問題描述
我正在嘗試在 three.js 中制作一個彎曲的 3D 箭頭.為了完成這項任務，我創建了一個
我正在嘗試將箭頭(圓柱形狀為錐形)定位在管的末端，如下所示:(Photoshopped)
我的數學不是特別強，而且對 three.js 很陌生.有人可以幫助我了解如何將兩者聯系起來嗎?
這是我當前的代碼:
import T from 'three';var findY = 函數(r，x){返回 Math.sqrt((r * r) - (x * x));}變量半徑 = 25;變量 x = 0;變量 z = 0;var numberOfPoints = 10;var 間隔 = (半徑/numberOfPoints);變量點 = [];for (var i = numberOfPoints; i >= 0; i--){var y = findY(半徑, x);points.push(new T.Vector3(x, y, z))x = x + 區間；}x = x - 間隔；for (var i = numberOfPoints - 1 ; i >= 0; i--){y = findY(半徑, x) * -1;points.push(new T.Vector3(x, y, z));x = x - 間隔；}var path = new T.CatmullRomCurve3(points);var tubeGeometry = new T.TubeGeometry(路徑，//路徑10,//段radius/10,//半徑8,//radiusSegmentsfalse//關閉);var coneGeometry = new T.CylinderGeometry(半徑頂部 = 0.1，radiusBottom = 半徑/5，高度 = 10，徑向段 = 10，高度段 = 10，開放式 = 1);var 材料 = 新 T.MeshBasicMaterial( { 顏色: 0x00ff00 } );var tube = new T.Mesh(tubeGeometry, material);var cone = new T.Mesh(coneGeometry, material);//平移和旋轉圓錐?如果有人可以嘗試簡單地解釋一下數學和編程完成的必要條件，我將不勝感激
- 找到位于管末端的法線
- 將圓錐體移動到正確的位置
感謝任何幫助！
當您可以直接在原地創建箭頭時，不要為此使用旋轉.同樣，彎管也可以這樣做.您唯一需要的是由 A,B 端點定義的最后一條線段.
令A為尖點，B為圓盤基中心.要創建箭頭，您需要 2 個額外的基礎向量，讓它們稱為基礎圓盤的 U,V 和半徑 r.從它們中，您可以使用如下簡單的圓形公式創建圓盤點:
獲取
AB端點計算
U,V基向量U,V應位于箭頭的圓盤底部，且相互垂直.箭頭的方向(行|BA|)是圓盤基法線，因此利用叉積將垂直向量返回到相乘的向量，因此:W = B-A;W/= |W|;//單位向量T = (1,0,0);//temp 任何不平行于 W 的非零向量如果 ( |(W.T)|>0.75 ) T = (0,1,0);//如果 T 和 W 的絕對點積接近 1，則表示它們接近平行，因此選擇不同的 TU = (T x W)//U 垂直于 T,WV = (U x W)//V 垂直于 U,W創建/渲染箭頭幾何
即簡單攤位
A,B是三角扇的中心(需要2個)，圓盤基點計算如下:P(ang) = B + U.r.cos(ang) + V.r.sin(ang)因此，只需將
ang循環遍歷整個圓圈，然后您就可以獲得足夠的點數(通常 36 點就足夠了)并從它們中進行兩個三角形扇形.不要忘記最后一個圓盤點必須與第一個圓盤點相同，否則您會在ang = 0或360度數上出現丑陋的外觀或孔.
