polar/vector input???

[LarryO]

Why do neither of these Poly(); not create an equilateral triangle with 15mm sides?

first point is cartesian, second and third are intended to be polar.

 

d1:=15.000000mm;

ClosePoly;
    Poly(0,0,  d1,#0.000000d,  d1,#150.000000d);

ClosePoly;
    Poly(0,0,  #0.000000d,d1,  #150.000000d,d1);

[Pat Stanford]

You need a Relative; before your code.

ClosePoly; sets the mode to close polygons, it does not close the polygon just drawn. It is a toggle.

Your angels for equilateral should be 120° instead of 150.

 

Try this:

 

Procedure Test;

VAR	d1	:Real;

Begin

Relative;
ClosePoly;
d1:=15.000000mm;
{ClosePoly;}
	Moveto(0,0);
    Poly(d1, #0.000000d,  d1, #120.000000d);
{    Poly(0,0,  #0.000000d, d1,  #150.000000d, d1);
 }   
End;

Run(Test);

 

[LarryO]

omg! Thanks.

As soon as you mentioned Relative; I recalled that I had done this once before a few years ago. I think that I am missing the obvious path to finding out these basics.

I am always expecting the function and procedure reference to give me that insight into those basic aspects of the compiler but never find them. The old help used to have compiler basics in it but I'm not sure where it has migrated to these days.

 

I was fumbling around trying to close the result of RegularPolygon and was getting to the point of wanting a workaround by using polar vectors in the Poly procedure.

In the end I was able to figure out how to close the result of RegularPolygon and eliminate the extra point at the start vertex.