig 4-7, Corking a bat the traditional way is a relatively easy thing to do.
You just drill a hole in the end of the bat, about 1-inch in diameter, and about 10-inches deep.
You fill the hole with cork, super balls, or styrofoam - if you leave the hole empty the bat
sounds quite different, enough to give you away. Then you glue a wooden plug, like a 1-inch
dowel, in to the end. Finally you sand the end to cover the evidence. Some sources suggest
smearing a bit of glue on the end of the bat and sprinkling sawdust over it so help camouflage
the work you have done.


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全国中学生数理化学科能力展示活动优秀论文精选
2) Situation studied：
Situation of the best hitting effect: vertical collision occurs between the bat and the ball,
and the energy loss of the collision is less than 10% and more than 90% of the momentum
transfers from the bat to the ball (the hitting point is the “sweet spot”).

3) Analysis of COR

After the collision the ball rebounded backwards and the bat rotated about its pivot. The
ratio of ball speeds (outgoing incoming) is termed the collision efficiency,
e
A
. A kinematic
factor
k
, which is essentially the effective mass of the bat, is defined as
m
ball
z
2
…………………………………………………………（4-6）
k?
I
bat
where
I
nat
is the moment-of-inertia of the bat as measured about the pivot point on the
handle, and
z
is the distance from the pivot point where the ball hits the bat. Once the
kinematic factor
k
has been determined and the collision efficiency
e
A
has been measured,
the
BBCOR
is calculated from
BBCOR?e
A
(1?k)?k
…………………………………………（4-7）
4) Physical parameters vary with the material:
The hitting effect of the “sweet spot” varies with the different bat material. It is related
with the mass of the ball
M
, the center-of-mass (
CM
), the location of the center-of-mass
d
,
the location of COP
L
, the coefficient of restitution
BBCOR
and the moment-of- inertia of
the bat
I
bat
.
4.2.2 Controlling variable method analysis

M
is the mass of the object；
d
is the location of the center-of-mass relative to the pivot
point；
g
is the gravitational field strength；
I
bat
is the moment-of- inertia of the bat as measured
about the pivot point on the handle;
z
is the distance from the pivot point where the ball hits
the bat；
v
inl
is the incoming ball speed；
v
bat
is the bat swing speed just before collision.

The following formulas are got by sorting the above variables
[1
]:
T?2
?
I
bat
L
?2
?
Mgdg ……………………………………………（4-8）

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全国中学生数理化学科能力展示活动优秀论文精选
I
bat
T
2
g
COP?L??
………………………………………………（4-9）
Md
4
?
2
BBS?e
A
v
in
?
?
1?e
A
?
v
bat
…………………………… ……………（4-10）
Associating the above three formulas with formula (4-6) and (4-7), the formulas
among
BBS
, the mass
M
, the center-of-mass (
CM
), the location of COP, the coefficient
of restitution
BBCOR
and the moment-of-inertia of the bat
I
bat
are:
BBS?
BBCOR?kBBCOR?kv
in
?(1?)v
bat
………………………（4-11）
1?k1?k
I
bat
T
2
Mgd
?
…………… ………………………………………………（4-12）
2
4
?
m
b all
………………………………………………………………（4-13）
M
It can be known form formula (4-11), (4-12) and (4-13):
1) When the coefficient of restitution
BBCOR
and mass
M
of the material changes,
BBS will change;
k?
2) When mass
M
and the location of center-of mass
CM
changes,
I
bat
changes,
which is the dominant factor deciding the swing speed.

4.2.3 Analysis of corked bat and wood bat
[5][6]

Table 4-3
coefficie
initial swing ball
swing bat Bat CM nt of
velocity speed mass
period mass length position restitutio
v
in

v
bat

m
ball

S

d

T

M

n
BBCOR
876.015.3 850.5
wood 0.12s 86.4 cm 41.62cm 0.443 27.7ms
15g ms g
833.415.3 850.6
cork 0.12s 86.5 cm 41.63cm 0.438 27.7ms
9g ms g
The hitting effect reflects by above physical parameters:
Apply these values in formula (4-11) and (4-13)
1) Calculate respectively：
BBS(1)?26.8ms

BB(S2)?27.4ms

Thus
BBS(1)?BBS(2)
.

