The angle of a parallelogram

DFLY

I'm trying to make a parametric drawing where struts find their correct angle. Given the attached diagram. Does anyone know how to mathematically find the angle from x,y and w ?

Anthony Apostolaros

That's a tough one. I couldn't think of any way to draw that geometry given only lengths x, y, and w -- except by drawing a circle and then a line tangent to it. But I don't know how to write a mathematical expression for what the Tangent osnap does.

I do, however, think it would be pretty easy to use Lisp to draw the geometry and measure the angle Omega, if that helps. I don't know of any built-in way to measure an angle except DimAngular, but Stig Madsen's AngleM function does that.
(setq pt6 (inters pt2 pt3 pt1 pt5))
(command "AngleM" pt6 pt2 pt1)

ScottS

You can use parallel and coincident 2D geometric constraints plus 2D dimensional constraints with parameters w,x, and y to fully define what you've drawn in the diagram. To solve for the angle Omega, the trigonometric identity sin(a-b)=sin(a)cos(b)-cos(a)sin(b) helps, or break it down as the sum of two angles.
Omega=arctan(y/x)+arcsin(w/d)
where d^2=x^2+y^2.

Anthony Apostolaros

ScottS said:

.... Omega=arctan(y/x)+arcsin(w/d)
where d^2=x^2+y^2.

Shouldn't that be w/e?

lgxavier

Please see attached file.

Anthony Apostolaros

lgxavier said:

Good work, Igxavier. I could only check it as far as x sin Ω - y cos Ω = w. I didn't know how to solve for Ω, and couldn't even figure out how to use WolframAlpha to do it for me. But up to that point it looks good, and I assume the solution you got from them is correct.

lgxavier

Hi Anthony, I did some tests with numerical values and it seems OK.
In WolframAlpha, normally type the expression and indicate the variable to be evaluated. For example:
xsinø-ycosø=w, solve ø (and press

lgxavier

Hi Anthony, I did some tests with numerical values and it seems OK.
In WolframAlpha, normally type the expression and indicate the variable to be evaluated. For example:
xsinø-ycosø=w, solve ø ( and press = ).

Anthony Apostolaros

Yes, I see. Thank you. That's very useful. Though when I paste it in that way (xsinø-ycosø=w, solve ø), without the spaces, I get a very different answer.

lgxavier

Well, I tested it now with and without spaces and, indifferent, same results for me. But the important thing is that it worked!

DFLY

Thanks Igxavier.
That is invaluable.
But now I'm going to be spending (and already have spent quite a few) hours trying to figure out why it is so.

ScottS

Anthony Apostolaros said:

show previous quotes

ScottS said:

.... Omega=arctan(y/x)+arcsin(w/d)
where d^2=x^2+y^2.

Shouldn't that be w/e?

No. In the dashed right triangle with hypotenuse e in your diagram, the length opposite angle phi is close to (for small phi) but not quite w.
phi = arcsin(w/d)

lgxavier

ScottS, your result for omega is correct and is much simpler. Congratulations!

Anthony Apostolaros

Yes, I see it now -- w is perpendicular to the blue lines, not to d. So d is the hypotenuse for angle phi.