如果您仍想進行輪換，則可以這樣做.以與上述相同的方式計算 U,V,W 并從中構造變換矩陣.原點 O 將是點 B 軸 X,Y,Z 將是 U,V,W順序取決于您的箭頭型號.W 應該與模型軸匹配.U,V 可以是任意順序.所以只需將所有向量復制到它們的位置并使用此矩陣進行渲染.欲了解更多信息，請參閱:
void glArrowRoundxy(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat a2){常量 int _glCircleN=50;//每圈點數常量 int n=3*_glCircleN;整數 i,j,ix,e;浮動 x,y,z,x1,y1,z1,a,b,da,db=pi2/(_glCircleN-1);浮動 ux,uy,uz,vx,vy,vz,u,v;//緩沖區GLfloat ptab[6*_glCircleN],*p0,*p1,*n0,*n1,*p;p0=ptab+(0*_glCircleN);//上一個管段圓點p1=ptab+(3*_glCircleN);//實際管段圓點大=+分貝；如果(a0>a1)da=-db；//主角度步長方向ux=0.0；//U 垂直于箭頭平面uy=0.0；uz=1.0；//圓弧插值 a=<a0,a1>對于 (e=1,j=0,a=a0;e;j++,a+=da){//結束條件如果 ((da>0.0)&&(a>=a1)) { a=a1;e=0;}if ((da<0.0)&&(a<=a1)) { a=a1;e=0;}//計算實際的管中心x1=x0+(r*cos(a));y1=y0+(r*sin(a));z1=z0；//V 是從 (x0,y0,z0) 到 (x1,y1,z1) 的方向vx=x1-x0；vy=y1-y0；vz=z1-z0；//和粗略的單位b=sqrt((vx*vx)+(vy*vy)+(vz*vz));如果(b＞1e-6)b=1.0/b；否則 b=0.0;vx*=b;vy*=b;vz*=b;//管段對于 (ix=0,b=0.0,i=0;i<_glCircleN;i++,b+=db){u=r0*cos(b)；v=r0*sin(b)；p1[ix]=x1+(ux*u)+(vx*v);九++；p1[ix]=y1+(uy*u)+(vy*v);九++；p1[ix]=z1+(uz*u)+(vz*v);九++；}如果 (!j){glBegin(GL_TRIANGLE_FAN);glVertex3f(x1,y1,z1);對于 (ix=0;ix這會在 XY 平面上渲染彎曲的箭頭，其中心為
x,y,z，半徑為r.r0是管半徑，r1是箭頭基部半徑.因為我沒有你的曲線定義，所以我選擇 XY 平面中的圓.a0,a1,a2是箭頭開始 (a0)、箭頭開始 (a1) 和結束 (a2).pi2只是常量pi2=6.283185307179586476925286766559.這個想法是記住實際和之前的管段圓點，以便
ptab,p0,p1存在，否則您需要計算所有內容兩次.當我直接選擇 XY 平面時，我知道一個基向量是垂直于它的.第二個是垂直于它和箭頭方向幸運的是圓形屬性提供了它自己，因此在這種情況下不需要叉積.
如果不評論我，希望它足夠清楚.
我需要將它添加到我的引擎中，所以這里是 3D 版本(不僅綁定到軸對齊的箭頭，而且圓錐體也彎曲了).除了基礎向量計算之外它是相同的，我還在標題中稍微改變了角度
<a0,a1>是整個間隔，aa是箭頭大小但后者在代碼中被轉換為原始約定.我還添加了用于照明計算的法線.我還添加了線性箭頭，其中基向量的計算沒有利用圓屬性，以防你得到不同的曲線.這里的結果://---------------------------------------------------------------------------常量 int _glCircleN=50;//每圈點數//--------------------------------------------------------------------------void glCircleArrowxy(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat aa){雙 pos[3]={ x0, y0, z0};雙倍nor[3]={0.0,0.0,1.0}；雙 bin[3]={1.0,0.0,0.0};glCircleArrow3D(pos,nor,bin,r,r0,r1,a0,a1,aa);}//--------------------------------------------------------------------------void glCircleArrowyz(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat aa){雙 pos[3]={ x0, y0, z0};雙nor[3]={1.0,0.0,0.0}；雙倉[3]={0.0,1.0,0.0}；glCircleArrow3D(pos,nor,bin,r,r0,r1,a0,a1,aa);}//--------------------------------------------------------------------------void glCircleArrowxz(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat aa){雙 pos[3]={ x0, y0, z0};雙倍nor[3]={0.0,1.0,0.0}；雙倉[3]={0.0,0.0,1.0}；glCircleArrow3D(pos,nor,bin,r,r0,r1,a0,a1,aa);}//--------------------------------------------------------------------------void glCircleArrow3D(double *pos,double *nor,double *bin,double r,double r0,double r1,double a0,double a1,double aa){//常量 int _glCircleN=20;//每圈點數整數 e,i,j,N=3*_glCircleN;雙 