CVector dir = dest - pos;
dir.z = 0.0f; // ignore height difference for now
float d = dir.length();
dir = dir.normalize();
float s = 0.0f; // height difference between the two points
float v0 = 35.0f; // v0 = velocity of projectile
float g = 9.0f; // gravity
float t = d / v0; // time to hit target
float y = (-s - 0.5f * g * t * t) / t;
float x2 = (v0 * v0) - (y * y); // x ^ 2
if( x2 < 0.0f )
{
// out of range!
}
else
{
float x = sqrtf( x2 );
traj = CVector( dir.x * x, dir.y * x, y );
}
Initial trajectory for a mortar-like weapon
Author
I''m currently writing a 3D RTS game, and I have a mortar unit, but I''m having a few problems with the maths involved in calculating the initial trajectory for the weapon, so that it will land where I am aiming it. Say the unit is at position pos, and I want the mortar to hit position dest. My current code is:
Author
No one has any ideas?
One other thing of note is that if there''s another way to describe the trajectory than with my current method of just adding a gravity vector to the velocity each frame (for example, I was thinking of maybe making it follow part of a circle or ellipse) then that''s cool. I just need to get it to look at least somewhat parabolic - it doesn''t need to be perfectly realistic.
One other thing of note is that if there''s another way to describe the trajectory than with my current method of just adding a gravity vector to the velocity each frame (for example, I was thinking of maybe making it follow part of a circle or ellipse) then that''s cool. I just need to get it to look at least somewhat parabolic - it doesn''t need to be perfectly realistic.
Gravity (on earth) is 9.8 meters/(sec^2). Here''s an equation to find change in distance over time:
Distance = Initial_Velocity*Time + 0.5*(Acceleration*Time^2)
Time is in seconds as long as your acceleration (gravity in this case) is in units of seconds (it is). You''d probably use -9.8 to make the object slow down and fall instead of flying away. For the horizontal direction, it isn''t affected over time unless there''s a wind or something.
Remember that this is how you get the velocity for the vertical direction:
http://druidgames.cjb.net/
Distance = Initial_Velocity*Time + 0.5*(Acceleration*Time^2)
Time is in seconds as long as your acceleration (gravity in this case) is in units of seconds (it is). You''d probably use -9.8 to make the object slow down and fall instead of flying away. For the horizontal direction, it isn''t affected over time unless there''s a wind or something.
Remember that this is how you get the velocity for the vertical direction:
. |\ |_\ | \ | \ | \ <-- Velocity of launch sin(AoS) | \ * | \ VoL | \ | \ | \ |_ \ |_|_______/_\ <-- Angle of shot http://druidgames.cjb.net/
Author
Thanks, Null and Void, it actually turned out that I was just being silly. I did get my maths right (I had fiddled with my code trying to get the best result, so it''s not the same maths as I came up with originally).
My problem was that I had three variables, vx, vy and t, but I could only think of two equations, namely vx = d / t and vy = (1/2 * g * t * t - s) / t (which is the one you gave, re-arranged to make vy the subject). But then I remembered that v^2 = vx^2 + vy^2, and since both vx^2 and vy^2 have a t in them, I can solve that for t. So, my final code (if anyone is interested) is:
where d, g, and dir are defined as in the original post.
My problem was that I had three variables, vx, vy and t, but I could only think of two equations, namely vx = d / t and vy = (1/2 * g * t * t - s) / t (which is the one you gave, re-arranged to make vy the subject). But then I remembered that v^2 = vx^2 + vy^2, and since both vx^2 and vy^2 have a t in them, I can solve that for t. So, my final code (if anyone is interested) is:
float t = sqrtf( 2 * d / g ); float y = 0.5f * g * t; // still ignoring s, the height difference for now float x = d / t; CVector traj = CVector( dir.x * x, dir.y * x, y ); where d, g, and dir are defined as in the original post.
The important thing to remember for projectiles is that the hrizontal and vertical components are independent, so you specify equations of motion for each.
In this case, you are specifying a final position, and an initial velocity in the horizontal plane, v0. From this you correctly calculate the time to reach the destination
t = v0/d, where d is the horizontal distance.
You can then use
s = v_h0*t + 0.5*g*t*t
to calculate your initial horizontal velocity, v_h0. Since s is zero in the horizontal plane, you have
v_h0 = -0.5*g*t
(be careful about the sign of gravity, I''m assuming it''s negative).
Mike.
In this case, you are specifying a final position, and an initial velocity in the horizontal plane, v0. From this you correctly calculate the time to reach the destination
t = v0/d, where d is the horizontal distance.
You can then use
s = v_h0*t + 0.5*g*t*t
to calculate your initial horizontal velocity, v_h0. Since s is zero in the horizontal plane, you have
v_h0 = -0.5*g*t
(be careful about the sign of gravity, I''m assuming it''s negative).
Mike.
The important thing to remember for projectiles is that the hrizontal and vertical components are independent, so you specify equations of motion for each.
In this case, you are specifying a final position, and an initial velocity in the horizontal plane, v0. From this you correctly calculate the time to reach the destination
t = v0/d, where d is the horizontal distance.
You can then use
s = v_h0*t + 0.5*g*t*t
to calculate your initial horizontal velocity, v_h0. Since s is zero in the horizontal plane, you have
v_h0 = -0.5*g*t
(be careful about the sign of gravity, I''m assuming it''s negative).
Mike.
In this case, you are specifying a final position, and an initial velocity in the horizontal plane, v0. From this you correctly calculate the time to reach the destination
t = v0/d, where d is the horizontal distance.
You can then use
s = v_h0*t + 0.5*g*t*t
to calculate your initial horizontal velocity, v_h0. Since s is zero in the horizontal plane, you have
v_h0 = -0.5*g*t
(be careful about the sign of gravity, I''m assuming it''s negative).
Mike.
This topic is closed to new replies.
Advertisement
Popular Topics
Advertisement