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全国中学生数理化学科能力展示活动优秀论文精选
BBS(1)
represents the maximum batted-ball-speed of a corked bat, and
BBS(2)
represents the maximum batted-ball-speed of a wood bat.
Conclusion: when the swing speed (
v
bat
) and the initial velocity of the ball (
v
in
) remain
the same, the initial velocity of the wood bat is higher than the corked bat. That is to say that
the corked bat does not enhance the “sweet spot” effect. Hence, it is not the reason why Major
League Baseball prohibits “corking”. But the change of the moment-of-inertia caused by the
material can explain the prohibit.
2) Calculation of the moment-of-inertia of two different bats:
I
bat
(1)?142g*cm
2
,
I
bat
(2)?159.92g*cm
2

Thus
I
bat
(1)?I
bat
(2)
.
I
bat
(1)
is the moment-of-inertia of the corked bat, and
I
bat
(2)
is the
moment-of-inertia of the wood bat.
Conclusion: the mass and the moment-of-inertia of the bat reduces after corking, so
then swing speed gets faster, which means the professional players are able to watch the ball
travel an additional 5-6 feet before having to commit to a swing. It makes the game unfair to
increase the hitting accuracy by corking the bat.
4.2.4 Reason for prohibiting corking
[4]

If the swing speed is unchanged, the corked bat cannot hit the ball as far as the wood bat,
but it grants the player more reaction time and increases the accuracy. Influenced by a
multitude of random factors, vertical collision cannot be assured in each hitting. The
following figure shows the situation of vertical collision between the bat and the ball:

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全国中学生数理化学科能力展示活动优秀论文精选

Fig 4-8
In order to realize the best hitting effect, all of the BBS drawn from the above calculating
results are assumed to be vertical collision. But in a professional baseball game, because the
hitting accuracy is also one of the decisive factors, increasing the hitting accuracy equals to
enhance the hitting effect.
After cording the bat, the moment-of-inertia of the bat reduces, which improves the
player’s capability of controlling the bat. Thus, the hitting is more accurate, which makes the
game unfair.
To sum up, in order to avoiding the unfairness of a game, Major League Baseball
prohibits “corking”
4.3 Modeling and Solution to Problem III
According to the model developed in Problem I, the hitting effect of the “sweet spot”
depends on the mass of the ball
M
, the center-of-mass (
CM
), the location of CM
d
, the
location of COP
L
, the coefficient of restitution
BBCOR
and the moment-of-inertia of the
bat
I
bat
. An analysis of metal bat and wood bat is made.
4.3.1 Analysis of metal bat and wood bat

[8][9]

Table 4-4

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全国中学生数理化学科能力展示活动优秀论文精选
coefficient initial
Bat CM
of velocity
length position
restitution
v
in

S

d

BBCOR
876.086.4 41.62c
wood 0.12s 0.443 27.7ms
15g cm m
827.886.6 41.64c
metal 0.12s 0.496 27.7ms
2g cm m
Apply these values in formula（4-11）和（4-13）
1) Calculating BBS of the two different bats:
swing
period
T