U[3],V[3],u,v;雙a,b,da,db=pi2/double(_glCircleN-1),a2,rr;雙 *ptab,*p0,*p1,*n0,*n1,*pp,p[3],q[3],c[3],n[3],tan[3];//緩沖區ptab=新雙 [12*_glCircleN];如果 (ptab==NULL) 返回；p0=ptab+(0*_glCircleN);n0=ptab+(3*_glCircleN);p1=ptab+(6*_glCircleN);n1=ptab+(9*_glCircleN);//準備角度a2=a1;大=分貝；aa=晶圓廠(aa)；如果(a0>a1){da=-da;aa=-aa;}a1-=aa；//計算缺失的基向量vector_copy(U,nor);//U 垂直于箭頭平面vector_mul(tan,nor,bin);//切線垂直于法線和副法線//圓弧插值 a=<a0,a2>對于 (e=0,j=0,a=a0;e<5;j++,a+=da){//結束條件if (e==0)//e=0{如果 ((da>0.0)&&(a>=a1)) { a=a1;e++;}if ((da<0.0)&&(a<=a1)) { a=a1;e++;}rr=r0;}else{//e=1,2,3,4如果 ((da>0.0)&&(a>=a2)) { a=a2;e++;}if ((da<0.0)&&(a<=a2)) { a=a2;e++;}rr=r1*fabs(除法(a-a2,a2-a1));}//計算實際管段中心 c[3]u=r*cos(a);v=r*sin(a);vector_mul(p,bin,u);矢量_mul(q,tan,v);vector_add(c,p,q);vector_add(c,c,pos);//V是從箭頭中心到管段中心的單位方向vector_sub(V,c,pos);矢量_一個(V，V)；//管段插值對于 (b=0.0,i=0;i用法:
glColor3f(0.5,0.5,0.5);glCircleArrowyz(+3.5,0.0,0.0,0.5,0.1,0.2,0.0*deg,+270.0*deg,45.0*deg);glCircleArrowyz(-3.5,0.0,0.0,0.5,0.1,0.2,0.0*deg,-270.0*deg,45.0*deg);glCircleArrowxz(0.0,+3.5,0.0,0.5,0.1,0.2,0.0*deg,+270.0*deg,45.0*deg);glCircleArrowxz(0.0,-3.5,0.0,0.5,0.1,0.2,0.0*deg,-270.0*deg,45.0*deg);glCircleArrowxy(0.0,0.0,+3.5,0.5,0.1,0.2,0.0*deg,+270.0*deg,45.0*deg);glCircleArrowxy(0.0,0.0,-3.5,0.5,0.1,0.2,0.0*deg,-270.0*deg,45.0*deg);glColor3f(0.2,0.2,0.2);glLinearArrow3D(vector_ld(+2.0,0.0,0.0),vector_ld(+1.0,0.0,0.0),0.1,0.2,2.0,0.5);glLinearArrow3D(vector_ld(-2.0,0.0,0.0),vector_ld(-1.0,0.0,0.0),0.1,0.2,2.0,0.5);glLinearArrow3D(vector_ld(0.0,+2.0,0.0),vector_ld(0.0,+1.0,0.0),0.1,0.2,2.0,0.5);glLinearArrow3D(vector_ld(0.0,-2.0,0.0),vector_ld(0.0,-1.0,0.0),0.1,0.2,2.0,0.5);glLinearArrow3D(vector_ld(0.0,0.0,+2.0),vector_ld(0.0,0.0,+1.0),0.1,0.2,2.0,0.5);glLinearArrow3D(vector_ld(0.0,0.0,-2.0),vector_ld(0.0,0.0,-1.0),0.1,0.2,2.0,0.5);和箭頭的概述(在圖像的右側):
我正在使用我的矢量庫，所以這里有一些解釋:
vector_mul(a[3],b[3],c[3])是叉積a = b x cvector_mul(a[3],b[3],c)是簡單的標量乘法a = b.ca = vector_mul(b[3],c[3])是點積a = (b.c)vector_one(a[3],b[3])是單位向量a = b/|b|vector_copy(a[3],b[3])只是復制a = bvector_add(a[3],b[3],c[3])正在添加a = b + cvector_sub(a[3],b[3],c[3])正在減去a = b - cvector_neg(a[3],b[3])是否定a = -bvector_ld(a[3],x,y,z)正在加載a = (x,y,z)pos是圓箭頭的中心位置，nor是箭頭所在平面的法線.bin是雙法線，角度從這個軸開始.應該垂直于nor.r,r0,r1是箭頭的半徑(彎、管、錐)線性箭頭類似于
dir是箭頭的方向，l是箭頭大小，al是箭頭大小.I am attempting to make a curved 3D arrow in three.js. To accomplish this task, I have created a Tube that follows a curved path and a Cylinder shaped as a cone (by setting radiusTop to be tiny). They currently look like so:
I am attempting to position the Arrow Head (Cylinder shaped as a cone) at the end of the Tube like so: (Photoshopped)
I am not terribly strong in math and pretty new to three.js. Could someone help me understand how to connect the two?
Here is my current code:
