I''m trying to make an actor jump, using this formula:
a = a + (Cos(b) * (0.01 * T)) * b
where:
a = the total jump (in height) so far
b = the height the jump has to reach
T = time
now, It''s half working. it seems the player does this in the jump:
/\
not: .-.
if that makes any sense. Any help would be greatly appreciated.
Making an Actor "Jump"
Author
Hi again all. Many thanks to you that helped me on the set distance movement problem i had, now a new one
I''m trying to make an actor jump, using this formula:
a = a + (Cos(b) * (0.01 * T)) * b
where:
a = the total jump (in height) so far
b = the height the jump has to reach
T = time
now, It''s half working. it seems the player does this in the jump:
/\
not: .-.
if that makes any sense. Any help would be greatly appreciated.
I''m trying to make an actor jump, using this formula:
a = a + (Cos(b) * (0.01 * T)) * b
where:
a = the total jump (in height) so far
b = the height the jump has to reach
T = time
now, It''s half working. it seems the player does this in the jump:
/\
not: .-.
if that makes any sense. Any help would be greatly appreciated.
Well he should jump following a parabolic path rather than a cosine...
I''d use the formulae for motion with constant acceleration...
a = a
v = u + at (u = velocity at t=0), integrate above equation to get this result
r = ut + 0.5 * a * t * t
where r is the relative displacement from time t. Again, it''s the integral of the equation above.
Use the third formula, with a constant downwards (negative) acceleration (9.81 m/s or whatever). Set off with a velocity u (upwards) and then apply the 3rd equation until r becomes negative (under the floor). Bingo, and Newton would be happy.
"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley
I''d use the formulae for motion with constant acceleration...
a = a
v = u + at (u = velocity at t=0), integrate above equation to get this result
r = ut + 0.5 * a * t * t
where r is the relative displacement from time t. Again, it''s the integral of the equation above.
Use the third formula, with a constant downwards (negative) acceleration (9.81 m/s or whatever). Set off with a velocity u (upwards) and then apply the 3rd equation until r becomes negative (under the floor). Bingo, and Newton would be happy.
"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley
Author
Thanks shifter, I''m a little when you say stuff like this though:
"v = u + at (u = velocity at t=0)"
when you say "at", what exactly do you mean?
"v = u + at (u = velocity at t=0)"
when you say "at", what exactly do you mean?
No, when he says "at" he really means a*t, where ''t'' is the time increment from the moment that the object had velocity ''u'' and ''a'' is the acceleration (assumed constant).
Cheers,
Timkin
Cheers,
Timkin
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