第五十四章 国奖开赛
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果然，才开始没多久。房间里喊选B的就多了许多。吴哲则是撇了下嘴，他都不用仔细考虑，最后他们得出的答案估计以后都不会再买彩票了。
吴哲选了A题，也没去打扰黄明海他们选题。
吴哲现在必须解决三个难题：（1）在满足该设计规范的条件下，计算线光源长度，使线光源的功率最小。
（2）对得到光源长度，在有标尺的坐标系中画出测试屏上反射光的亮区。
（3）讨论该设计规范的合理性。
已知：其开口半径36毫米，深度21.6毫米。经过车灯的焦点，在与对称轴相垂直的水平方向，对称地放置一定长度的均匀分布的光源。
那么在该直线A点的同侧取B点和C点，使AC=2AB=2.6米。要求C点的光强度不小于某一额定值（可取为1个单位），B点的光强度不小于该额定值的两倍（只须考虑一次反射）。
建立坐标系如下，记线光源长度为l，功率为W，B，C点的光强度分别为W和W，先求和的表达式，再建立整个问题的数学模型.
以下均以毫米为单位，由所给信息不难求出车灯反射面方程为，焦点坐标为（0，0，15）。位于点P(0，w，15)的单位能量的点光源反射到点C(0，2600，25015)的能量
设反射点的坐标为Q.记入射向量为，该点反射面外法线方向为，不难得到反射向量，满足K为常量。
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其中e为起亮值，l为最大值点，为考察的最大范围，例如取为20mm。也有类似的性质，且起亮值和最大值点均相应地右移.数值求解
记其解为，再求出，不难看出e落在l之中。
这就证明了e的确是最小值。最后计算得出e=3.294
下来就是求放射光亮区的问题了。分别将线光源和车灯反射面离散化为点光源和面元的集合，计算每一点光源关于每一车灯反射面元的反射光线，判断其是否与车灯反射面相交，若相交，一次反射光不能到达测试屏，否则求出该反射光线与反射屏平面的交点，即为反射亮点。
所有这些亮点的集合即为反射光亮区。亮区的上半部分由下图所示(横坐标为x轴，纵坐标为y轴，单位为mm)，下半部分与上半部分是关于x轴对称的.
计算的另一方法是建立问题的数值模型用数值模拟的方法加以解决.具体的做法是:在得到反射光线和反射到测试屏上能量的数学模型后，分别将线光源和车灯反射面离散化为点光源和面元的集合，在测试屏B(或C)点附近取一微小面元.计算每一点光源关于每一车灯反射面元的反射光线，将所有能到达该面元的反射光线的反射能量迭加起来，除以面元的面积即为B(或C)点的反射能量密度.
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