import T from 'three'; var findY = function(r, x) { return Math.sqrt((r * r) - (x * x)); } var radius = 25; var x = 0; var z = 0; var numberOfPoints = 10; var interval = (radius/numberOfPoints); var points = []; for (var i = numberOfPoints; i >= 0; i--) { var y = findY(radius, x); points.push(new T.Vector3(x, y, z)) x = x + interval; } x = x - interval; for (var i = numberOfPoints - 1 ; i >= 0; i--) { y = findY(radius, x) * -1; points.push(new T.Vector3(x, y, z)); x = x - interval; } var path = new T.CatmullRomCurve3(points); var tubeGeometry = new T.TubeGeometry( path, //path 10, //segments radius / 10, //radius 8, //radiusSegments false //closed ); var coneGeometry = new T.CylinderGeometry( radiusTop = 0.1, radiusBottom = radius/5, height = 10, radialSegments = 10, heightSegments = 10, openEnded = 1 ); var material = new T.MeshBasicMaterial( { color: 0x00ff00 } ); var tube = new T.Mesh( tubeGeometry, material ); var cone = new T.Mesh( coneGeometry, material ); // Translate and Rotate cone?I would greatly appreciate if someone could attempt a simple explanation of what is necessary mathematically and programmatically accomplish
- Finding the normal located at the end of the tube
- Shifting the Cone to the correct location
Any help is appreciated!
解決方案Do not use rotation for this when you can create the arrowhead directly in place. Similarly the bended tube can be done this way too. Only thing you need for it is the last line segment defined by
A,Bendpoints.Let
Abe the sharp point andBthe disc base center. To create arrowhead you need 2 additional basis vectors let call themU,Vand radiusrof base disc. From them you can create disc points with simple circle formula like this:obtain
ABendpointscompute
U,Vbasis vectorsThe
U,Vshould lie in the disc base of arrowhead and should be perpendicular to each other. direction of the arrowhead (line|BA|) is the disc base normal so exploit cross product which returns perpendicular vector to the multiplied ones so:W = B-A; W /= |W|; // unit vector T = (1,0,0); // temp any non zero vector not parallel to W if ( |(W.T)|>0.75 ) T = (0,1,0); // if abs dot product of T and W is close to 1 it means they are close to parallel so chose different T U = (T x W) // U is perpendicular to T,W V = (U x W) // V is perpendicular to U,Wcreate/render arrowhead geometry
That is easy booth
A,Bare centers of triangle fan (need 2) and the disc base points are computed like this:P(ang) = B + U.r.cos(ang) + V.r.sin(ang)So just loop
angthrough the whole circle with some step so you got enough points (usually 36 is enough) and do both triangle fans from them. Do not forget the last disc point must be the same as the first one otherwise you will got ugly seems or hole on theang = 0or360deg.
