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 Solving Simple tank problem with matrices  (Read 1555 times) 0 Members and 1 Guest are viewing this topic.
dranonymous

Junior Devvie

Hoping to become a Java Titan someday!

 « Posted 2006-10-20 00:12:03 »

Ok, so I've tried a few times to figure this out and keep getting the answer wrong.  I've attached a picture to make it more clear.

You have a target point, which is relative to a missile.  I want to find the target location relative to the tank turret's coordinate system.

target location relative to missile coordinate system = (8, -4)
tank location relative to missile coordinate system = (4, 0)
turret is rotated 45 degrees from the front of the tank
The coordinate system is X out the front of a vehicle and Y to the left

Using basic algebra, you can find the answer almost trivially.  I want to understand the process when using rotation matrices.

I thought the answer should be -

R(-45)R(90)(target pos - tank pos)

Where R(x) represents the rotation matrix of x degrees.

I seem to end up getting the correct number (4* sqrt(2), 0), but I end up with it being negative, which isn't correct.

So what is the correct way of doing this?

Cheers,
Dr. A>

PS - I was using column vectors, so my rotation matix looked like:

 1  2  3  4  5  6  7 `cos x  -sin xsin x  cos xso P2 = RP1| x2 |  =  | cos x  -sin x | | x1 || y2 |     | sin x   cos x | | y1 |`

noblemaster

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 « Reply #1 - Posted 2006-10-20 03:22:45 »

your Y goes up for the screenshot - but the actual physical screen coordinates point down. could this be the problem?

dranonymous

Junior Devvie

Hoping to become a Java Titan someday!

 « Reply #2 - Posted 2006-10-20 05:41:16 »

It shouldn't be the problem.  The problem is wholly independent of 2d drawing in java.  So, my Y going left and X going out the front of a vehicle should be acceptable.  It might be a tad confusing, but I believe its fine.

I tried to make this problem independent of right and left hand coordinate systems, by reducing it to a 2d one.

Other thoughts?

Dr. A>
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