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#1 2008-11-09 16:38:59
Some trig help.
I need some help coming up with a formula that:
Will set an x postion according to an angle so say
Angle=50,90 and 40. The sprite is at the 50 angle, or at the top. I can get the y postion. Anyway to tell where it needs to be so it always goes to the right x postion?
Sprite 1
:
: :
:50:
: :
Unkown : : Y of (sprite 1) - Y of (Sprite 2)
: :
: :
:40 90:
*****************
X(Also Unknown)
What I basically want is a formula that positions sprite one so it is always at this ratio thingy from sprite one.
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#2 2008-11-09 17:28:11
- Paddle2See
- Scratch Team
- Registered: 2007-10-27
- Posts: 1000+
Re: Some trig help.
The tangent of the 40 degree angle is the ratio of the perpendicular sides of the right triangle so
[(Y of Sprite 1) - (Y of Sprite 2)]
tangent 40 = --------------------------------------------
[(X of Sprite 1) - (X of Sprite 2)]
So, you can then solve for what you want
X of Sprite 1 = (X of Sprite 2) + [(Y of Sprite 1) - (Y of Sprite 2)} / tangent 40
Note: Because Scratch uses an odd definition of zero degrees (straight up) which differs from the usual math book definition (zero degrees is off to the right), I might have screwed this up a little...if it doesn't work, try switching the roles of X and Y. Sorry I can't do any better right now, gotta run!
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#3 2008-11-09 17:41:20
Re: Some trig help.
Thanks Man! That is exactly what I was looking for and it works pretty well. One question though. When did you join the scratch team?
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#4 2008-11-09 20:39:10
- Paddle2See
- Scratch Team
- Registered: 2007-10-27
- Posts: 1000+
Re: Some trig help.
TheSaint wrote:
Thanks Man! That is exactly what I was looking for and it works pretty well. One question though. When did you join the scratch team?
Well, look at that! I wasn't aware that I had joined the team. Thanks for pointing it out. I'm not sure why my title is showing up that way...
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