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ending location from angle and speed - Printable Version

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ending location from angle and speed - Kazooless - Mar 18, 2009 03:56 AM

I'm trying to calculate the ending x,y of my object. I have the starting x,y and the angle in radians and the speed

this is what I'm trying:

Code:
deltaX = cos( rotation ) * speed;
deltaX = startingX * deltaX;
deltaY = sin( rotation ) * speed;
deltaX = startingY * deltaY;

the result is almost as if everything is wildly exagerated.

does anyone know how to calculate this correctly?


ending location from angle and speed - ThemsAllTook - Mar 18, 2009 06:07 AM

Seems like you'd want to add deltaX/Y to startingX/Y instead of multiplying them together.


ending location from angle and speed - Ingemar - Mar 26, 2009 02:18 AM

Also, why multiply the *speed* by cos and sin of the angle? What are you trying to do there? Should the translation be affected by rotation?

I usually express this kind of things by matrix multiplication. It scales better to more complex situations.


ending location from angle and speed - kendric - Mar 30, 2009 10:52 AM

I have a facing vector instead of a facing angle. The plus side is you don't need to cos and sin constantly, only when they turn. Then when you want to move somebody you just do
movement=[facing scale:speed];
[position add:movement];

I just made up arbitrary function names but you can probably get the point.
When you want to turn all you need to do is rotate the facing vector by X radians negative for a left turn, positive for a right turn. You can rotate a vector using CGAffineTransform functions.


ending location from angle and speed - masdest - Mar 30, 2009 07:27 PM

Kazooless Wrote:I'm trying to calculate the ending x,y of my object. I have the starting x,y and the angle in radians and the speed

this is what I'm trying:

Code:
deltaX = cos( rotation ) * speed;
deltaX = startingX * deltaX;
deltaY = sin( rotation ) * speed;
deltaX = startingY * deltaY;

the result is almost as if everything is wildly exagerated.

does anyone know how to calculate this correctly?

I assume this is so that you can move an object on an angle? if so you can use the following code
Code:
deltaY = sin (rotation) * speed; //this finds the y travel distance
finalY = startingY + deltaY; // this is the location of object on the y plane
deltaX = cos (rotation)  * speed; //this finds the x travel distance
finalX = startingY + deltaY;

It's basic trigonometry where the speed is the hypoteneuse and the y is "Opposite" and x is "Adjacent".
looking at a clock 3 o'clock would be 0 (or 360) degree, 12 - 90 (or -270) degrees, 9 - 180 (or -180) degrees, 6 - 270 (or -90) degrees.
speed * cos (rotation)


ending location from angle and speed - Gillissie - Apr 3, 2009 02:40 PM

masdest Wrote:I assume this is so that you can move an object on an angle? if so you can use the following code
Code:
deltaY = sin (rotation) * speed; //this finds the y travel distance
finalY = startingY + deltaY; // this is the location of object on the y plane
deltaX = cos (rotation)  * speed; //this finds the x travel distance
finalX = startingY + deltaY;

It's basic trigonometry where the speed is the hypoteneuse and the y is "Opposite" and x is "Adjacent".
looking at a clock 3 o'clock would be 0 (or 360) degree, 12 - 90 (or -270) degrees, 9 - 180 (or -180) degrees, 6 - 270 (or -90) degrees.
speed * cos (rotation)

That last line should read:
Code:
finalX = startingX + deltaX;