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HDZJY2009

木虫 (正式写手)

[求助] 使各个圆相交部分面积相等,求纵坐标

如下图,有8个半径为5的圆沿Y轴罗列,已知圆1的圆心坐标为(5,5),求其他圆的坐标分别为多少时,能使圆1所截的各部分面积是相等的?即圆8与圆1的相交面积为圆面积的pi*5*5的1/8,圆7与圆1的相交面积为圆面积的2/8,这样圆8与圆7之间的面积也是圆面积的1/8,依次类推。
     例子中是8个圆的,我要做的是35个圆的,能帮我得到做35个圆时,圆心纵坐标各自的值吗?

所夹的面积相等-3.jpg

[ Last edited by HDZJY2009 on 2013-3-8 at 21:06 ]
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leedobb

金虫 (正式写手)

【答案】应助回帖

★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ...
感谢参与,应助指数 +1
HDZJY2009: 金币+150, ★★★★★最佳答案, 非常感谢,居然连图也给了,图是用matlab做的吗?顺便把程序给我吧 2013-03-09 14:29:23
首先应该无解析解
8个的情形下的y坐标分别为
12.733898602563759280736579023416
11.347045926489442402527056525453
   10.1458423138090761429485989637
  9.039727510571245179500385832818
7.9904317306745460526790641141417
6.9764395001682878743743313503921
5.9833346959308628632804819059461
                                 5
它们是方程
2*arccos((yi-5)/2/5)*5^2 - 2* (25-(yi-5)^2/4)^(0.5) *(yi-5)/2 = (N-i+1)/N *pi *25
的解。

35个圆时
[  1, 14.165537622457287366799186533746]
[  2, 13.668528743815480457269196244901]
[  3,  13.24752297250298765894719002992]
[  4, 12.868408096361129663480370000015]
[  5, 12.517081838974052214423900375613]
[  6, 12.185988992310452299124047418117]
[  7, 11.870488250033117769672469830192]
[  8, 11.567470741845687876233937653017]
[  9, 11.274724893453649882935569774775]
[ 10, 10.990605790555390961310020823067]
[ 11, 10.713847109073235682684998706036]
[ 12, 10.443446748832963085535806099172]
[ 13, 10.178593549887409133624472471737]
[ 14, 9.9186183087893783295609191829997]
[ 15, 9.6629598761480137244981067044324]
[ 16, 9.4111409941769388615577892746821]
[ 17, 9.1627506390029620260206093438695]
[ 18, 8.9174308331199258797934047970661]
[ 19, 8.6748666064888354788137719420696]
[ 20, 8.4347782235647880749325053042479]
[ 21, 8.1969150719264140771542029718434]
[ 22, 7.9610507896406753370124822864297]
[ 23, 7.7269793296186974568033540724253]
[ 24, 7.4945117417972033347053269643881]
[ 25, 7.2634735113535584092502476772318]
[ 26, 7.0337023317013247270031880072913]
[ 27, 6.8050462200898331849674962699517]
[ 28, 6.5773619047636664255967319736435]
[ 29, 6.3505134281720696158979627214228]
[ 30, 6.1243709222441346670493526815346]
[ 31, 5.8988095203561772860239157165265]
[ 32, 5.6737083770735489733221395621439]
[ 33, 5.4489497715841969944735282616306]
[ 34, 5.2244182743324640640883487380682]
[ 35,                                 5]
有一天,我打了个瞌睡就到了这里,但我知道我掉入了时光的循环中,虽得以永生,但只有第一个循环有意义。
2楼2013-03-08 23:21:43
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leedobb

金虫 (正式写手)

引用回帖:
2楼: Originally posted by leedobb at 2013-03-08 23:21:43
首先应该无解析解
8个的情形下的y坐标分别为
12.733898602563759280736579023416
11.347045926489442402527056525453
   10.1458423138090761429485989637
  9.039727510571245179500385832818
7.99043173 ...

相应计算图如下

untitled.jpg



untitled2.jpg

有一天,我打了个瞌睡就到了这里,但我知道我掉入了时光的循环中,虽得以永生,但只有第一个循环有意义。
3楼2013-03-08 23:28:08
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leedobb

金虫 (正式写手)

%% emuch2 我写的程序比较土,可以用eval把它写简洁点
%% http://muchong.com/bbs/viewthread.php?tid=5591194&fpage=1
syms y
kk = 1 ; % 1 for N=8, 0 for N=35
if(kk==1)
Y(1)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-1/8*3.14159267*25/2');
Y(2)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-2/8*3.14159267*25/2');
Y(3)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-3/8*3.14159267*25/2');
Y(4)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-4/8*3.14159267*25/2');
Y(5)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-5/8*3.14159267*25/2');
Y(6)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-6/8*3.14159267*25/2');
Y(7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/8*3.14159267*25/2');
Y(8)=5;

Cx(1:360)= 5*cos((1:360)/360*2*pi);
Cy(1:360)= 5*sin((1:360)/360*2*pi);
for k=1:8
plot(Cx+5,Cy+Y(k));
hold on
end
axis equal
end

if(kk==0)
Y(1)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-1/35*3.14159267*25/2');
Y(2)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-2/35*3.14159267*25/2');
Y(3)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-3/35*3.14159267*25/2');
Y(4)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-4/35*3.14159267*25/2');
Y(5)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-5/35*3.14159267*25/2');
Y(6)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-6/35*3.14159267*25/2');
Y(7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2');

