cnreader Posted March 2, 2009 Share Posted March 2, 2009 Illustration bellow, where point a is outside of plane S1 and line ab is perpendicular to line bc, which is on the S1. another plane s2 is under S1. line gd through vertex g of s2 and is perpendicular to cd, which is on plane S1. ag and ah is two ray, from point a to vertex g and h individually. How to define the cross points of these two ray to plane S1? By which way we can get the two cross points? o Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 2, 2009 Author Share Posted March 2, 2009 pic l Quote Link to comment Share on other sites More sharing options...
Cad64 Posted March 2, 2009 Share Posted March 2, 2009 Pictures are not working. Refer to THIS thread for instructions on how to add images to your post. Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 3, 2009 Author Share Posted March 3, 2009 try it again. and thanks Quote Link to comment Share on other sites More sharing options...
PS_Port Posted March 3, 2009 Share Posted March 3, 2009 This sounds rather like a maths question.. any more information ? points? vectors or plane equations? Also lines ab & gd are parallel, as they are orthogonal to vectors contained in the plane S1. Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 3, 2009 Author Share Posted March 3, 2009 This sounds rather like a maths question..any more information ? points? vectors or plane equations? Also lines ab & gd are parallel, as they are orthogonal to vectors contained in the plane S1. yes, its a math question. geometry drawing. you are right, ab//gd, as they are all perpendular to plane S1, so they are parallel. As its belong to a goemetry drawing, so no equation. of cause you can get the cross points by hide render. but I hope not to use this method. Quote Link to comment Share on other sites More sharing options...
PS_Port Posted March 3, 2009 Share Posted March 3, 2009 Ahhh I get the question, wouldn't you just view from the side and end of plane S1 and note where the rays pass through S1 (x,y directions). I would much prefer a pure mathematical solution. Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 3, 2009 Author Share Posted March 3, 2009 no, your answer is wrong. a little cute: try to draw a plane adg, it will be perpendicular to S1 (can you prove?) then find cross line of S1 and adg, which will cross ag, the cross point is one we wanted. (can you prove?) next ah... the pure math solution is simple very much, just find the plane equation as well as lines; solve can get the point. go on... Quote Link to comment Share on other sites More sharing options...
eldon Posted March 3, 2009 Share Posted March 3, 2009 Have you drawn the figure in 3D? If you do, then you can take a front view, so that the plane S1 is seen as a line, then trim the lines ah and ag where they cross that line. Then when you go back to your view, then you can see the end of the lines on plane S1. I would prefer the graphical approach, where you can if what you are doing makes sense. Quote Link to comment Share on other sites More sharing options...
JD Mather Posted March 3, 2009 Share Posted March 3, 2009 Attach your dwg file here. Quote Link to comment Share on other sites More sharing options...
PS_Port Posted March 3, 2009 Share Posted March 3, 2009 try to draw a plane adg, it will be perpendicular to S1 (can you prove?) yes the dot product of the normals of the planes will equal zero... the pure math solution is simple very much, just find the plane equation as well as lines; solve can get the point. go on... You have points 'a' and 'g' so you can get the equation for the line. You have 2 points 'b' and 'c' on plane S1 and the normal 'ba' so you can get the equation for the plane S1 solve the 2 equations simultaneuosly to get the point that satisfies S1 and 'ag'. But this being a CAD site post the drawing so we can have a look at a graphical solution. Quote Link to comment Share on other sites More sharing options...
SEANT Posted March 3, 2009 Share Posted March 3, 2009 If the objective is to find where a linear element would intersect a plane described by a planar element then this AutoCAD assist may help. The first two picks must be on the linear element, and the following three must be on the planar. C^C^_Point;'cal;ilp(nea,nea,nea,nea,nea);pdmode;34; Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 4, 2009 Author Share Posted March 4, 2009 yes the dot product of the normals of the planes will equal zero... You have points 'a' and 'g' so you can get the equation for the line. You have 2 points 'b' and 'c' on plane S1 and the normal 'ba' so you can get the equation for the plane S1 solve the 2 equations simultaneuosly to get the point that satisfies S1 and 'ag'. But this being a CAD site post the drawing so we can have a look at a graphical solution. quit right, you have a good math foundation. Its easy for you to get the graphic solution in theacad. I m sure. Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 4, 2009 Author Share Posted March 4, 2009 If the objective is to find where a linear element would intersect a plane described by a planar element then this AutoCAD assist may help. The first two picks must be on the linear element, and the following three must be on the planar. C^C^_Point;'cal;ilp(nea,nea,nea,nea,nea);pdmode;34; yes if object is to find. all is easy. Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 4, 2009 Author Share Posted March 4, 2009 Have you drawn the figure in 3D? If you do, then you can take a front view, so that the plane S1 is seen as a line, then trim the lines ah and ag where they cross that line. Then when you go back to your view, then you can see the end of the lines on plane S1. I would prefer the graphical approach, where you can if what you are doing makes sense. if you use 3D viewport. its easy to solve. but I try to get it in 2D space. see PS_PORT's answer. not bad. Yesterday, 11:44 am #7 PS_Port vbmenu_register("postmenu_218101", true); Quote Link to comment Share on other sites More sharing options...
cnreader Posted March 4, 2009 Author Share Posted March 4, 2009 thank you all, go on... Quote Link to comment Share on other sites More sharing options...
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