计算加热杆形变距离-优快云博客

本文介绍了一种计算方法，当细长杆受热膨胀并固定于两墙间时，如何计算其中心位移距离。文章给出了详细的数学推导过程，并提供了一段使用C++实现的示例代码。

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Description

When a thin rod of length L is heated n degrees, it expands to a new length L'=(1+n*C)*L, where C is the coefficient of heat expansion.
When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment.

Your task is to compute the distance by which the center of the rod is displaced.

Input

The input contains multiple lines. Each line of input contains three non-negative numbers: the initial lenth of the rod in millimeters, the temperature change in degrees and the coefficient of heat expansion of the material. Input data guarantee that no rod expands by more than one half of its original length. The last line of input contains three negative numbers and it should not be processed.

Output

For each line of input, output one line with the displacement of the center of the rod in millimeters with 3 digits of precision.

Sample Input

1000 100 0.0001
15000 10 0.00006
10 0 0.001
-1 -1 -1

Sample Output

61.329
225.020
0.000

题意：

弧L'=(1+N*C)*L；求弧到线的最短距离

分析：
先求出各个方程

L’=(1+n*C)*L

θr = 1/2*L'

sinθ= 1/2*L/r

r*r = (1/2*L)*(1/2*L) + (r-x)*(r-x)

导出一个大工式：

r=(L*L+4*x*x)/(8*x);

L'=2*θ*r=2*θ*((L*L+4*x*x)/(8*x))

再代入sinθ= 1/2*L/r;

利用asin函数

代码：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<stdio.h>
using namespace std;
double L, N, C, S;
bool check(double mid)
{
    double ans=2*asin((L/2)/((L*L+4*mid*mid)/(8*mid)))*((L*L+4*mid*mid)/(8*mid));
    return ans >= S;
}
int main()
{
    double left, right, mid;
    while (cin>>L>>N>>C)
    {
        if (L + N + C < 0 )
            break;
        S = (1 + N*C)*L;
        left = 0.0;
        right = L/2;
        while (right-left> 1e-5)
        {
            mid=(right+left)/2;
            if (check(mid))
            {
                right=mid;
            }
            else
            {
                left=mid;
            }
        }
        printf("%.3f\n",right);
    }
    return 0;
}

感受：
asin（）函数我当然不知道了，嘻嘻，偷偷百度的