Y(1+7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2-1/35*3.14159267*25/2');
Y(2+7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2-2/35*3.14159267*25/2');
Y(3+7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2-3/35*3.14159267*25/2');
Y(4+7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2-4/35*3.14159267*25/2');
Y(5+7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2-5/35*3.14159267*25/2');
Y(6+7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2-6/35*3.14159267*25/2');
Y(7+7)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-7/35*3.14159267*25/2-7/35*3.14159267*25/2');


Y(1+14)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-14/35*3.14159267*25/2-1/35*3.14159267*25/2');
Y(2+14)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-14/35*3.14159267*25/2-2/35*3.14159267*25/2');
Y(3+14)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-14/35*3.14159267*25/2-3/35*3.14159267*25/2');
Y(4+14)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-14/35*3.14159267*25/2-4/35*3.14159267*25/2');
Y(5+14)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-14/35*3.14159267*25/2-5/35*3.14159267*25/2');
Y(6+14)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-14/35*3.14159267*25/2-6/35*3.14159267*25/2');
Y(7+14)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-14/35*3.14159267*25/2-7/35*3.14159267*25/2');


Y(1+21)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-21/35*3.14159267*25/2-1/35*3.14159267*25/2');
Y(2+21)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-21/35*3.14159267*25/2-2/35*3.14159267*25/2');
Y(3+21)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-21/35*3.14159267*25/2-3/35*3.14159267*25/2');
Y(4+21)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-21/35*3.14159267*25/2-4/35*3.14159267*25/2');
Y(5+21)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-21/35*3.14159267*25/2-5/35*3.14159267*25/2');
Y(6+21)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-21/35*3.14159267*25/2-6/35*3.14159267*25/2');
Y(7+21)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-21/35*3.14159267*25/2-7/35*3.14159267*25/2');


Y(1+28)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-28/35*3.14159267*25/2-1/35*3.14159267*25/2');
Y(2+28)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-28/35*3.14159267*25/2-2/35*3.14159267*25/2');
Y(3+28)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-28/35*3.14159267*25/2-3/35*3.14159267*25/2');
Y(4+28)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-28/35*3.14159267*25/2-4/35*3.14159267*25/2');
Y(5+28)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-28/35*3.14159267*25/2-5/35*3.14159267*25/2');
Y(6+28)=solve('acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-28/35*3.14159267*25/2-6/35*3.14159267*25/2');

Y(35)=5;

Cx(1:360)= 5*cos((1:360)/360*2*pi);
Cy(1:360)= 5*sin((1:360)/360*2*pi);
for k=1:35
plot(Cx+5,Cy+Y(k));
hold on
end
axis equal

end
有一天,我打了个瞌睡就到了这里,但我知道我掉入了时光的循环中,虽得以永生,但只有第一个循环有意义。
4楼2013-03-09 19:10:52
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wgdxidname

木虫 (著名写手)

这个问题很简单, 事实上你已经知道两个圆的相交面积了。 两个圆是一样的,相交面积可以分成相等的两部分。就是两个球冠。 所以就是一个简单的计算球冠面积与球冠的高的关系。知道了球冠的高度,就知道了球的位置。球的的位置其实就是球心的距离决定的。
四大皆空
5楼2013-03-10 13:53:57
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leedobb

金虫 (正式写手)

[  1, 13.176027151195730887376037511169]
[  2, 12.069645466497588461407977633746]
[  3, 11.118769462055524345291348660186]
[  4, 10.250316519654125503870780330281]
[  5, 9.4338861837026170888342452698181]
[  6, 8.6529420407157626908937775258771]
[  7, 7.8970435547786043823469930500081]
[  8, 7.1588843487875437210536890928014]
[  9, 6.4329153888473456695247094804758]
[ 10, 5.7146069654980091268508018372835]
[ 11,                                 5]

还是改成容易执行的给你吧,

%% emuch2
clear all
syms y
NN = 11;  %%% number of circles
NNstr = int2str(NN);

for k=1: NN-1
    kstr = int2str(k);
    expstr = strcat('Y(',kstr,')=solve(acos((y-5)/2/5)*5^2-(25-((y-5)^2/4))^0.5*(y-5)/2-',kstr,'/',NNstr,'*3.14159267*25/2);');
    eval(expstr);
end
clear expstr
Y(NN)=5;
[[1:NN]',Y']
Cx(1:360)= 5*cos((1:360)/360*2*pi);
Cy(1:360)= 5*sin((1:360)/360*2*pi);
for k=1:NN
plot(Cx+5,Cy+Y(k));
hold on
end
axis equal
有一天,我打了个瞌睡就到了这里,但我知道我掉入了时光的循环中,虽得以永生,但只有第一个循环有意义。
6楼2013-03-11 12:00:59
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HDZJY2009

木虫 (正式写手)

引用回帖:
6楼: Originally posted by leedobb at 2013-03-11 12:00:59
还是改成容易执行的给你吧,

%% emuch2
clear all
syms y
NN = 11;  %%% number of circles
NNstr = int2str(NN);

for k=1: NN-1
    kstr = int2str(k);
    expstr = strcat ...

非常感谢!!!!
感恩知足
7楼2013-03-11 15:06:33
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