bat
mass
M

swing
speed
ball
mass
v
bat

15.3
ms
15.3
ms
m
ball

850.5
g
850.7
g
BBS(1)?29.6ms
, and
BBS(2)?27.4ms
, thus
BBS(1)?BBS(2)
.
BBS(1)
refers to the maximum initial velocity of the metal bat, and
BBS(2)
refers to
the maximum initial velocity of the wood bat.
Conclusion: when the swing speed (
v
bat
) and the initial velocity of the ball (
v
in
)
remain the same, the initial velocity of the metal bat is higher than the wood bat. It is also
found in the calculation that the BBCOR of the metal bat is higher than the wood bat, which
enhances the hitting effect of the metal bat and makes the game unfair.
2) Calculation of the moment-of- inertia of two different bats:
I
bat
(1)?128g*cm
2
, and
I
bat
(2)?159.92g*cm
2
,
thus
I
bat
(1)?I
bat
(2)
.
I
bat
(1)
is the moment-of-inertia of the metal bat, and
I
bat
(2)
is moment-of-inertia of
the wood bat.
Conclusion: Because the hitting part is hollow for the metal bat, the CM is closer to
the handle of bat for an aluminum bat than a wood bat.
I
bat
of metal bat is less than
I
bat
of
the wood bat, which increases the swing speed. It means the professional players are able to
watch the ball travel an additional 5-6 feet before having to commit to a swing, which makes
the hitting more accurate to damage the fairness of the game.
4.3.2 Reason for prohibiting the metal bat

[4]

Through the studies on the above models:
【4.3.1-(1)】proves the best hitting effect of a metal bat is better than a wood bat.
【4.3.1-(2）】proves the hitting accuracy of a metal bat is better than a wood bat.
To sum up，the metal bat is better than the wood bat in both the two factors, which makes
the game unfair. And that’s why Major League Baseball prohibits metal bat.


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全国中学生数理化学科能力展示活动优秀论文精选
ths and Weaknesses of the Model
ths
1) The model, with the help of the single-pendulum theory and the analysis of the
vibration of the bat and the ball, locates the COP and vibrational node of bats
respectively, and locates the “sweet spot” influenced by multitudes of factors
utilizing the integrated optimization method. The overall optimized solution makes
the “sweet spot” more persuasive.
2) The Model analyzes the integrated performance of different material bats from the
aspects of the maximum initial velocity (BBS) and the hitting accuracy, and explains
why corked bats are prohibited successfully.
3) Deriving results from the controlling variable method analysis and taking the Law of
Energy Conservation and the theories of structural dynamics as foundation enable to
avoid the complicated mechanical analysis and derivation.
5.2 Weaknesses
The model fails to evaluate the performance of bats exactly from the angle of the
game relationship between the maximum initial velocity (BBS) and the accuracy when
evaluating the hitting effect.






nces
[1]http:~

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全国中学生数理化学科能力展示活动优秀论文精选
[2] Mathematical Modeling Contest: Selection and Comment on Award-winning Papers
[3]http:~
[4]Adair, Physics of Baseball. New York: Harper Perennial.
[5]D.A. Russell,
of Sport 5 Vol. 2, pp. 641-647 (International Sports Engineering Association, 2004).
Proceedings of the 5th International Conference on the Engineering of Sport, UC Davis,
September 11-15, 2004.
[6] A. M. Nathan,
[7] ESPN Baseball Tonight, on June 3, 2003 aired a nice segment in which Buck Showalter
showed how to cork a bat, drilling the hole, filling it with cork, and plugging the end.
[8] R.M. Greenwald R.M., L.H. Penna , and J.J. Crisco,
with Wood and Aluminum Baseball Bats: A Batting Cage Study,
241-252 (2001).
[9] J.J. Crisco, R.M. Greenwald, J.D. Blume, and L.H. Penna,
and metal baseball bats,
[10] Robert K. Adair, The Physics of Baseball, 3rd Ed., (Harper Collins, 2002)
[11] R. Cross,
[12] A. M. Nathan,
979-990 (2000)
[13] K. Koenig, J.S. Dillard, D.K. Nance, and D.B. Shafer,
on baseball bat testing,
Engineering Association, 2004). Proceedings of the 5th International Conference on the
Engineering of Sport, UC Davis, September 11-15, 2004.
[14] Smith, L, Broker, J, Nathan, A, 揂 Study of Softball Player Swing Speed,?Sports
Dynamics Discovery and Application, Edited by A. Subic, P. Trivailo, and F. Alam,
RMIT University, Melbourne Australia, pp. 12-17 (2003)
[15] L. Noble and H. Walker,
discomfort following ball-bat impacts,
(1994).
[16] R. Cross,
771-779 (1998).
[17] R.K. Adair,
Physics, 69(2), 229-230 (2001);R. Cross,
baseball bat,


77