Well, I see there is another message thread which may make this all unneeded.
But here are some comments.
First about the length. The points x,y do depend on length L.
But for plotting purposes everything is then scaled to fit the screen.
After all 16 pixels actual size doesn't look like much.
I have generated these using the 22.5 degree change in the 'central angle' rather than the edge which approximates the spiral. It may not be obvious but this is the same angle by which the edge rotates.
Try my code with the call to DrawSpiral commented out. Study the triangles.
Each succeeding triangle is exactly like the its predecessor, but with all sides scaled by the same amount.
It is also rotated, in this case by 22.5 degrees. That means all three sides are rotated by this same amount, including the white edge.
It starts at x=A, y=0 and is not in the direction you want.
But this easily fixed by subtracting an offset, just as Blitz's Origin command does.
If you have an equation y = f(x) and you want a point (p,q) to be the new origin then the new equation is (y-q) = f(x-p).
The same is true in the polar case. If you need to rotate the entire graph then simply replace theta with (theta-offset).
But here are some comments.
First about the length. The points x,y do depend on length L.
But for plotting purposes everything is then scaled to fit the screen.
After all 16 pixels actual size doesn't look like much.
I have generated these using the 22.5 degree change in the 'central angle' rather than the edge which approximates the spiral. It may not be obvious but this is the same angle by which the edge rotates.
Try my code with the call to DrawSpiral commented out. Study the triangles.
Each succeeding triangle is exactly like the its predecessor, but with all sides scaled by the same amount.
It is also rotated, in this case by 22.5 degrees. That means all three sides are rotated by this same amount, including the white edge.
It starts at x=A, y=0 and is not in the direction you want.
But this easily fixed by subtracting an offset, just as Blitz's Origin command does.
If you have an equation y = f(x) and you want a point (p,q) to be the new origin then the new equation is (y-q) = f(x-p).
The same is true in the polar case. If you need to rotate the entire graph then simply replace theta with (theta-offset).