Jumping Formulas

[Tenry]

If you are developing a platformer game, you typically need some gravity on your character and the start velocity on jumping. I think people often test some values out until they have a jump (duration and height) they like. But I've worked on a more efficient way:
You decide the jump duration and maximum height the character can reach and calculate the needed gravity and initial velocity.
I've worked on some formulas you can easily use. Tell me whether you find it useful or not, or if there are any mistakes or confusion.

And I hope this toptic fits here.


EDIT:
Here are the formulas without the need of any download.

For a jump'n run alias platformer you might want the protagonist jumping a very certain height and time, but you don't know which gravity and jump velocity to set. Here are some useful formulas with that you can calculate the needed parameters for your wished jump.

T₀ is the duration from jump start till maximum height, where v (vertical velocity) is 0 and changes sign. Example: 10
T0 = k / g

hmax is the max height the player jumps. Example: 55
hmax = (k + k * T0) / 2

k is the jump velocity (should be > 0). That is the initial velocity on jump. Example: 10 in the attachment it's -10. that's wrong!
k = (2 * h) / (1 + T0)

g is the gravity (should be > 0). That is the value, that is subtracted from your current velocity. Example: 0.5
g = k / T0

[attachment deleted by admin]

[Nexus]

Most important, you should declare what the variables mean. And if I were you, I'd only provide the linearly independent equalities, i.e. not both T0=k/g and g=k/T0. People should be able to rearrange the terms. Once, you write k = 2h / (1 + T0), the units in the denominator are inconsinstent. Generally, I think your formulas are incorrect...


Let's find the formulas. First, we define some variables:
h(t) = height of the object at time t
T = total jump time
H = maximal height

When we assume h(t) to be a parabola, it has a quadradic equation:
h(t) = a*t² + b*t + c

Some relations at begin, middle and end of the jump are known:
h(0) = 0
h(T/2) = H
h(T) = 0

By inserting them in the equation, we can compute a, b and c:
a = -4H/T²
b = 4H/T
c = 0

Therefore, we have:
h(t) = -4H/T² * t² + 4H/T * t

A constantly accelerated object with acceleration g and initial velocity v₀ travels the following distance during time t:
s(t) = g/2 * t² + v₀ * t

Thus, comparing the coefficients of h(t) and s(t) yields:
g = -8H/T²
v₀ = 4H/T

Note however that I'm talking about the continuous case. In a game, you have to consider a certain time step, to which your gravity and velocity need to be adapted.

[Tenry]

Most important, you should declare what the variables mean. And if I were you, I'd only provide the linearly independent equalities, i.e. not both T0=k/g and g=k/T0. People should be able to rearrange the terms. Once, you write k = 2h / (1 + T0), the units in the denominator are inconsinstent. Generally, I think your formulas are incorrect...


Let's find the formulas. First, we define some variables:
h(t) = height of the object at time t
T = total jump time
H = maximal height

When we assume h(t) to be a parabola, it has a quadradic equation:
h(t) = a*t² + b*t + c

Some relations at begin, middle and end of the jump are known:
h(0) = 0
h(T/2) = H
h(T) = 0

By inserting them in the equation, we can compute a, b and c:
a = -4H/T²
b = 4H/T
c = 0

Therefore, we have:
h(t) = -4H/T² * t² + 4H/T * t

A constantly accelerated object with acceleration g and initial velocity v₀ travels the following distance during time t:
s(t) = g/2 * t² + v₀ * t

Thus, comparing the coefficients of h(t) and s(t) yields:
g = -8H/T²
v₀ = 4H/T

Note however that I'm talking about the continuous case. In a game, you have to consider a certain time step, to which your gravity and velocity need to be adapted.

I've checked my formulas and they fully worked.
But it might depend on how people interpret the variables, about like you said I think.

For me: T₀ (it's about like your T/2) is the time the object is at the highest point.
k (your v₀, yeah it's a better name for it) is always positive and means the velocity to the upwards.
g is always positive as well but means the acceleration downwards.

If I for example want the object reaching a height of 6 pixels (H=6) in 3 frames (T₀=3), I get v₀=3 and g=1, and what's calculated then is:
3 + 2 + 1
Result is 6 (height) and there are 3 summands (T₀=3).


You are taking the total time being in air, which might be easier to use. I think our both formulas are correct, but we are interpreting the variables differently. Is it like that?