空間任意兩點(diǎn)之間槽盒及橋架的繪制

1、題目:在空間中以（x1,y1,z1）及（x2,y2,z2）為兩端下邊中點(diǎn)，繪制高為height，寬為width的槽盒。解法：所求兩點(diǎn)間的距離為distance=1. 以（0,0,0）及（distance,0,0）為兩端下邊中點(diǎn)繪制高為height，寬為width的槽盒。繪制函數(shù)如下：glPushMatrix();glBegin(GL_QUADS);glVertex3f(0,0,-width/2);glVertex3f(distance,0,-width/2);glVertex3f(distance,height,-width/2);glVertex3f(0,height,-width/2);g

2、lEnd();glBegin(GL_QUADS);glVertex3f(0,0,-width/2);glVertex3f(0,0,width/2);glVertex3f(distance,0,width/2);glVertex3f(distance,0,-width/2);glEnd();glBegin(GL_QUADS);glVertex3f(0,0,width/2);glVertex3f(distance,0,width/2);glVertex3f(distance,height,width/2);glVertex3f(0,height,width/2);glEnd();glPopMatr

3、ix();（x1,y1,z1）（x2,y2,z2）的向量vec1=（x2-x1,y2-y1,z2-z1）;求出向量vect1及向量的vect2=（distance,0,0）的夾角。求出求出vect3=vect1vect2。的求法:取三個(gè)點(diǎn)A（x2-x1,y2-y1,z2-z1）、B（0,0,0）、C（distance,0,0）由圖可知|AB|=distance |BC|=distance |AC|=依據(jù)三角形余弦定理可知依據(jù)向量vect1和vect2求向量Vect3 的求法：設(shè)vect3（x3,y3,z3）則glTransalted（x1,y1,z1），然后繞vec3逆時(shí)針旋轉(zhuǎn)，即glRota

4、tef（，vect3）;注:此處的vect3需將其三個(gè)坐標(biāo)值寫入已知向量v1(x1,y1,z1)及向量v2(x2,y2,z2),求它兩的法向量v3(x3,y3,z3)橋架的繪制和槽盒大致相同，只需將上面的第1步稍作修改就可，修該后為：綠色為橋架間隔，紅色為底邊。設(shè)間隔長度為gap，單位底邊長為len。如下圖所示:int x=0;glBegin(GL_QUADS);glVertex3f(0,0,-width/2);glVertex3f(distance,0,-width/2);glVertex3f(distance,height,-width/2);glVertex3f(0,height,-wi

5、dth/2);glEnd();glBegin(GL_QUADS);glVertex3f(0,0,width/2);glVertex3f(distance,0,width/2);glVertex3f(distance,height,width/2);glVertex3f(0,height,width/2);glEnd();for (;x<=distance;)if (x+len<distance)glBegin(GL_QUADS);glVertex3f(x,0,-widht/2);glVertex3f(x,0,width/2);glVertex3f(x+len,0,width/2);glVertex3f(x+len,0,-width/2);glEnd();elseglBegin(GL_QUADS);glVertex3f(x,0,-width/2);glVertex3f(x,0,width/2);g