If you still want to go for rotations instead then this is doable like this. compute
U,V,Win the same way as above and construct transformation matrix from them. the originOwill be pointBand axisesX,Y,Zwill beU,V,Wthe order depends on your arrowhead model.Wshould match the model axis.U,Vcan be in any order. So just copy all the vectors to their places and use this matrix for rendering. For more info see:- Understanding 4x4 homogenous transform matrices
[Notes]
If you do not know how to compute vector operations like cross/dot products or absolute value see:
// cross product: W = U x V W.x=(U.y*V.z)-(U.z*V.y) W.y=(U.z*V.x)-(U.x*V.z) W.z=(U.x*V.y)-(U.y*V.x) // dot product: a = (U.V) a=U.x*V.x+U.y*V.y+U.z*V.z // abs of vector a = |U| a=sqrt((U.x*U.x)+(U.y*U.y)+(U.z*U.z))[Edit1] simple GL implementation
I do not code in your environment but as downvote and comment suggest you guys are not able to put this together on your own which is odd considering you got this far so here simple C++/GL exmaple of how to do this (you can port this to your environment):
void glArrowRoundxy(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat a2) { const int _glCircleN=50; // points per circle const int n=3*_glCircleN; int i,j,ix,e; float x,y,z,x1,y1,z1,a,b,da,db=pi2/(_glCircleN-1); float ux,uy,uz,vx,vy,vz,u,v; // buffers GLfloat ptab[6*_glCircleN],*p0,*p1,*n0,*n1,*p; p0=ptab+(0*_glCircleN); // previous tube segment circle points p1=ptab+(3*_glCircleN); // actual tube segment circle points da=+db; if (a0>a1) da=-db; // main angle step direction ux=0.0; // U is normal to arrow plane uy=0.0; uz=1.0; // arc interpolation a=<a0,a1> for (e=1,j=0,a=a0;e;j++,a+=da) { // end conditions if ((da>0.0)&&(a>=a1)) { a=a1; e=0; } if ((da<0.0)&&(a<=a1)) { a=a1; e=0; } // compute actual tube ceneter x1=x0+(r*cos(a)); y1=y0+(r*sin(a)); z1=z0; // V is direction from (x0,y0,z0) to (x1,y1,z1) vx=x1-x0; vy=y1-y0; vz=z1-z0; // and unit of coarse b=sqrt((vx*vx)+(vy*vy)+(vz*vz)); if (b>1e-6) b=1.0/b; else b=0.0; vx*=b; vy*=b; vz*=b; // tube segment for (ix=0,b=0.0,i=0;i<_glCircleN;i++,b+=db) { u=r0*cos(b); v=r0*sin(b); p1[ix]=x1+(ux*u)+(vx*v); ix++; p1[ix]=y1+(uy*u)+(vy*v); ix++; p1[ix]=z1+(uz*u)+(vz*v); ix++; } if (!j) { glBegin(GL_TRIANGLE_FAN); glVertex3f(x1,y1,z1); for (ix=0;ix<n;ix+=3) glVertex3fv(p1+ix); glEnd(); } else{ glBegin(GL_QUAD_STRIP); for (ix=0;ix<n;ix+=3) { glVertex3fv(p0+ix); glVertex3fv(p1+ix); } glEnd(); } // swap buffers p=p0; p0=p1; p1=p; p=n0; n0=n1; n1=p; } // arrowhead a=<a1,a2> for (ix=0,b=0.0,i=0;i<_glCircleN;i++,b+=db) { u=r1*cos(b); v=r1*sin(b); p1[ix]=x1+(ux*u)+(vx*v); ix++; p1[ix]=y1+(uy*u)+(vy*v); ix++; p1[ix]=z1+(uz*u)+(vz*v); ix++; } glBegin(GL_TRIANGLE_FAN); glVertex3f(x1,y1,z1); for (ix=0;ix<n;ix+=3) glVertex3fv(p1+ix); glEnd(); x1=x0+(r*cos(a2)); y1=y0+(r*sin(a2)); z1=z0; glBegin(GL_TRIANGLE_FAN); glVertex3f(x1,y1,z1); for (ix=n-3;ix>=0;ix-=3) glVertex3fv(p1+ix); glEnd(); }This renders bended arrow in XY plane with center
x,y,zand big radiusr. Ther0is tube radius andr1is arrowhead base radius. As I do not have your curve definition I choose circle in XY plane. Thea0,a1,a2are angles where arrow starts (a0), arrowhead starts (a1) and ends (a2). Thepi2is just constantpi2=6.283185307179586476925286766559.The idea is to remember actual and previous tube segment circle points so there for the
ptab,p0,p1otherwise you would need to compute everything twice.As I chose XY plane directly then I know that one base vector is normal to it. and second is perpendicular to it and to arrow direction luckily circle properties provides this on its own therefore no need for cross products in this case.
