Rotating a polygon on graph paper

A

Aardvqrk

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Mar 2, 2024
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How do I change the ”correctional” rotation so that the angle of each line in the polygon stays relative.
got pretty far, but I couldn’t figure out how to compensate for changing the triangle,

In the attached link, if you change the triangle on the right, the angle of the sides of the third triangle don’t compensate. You can manually compensate by changing the values of f and q, but I want it to automatically compensate, can anyone help?

Link
 
Am I the only one not having a clue what this problem is about? Is "change correctional rotation" the same as rotation of the polygon? What does it mean to "stay relative" ?
 
blamocur said:
Am I the only one not having a clue what this problem is about? Is "change correctional rotation" the same as rotation of the polygon? What does it mean to "stay relative" ?
Click to expand...

I didn’t explain it very well, did you try opening the graph? it might be a little more obvious that way.

by “correctional rotation“ I am talking about the value which I add to the cos and sin functions which keeps the rotation of both line GI and line HI the same as the amount which the initial triangle DEF was rotated, since without this value, the lines will both head towards 0
 
Aardvqrk said:
I didn’t explain it very well, did you try opening the graph? it might be a little more obvious that way.

by “correctional rotation“ I am talking about the value which I add to the cos and sin functions which keeps the rotation of both line GI and line HI the same as the amount which the initial triangle DEF was rotated, since without this value, the lines will both head towards 0
Click to expand...
I think you need to tell us what you are actually trying to do, not just expect us to guess by looking at an erroneous implementation.

You've reflected ABC to make DEF; what is the intention for GHI? And what is each input (q, f, w) intended to do? And can you explain the formulas?
 
My intention for GHI is to be able to make a rotated version of DEF, q and f are kind of where I’m stuck with manually changing if I want to change the triangle I am rotating, and w is the rotation (in degrees).

hope that helps
 
Aardvqrk said:
My intention for GHI is to be able to make a rotated version of DEF, q and f are kind of where I’m stuck with manually changing if I want to change the triangle I am rotating, and w is the rotation (in degrees).

hope that helps
Click to expand...
So, you want to rotate DEF about I (that is, F) by angle w, and q and f are just there because it doesn't work? You can't mean they are used to change DEF, the triangle you are rotating, since they only affect GHI, the result of the rotation. What are they specifically meant to do?

How much have you learned about vectors, and specifically rotation of vectors? Why did you write this particular formula:

[math]G=\left(\operatorname{distance}\left(E,F\right)\cdot\cos\left(w\cdot\frac{\pi}{180}+\frac{q\pi}{180}\right)-h,\operatorname{distance}\left(E,F\right)\cdot\sin\left(w\cdot\frac{\pi}{180}+\frac{q\pi}{180}\right)+j\right)[/math]
I'd expect you to be using the angle of vector FD rather than a separately set angle q. Are you asking, ultimately, how to find that angle?
 
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