dedreux Posted February 2, 2012 Share Posted February 2, 2012 Sorry for the extremly beginner question. I would like to draw a line that is a specific length, lets say 16, that orginates from a specfic x,y location say 10,23, that terminates at exactly atleast the y cordinate 38. I dont care about the x or the angle. Thank you.. Quote Link to comment Share on other sites More sharing options...
ScribbleJ Posted February 2, 2012 Share Posted February 2, 2012 If you use Polar Tracking (F10) Object Snap Tracking (F11) and have Osnaps turned on (F3) (if your starting at a point on another entity) otherwise you can start a line and then drag your mouse in the direction you want to go then just type in the distance you want to draw to in the command line. Or type in the command line @0,20 where the 0 is the x value and the 20 is the y value. Quote Link to comment Share on other sites More sharing options...
bill_borec Posted February 3, 2012 Share Posted February 3, 2012 I would use geometry to accomplish...if you want this to be EXACT. 1. Draw a line from 0,38 to 50,38 using absolute coordinates. 2. Draw a circle with a center of 12,23 and a radius of 16. 3. Draw a line from the center of the circle to either of the intersections of the circle. (Resulting in a line from 10,23 to either point - there will be two - corresponding with the y-coordinate of 38. Hope this helps. BTW - Welcome to the forum! (Can I do that even though I am not a moderator?) Quote Link to comment Share on other sites More sharing options...
paulmcz Posted February 3, 2012 Share Posted February 3, 2012 There are 2 solutions: Y=38 1) X=10+sqrt(16^2-15^2) 2) X=10-sqrt(16^2-15^2) Quote Link to comment Share on other sites More sharing options...
dedreux Posted February 3, 2012 Author Share Posted February 3, 2012 thank you! thought maybe i could cheat it but the geometry worked out! Quote Link to comment Share on other sites More sharing options...
designerstuart Posted February 3, 2012 Share Posted February 3, 2012 BTW - Welcome to the forum! (Can I do that even though I am not a moderator?) of course you can - this is a community! Quote Link to comment Share on other sites More sharing options...
jgbgod Posted February 3, 2012 Share Posted February 3, 2012 Try this : @ Distance Quote Link to comment Share on other sites More sharing options...
eldon Posted February 4, 2012 Share Posted February 4, 2012 Try this : @ Distance I am not sure how that would work in this case. The distance is certainly known, but not the angle. The difference in the y distance is known, so there are two solutions, which your method does not seem to cover. Quote Link to comment Share on other sites More sharing options...
techberth Posted February 6, 2012 Share Posted February 6, 2012 use "dynamic cordinates" and press tab to enter different values Quote Link to comment Share on other sites More sharing options...
edwinprakoso Posted February 16, 2012 Share Posted February 16, 2012 Just use dynamic input... just like techbert said... Quote Link to comment Share on other sites More sharing options...
nestly Posted February 16, 2012 Share Posted February 16, 2012 That method is fine for length and/or angle but the question is how to draw a line 16 units long beginning at 10,23 and ending with a Y coordinate of 38 Quote Link to comment Share on other sites More sharing options...
Dadgad Posted February 16, 2012 Share Posted February 16, 2012 (edited) Use a circle's point of intersection. Start an orthogonal polyline of length 15 (the difference in Y values) and then create a horizontal xline from the endpoint of that. That line will intersect a circle of radius 16 (centered at 10,23) at two points, both of which will have the Y value of 38. Use your INTERSECTION or your APPARENT INTERSECTION snap to define the end point of your 16 length polyline. Whoops, this is pretty much a graphic rehash of Post # 3, which is spot on the mark, just saw it. Edited February 16, 2012 by Dadgad Quote Link to comment Share on other sites More sharing options...
eldon Posted February 16, 2012 Share Posted February 16, 2012 Just use dynamic input... just like techbert said... Your animation is very fast, and the length input appears to be 500. The burning question is, have you been able to achieve exactly what the OP desires with your method Perhaps you could extend your animation to prove your method, using the OP's figures. Quote Link to comment Share on other sites More sharing options...
JD Mather Posted February 16, 2012 Share Posted February 16, 2012 In 2012 you can parametrically control the solution(s). Quote Link to comment Share on other sites More sharing options...
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