Chaps, I have a 2D object(Over the top view), I wish to move it by the angle its facing, for instance 40 degrees. How can i calculate its new coordinate on my x,y grid by its angle and its X and Y velocity? I am so rusty on all the this trigonometry maths! Thanks for any help or guidence. pete. [edited by - BlackEye on August 8, 2003 5:26:40 AM]

Thanks i think!
So a function would just look like this i guess?

CPitchObj::CalculatePosition()
{
m_X += m_Velocity * cos(angle)
m_Y += m_Velocity * sin(angle)
}


Do i need to work about different velocitys for my X and Y ?
I mean should i be side steping the object and / XOR moving forward? If so what are the changes to the above with 2 different Velocitys? Man i wish i got this math bit. Where is that GCSE Maths book!

cool ill have to try this

quote:Original post by BlackEye
CPitchObj::CalculatePosition()
{
m_X += m_Velocity * cos(angle)
m_Y += m_Velocity * sin(angle)
}

Just one suggestion. If you want frame rate independence then pass in the time since the last frame and like this

CPitchObj::CalculatePosition( float deltaTime )
{
const float velThisFrame = deltaTime * mVelocity;
mY += velThisFrame * cos(angle);
mX += velThisFrame * sin(angle);
}

( Aside: Underscores are evil! )

If you want to move an object based on a certain angle in 2D you must find out how far to move the object on each axis, ie x and y.

"O" = Object to be moved.
"@" = Angle the object will be moving
"H" = Hypotenuse of a right triangle (or ship velocity)
"vy" = y component of velocity
"vx" = x component of velocity
| .
| /^
| H / |
| / | vy
| /@ |
| O---->
| vx


"EDITsorry about the picture I can't get it to look right)"

We know the velocity we want the ship fly at (or the hypotenuse of the triangle above) and we know the angle at which it will fly(@). We now need to find the vx and vy components so we can add them to the ship coordinates thereby moving the ship in the desired path.

From the diagram we end up will a right triangle. From the properties of a right triangle we know that the "vx side" is equal to cos(@) * H and the "vy side" is equal to sin(@) * H.
Once we have the "vx" and "vy" we simply add our position to the velocities (vx and vy) and our ship will move at the desired speed along the desired path.

vx = cos(@) * H;
vy = sin(@) * H;

x += vx;
y += vy;

I believe I got the math right there. If I have made mistakes I would appreciate the input.

[edited by - OneBitWonder on August 14, 2003 12:35:26 PM]

[edited by - OneBitWonder on August 14, 2003 12:37:19 PM]

ARGGH! I can't get the picture to look right!

[edited by - OneBitWonder on August 14, 2003 12:40:17 PM]

Thanks reana1. That is the exact picture I was thinking of.