Hope it is clear enough if not comment me.
[Edit2]
I needed to add this to my engine so here is the 3D version (not bound just to axis aligned arrows and the cone is bended too). It is the same except the basis vector computation and I also change the angles a bit in the header
<a0,a1>is the whole interval andaais the arrowhead size but latter in code it is converted to the original convention. I added also normals for lighting computations. I added also linear Arrow where the computation of basis vectors is not taking advantage of circle properties in case you got different curve. Here result://--------------------------------------------------------------------------- const int _glCircleN=50; // points per circle //--------------------------------------------------------------------------- void glCircleArrowxy(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat aa) { double pos[3]={ x0, y0, z0}; double nor[3]={0.0,0.0,1.0}; double bin[3]={1.0,0.0,0.0}; glCircleArrow3D(pos,nor,bin,r,r0,r1,a0,a1,aa); } //--------------------------------------------------------------------------- void glCircleArrowyz(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat aa) { double pos[3]={ x0, y0, z0}; double nor[3]={1.0,0.0,0.0}; double bin[3]={0.0,1.0,0.0}; glCircleArrow3D(pos,nor,bin,r,r0,r1,a0,a1,aa); } //--------------------------------------------------------------------------- void glCircleArrowxz(GLfloat x0,GLfloat y0,GLfloat z0,GLfloat r,GLfloat r0,GLfloat r1,GLfloat a0,GLfloat a1,GLfloat aa) { double pos[3]={ x0, y0, z0}; double nor[3]={0.0,1.0,0.0}; double bin[3]={0.0,0.0,1.0}; glCircleArrow3D(pos,nor,bin,r,r0,r1,a0,a1,aa); } //--------------------------------------------------------------------------- void glCircleArrow3D(double *pos,double *nor,double *bin,double r,double r0,double r1,double a0,double a1,double aa) { // const int _glCircleN=20; // points per circle int e,i,j,N=3*_glCircleN; double U[3],V[3],u,v; double a,b,da,db=pi2/double(_glCircleN-1),a2,rr; double *ptab,*p0,*p1,*n0,*n1,*pp,p[3],q[3],c[3],n[3],tan[3]; // buffers ptab=new double [12*_glCircleN]; if (ptab==NULL) return; p0=ptab+(0*_glCircleN); n0=ptab+(3*_glCircleN); p1=ptab+(6*_glCircleN); n1=ptab+(9*_glCircleN); // prepare angles a2=a1; da=db; aa=fabs(aa); if (a0>a1) { da=-da; aa=-aa; } a1-=aa; // compute missing basis vectors vector_copy(U,nor); // U is normal to arrow plane vector_mul(tan,nor,bin); // tangent is perpendicular to normal and binormal // arc interpolation a=<a0,a2> for (e=0,j=0,a=a0;e<5;j++,a+=da) { // end conditions if (e==0) // e=0 { if ((da>0.0)&&(a>=a1)) { a=a1; e++; } if ((da<0.0)&&(a<=a1)) { a=a1; e++; } rr=r0; } else{ // e=1,2,3,4 if ((da>0.0)&&(a>=a2)) { a=a2; e++; } if ((da<0.0)&&(a<=a2)) { a=a2; e++; } rr=r1*fabs(divide(a-a2,a2-a1)); } // compute actual tube segment center c[3] u=r*cos(a); v=r*sin(a); vector_mul(p,bin,u); vector_mul(q,tan,v); vector_add(c,p, q); vector_add(c,c,pos); // V is unit direction from arrow center to tube segment center vector_sub(V,c,pos); vector_one(V,V); // tube segment interpolation for (b=0.0,i=0;i<N;i+=3,b+=db) { u=cos(b); v=sin(b); vector_mul(p,U,u); // normal vector_mul(q,V,v); vector_add(n1+i,p,q); vector_mul(p,n1+i,rr); // vertex vector_add(p1+i,p,c); } if (e>1) // recompute normals for cone { for (i=3;i<N;i+=3) { vector_sub(p,p0+i ,p1+i); vector_sub(q,p1+i-3,p1+i); vector_mul(p,p,q); vector_one(n1+i,p); } vector_sub(p,p0 ,p1); vector_sub(q,p1+N-3,p1); vector_mul(p,q,p); vector_one(n1,p); if (da>0.0) for (i=0;i<N;i+=3) vector_neg(n1+i,n1+i); if (e== 3) for (i=0;i<N;i+=3) vector_copy(n0+i,n1+i); } // render base disc if (!j) { vector_mul(n,U,V); glBegin(GL_TRIANGLE_FAN); glNormal3dv(n); glVertex3dv(c); if (da<0.0) for (i=N-3;i>=0;i-=3) glVertex3dv(p1+i); else for (i= 0;i< N;i+=3) glVertex3dv(p1+i); glEnd(); } // render tube else{ glBegin(GL_QUAD_STRIP); if (da<0.0) for (i=0;i<N;i+=3) { glNormal3dv(n1+i); glVertex3dv(p1+i); glNormal3dv(n0+i); glVertex3dv(p0+i); } else for (i=0;i<N;i+=3) { glNormal3dv(n0+i); glVertex3dv(p0+i); glNormal3dv(n1+i); glVertex3dv(p1+i); } glEnd(); } // swap buffers pp=p0; p0=p1; p1=pp; pp=n0; n0=n1; n1=pp; // handle r0 -> r1 edge if (e==1) a-=da; if ((e==1)||(e==2)||(e==3)) e++; } // release buffers delete[] ptab; } //--------------------------------------------------------------------------- void glLinearArrow3D(double *pos,double *dir,double r0,double r1,double l,double al) { // const int _glCircleN=20; // points per circle int e,i,N=3*_glCircleN; double U[3],V[3],W[3],u,v; double a,da=pi2/double(_glCircleN-1),r,t; double *ptab,*p0,*p1,*n1,*pp,p[3],q[3],c[3],n[3]; // buffers ptab=new double [9*_glCircleN]; if (ptab==NULL) return; p0=ptab+(0*_glCircleN); p1=ptab+(3*_glCircleN); n1=ptab+(6*_glCircleN); // compute basis vectors vector_one(W,dir); vector_ld(p,1.0,0.0,0.0); vector_ld(q,0.0,1.0,0.0); vector_ld(n,0.0,0.0,1.0); a=fabs(vector_mul(W,p)); pp=p; t=a; a=fabs(vector_mul(W,q)); if (t>a) { pp=q; t=a; } a=fabs(vector_mul(W,n)); if (t>a) { pp=n; t=a; } vector_mul(U,W,pp); vector_mul(V,U,W); vector_mul(U,V,W); for (e=0;e<4;e++) { // segment center if (e==0) { t=0.0; r= r0; } if (e==1) { t=l-al; r= r0; } if (e==2) { t=l-al; r= r1; } if (e==3) { t=l; r=0.0; } vector_mul(c,W,t); vector_add(c,c,pos); // tube segment interpolation for (a=0.0,i=0;i<N;i+=3,a+=da) { u=cos(a); v=sin(a); vector_mul(p,U,u); // normal vector_mul(q,V,v); vector_add(n1+i,p,q); vector_mul(p,n1+i,r); // vertex vector_add(p1+i,p,c); } if (e>2) // recompute normals for cone { for (i=3;i<N;i+=3) { vector_sub(p,p0+i ,p1+i); vector_sub(q,p1+i-3,p1+i); vector_mul(p,p,q); vector_one(n1+i,p); } vector_sub(p,p0 ,p1); vector_sub(q,p1+N-3,p1); vector_mul(p,q,p); vector_one(n1,p); } // render base disc if (!e) { vector_neg(n,W); glBegin(GL_TRIANGLE_FAN); glNormal3dv(n); glVertex3dv(c); for (i=0;i<N;i+=3) glVertex3dv(p1+i); glEnd(); } // render tube else{ glBegin(GL_QUAD_STRIP); for (i=0;i<N;i+=3) { glNormal3dv(n1+i); glVertex3dv(p0+i); glVertex3dv(p1+i); } glEnd(); } // swap buffers pp=p0; p0=p1; p1=pp; } // release buffers delete[] ptab; } //---------------------------------------------------------------------------usage:
glColor3f(0.5,0.5,0.5); glCircleArrowyz(+3.5,0.0,0.0,0.5,0.1,0.2,0.0*deg,+270.0*deg,45.0*deg); glCircleArrowyz(-3.5,0.0,0.0,0.5,0.1,0.2,0.0*deg,-270.0*deg,45.0*deg); glCircleArrowxz(0.0,+3.5,0.0,0.5,0.1,0.2,0.0*deg,+270.0*deg,45.0*deg); glCircleArrowxz(0.0,-3.5,0.0,0.5,0.1,0.2,0.0*deg,-270.0*deg,45.0*deg); glCircleArrowxy(0.0,0.0,+3.5,0.5,0.1,0.2,0.0*deg,+270.0*deg,45.0*deg); glCircleArrowxy(0.0,0.0,-3.5,0.5,0.1,0.2,0.0*deg,-270.0*deg,45.0*deg); glColor3f(0.2,0.2,0.2); glLinearArrow3D(vector_ld(+2.0,0.0,0.0),vector_ld(+1.0,0.0,0.0),0.1,0.2,2.0,0.5); glLinearArrow3D(vector_ld(-2.0,0.0,0.0),vector_ld(-1.0,0.0,0.0),0.1,0.2,2.0,0.5); glLinearArrow3D(vector_ld(0.0,+2.0,0.0),vector_ld(0.0,+1.0,0.0),0.1,0.2,2.0,0.5); glLinearArrow3D(vector_ld(0.0,-2.0,0.0),vector_ld(0.0,-1.0,0.0),0.1,0.2,2.0,0.5); glLinearArrow3D(vector_ld(0.0,0.0,+2.0),vector_ld(0.0,0.0,+1.0),0.1,0.2,2.0,0.5); glLinearArrow3D(vector_ld(0.0,0.0,-2.0),vector_ld(0.0,0.0,-1.0),0.1,0.2,2.0,0.5);and overview of the arows (on the right side of image):
I am using my vector lib so here are some explanations:
vector_mul(a[3],b[3],c[3])is cross producta = b x cvector_mul(a[3],b[3],c)is simple multiplication by scalara = b.ca = vector_mul(b[3],c[3])is dot producta = (b.c)vector_one(a[3],b[3])is unit vectora = b/|b|vector_copy(a[3],b[3])is just copya = bvector_add(a[3],b[3],c[3])is addinga = b + cvector_sub(a[3],b[3],c[3])is substractinga = b - cvector_neg(a[3],b[3])is negationa = -bvector_ld(a[3],x,y,z)is just loadinga = (x,y,z)The
posis the center position of your circle arrow andnoris normal of the plane in which the arrow lies.binis bi-normal and the angles are starting from this axis. should be perpendicular tonor.r,r0,r1are the radiuses of the arrow (bend,tube,cone)The linear arrow is similar the
diris direction of the arrow,lis arrow size andalis arrowhead size.這篇關于根據管端法線向圓柱體應用旋轉的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！
【網站聲明】本站部分內容來源于互聯網,旨在幫助大家更快的解決問題，如果有圖片或者內容侵犯了您的權益，請聯系我們刪除處理，感謝您的支持！