The running multiplier not being factored into Running-jump distance has been bugging me, so I've decided to apply some Physics to get a better idea of how far these jumps would be. Here's what I've got:
Vo = Initial Velocity (in m/s)
The axis velocity components:
Vx = Vo*cos(a)
Vy = Vo*sin(a)
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In the x direction the acceleration is zero so we can find how long it takes to go a distance d:
t = d/Vx = d/(Vo*cos(a))
In the y direction, the projectile undergoes an acceleration -g and does a displacement of zero because it lands at the same height it was launched from. In the third equation below, substitute the value for t from above and then do some pretty serious trig to solve for d and simplify.
d = Vo * t + (1/2) * a * t^2
0 = (Vo * sin(a)) * t + (1/2) * (-g) * t^2
0 = (Vo * sin(a) / (d / Vo * cos(a))) - (1/2) * g * (d / Vo * cos(a))
0 = d * tan(a) - ((d^2) * g) / (2 * Vo^2 * cos^2(a))
((d^2) * g) / (2 * Vo^2 * cos^2(a)) = d * tan(a)
((d * g) / (2 * Vo^2 * cos^2(a)) = tan(a)
d = (2 * Vo^2 * cos^2(a) * tan(a)) / g
d = (2 * Vo^2 * cos^2(a) / (sin(a) / cos(a)) / g
d = (2 * Vo^2 * cos(a) * sin(a)) / g
Using Trig Idendity that: sin(a) * cos(a) = (1/2) * sin(2*a)
ANSWER: Distance, d = (Vo ^2 * sin(2*a)) / g
The time it takes for the projectile to get to its maximum height is half the time it takes to get to d because of the symmetry of projectile motion. Take the time we got at the top and replace with d and then do some more simplification:
t = (1/2) * d / (Vo * cos(a))
t = (Vo^2 * sin (2*a)) / 2 * g * (Vo^2 * cos(a))
t = (Vo/2 * g) * (sin(2*a) / cos(a))
Using Trig Idendity that: sin(2*a) / cos(a) = 2 * sin (a)
t = (Vo / 2 * g) * 2 * sin(a)
ANSWER: Time, t = (Vo * sin(a)) / g
Now we plug in this value for t in the equation of motion for the y direction:
d = Vo * t + (1/2) * a * t^2
h = t * (Vo * sin(a)) + (1/2) * (-g) * t^2
h = (Vo * sin(a) * Vo * sin(a)) / g - (1/2) * g * (Vo * sin(a) / g) ^ 2
h = (Vo ^2 * sin^2(a)) / g - (Vo^2 * sin^2(a)) / 2 * g
ANSWER: Height, h = (Vo^2 * sin^2(a)) / 2*g
Max_Distance, d = (Vo ^2 * sin(2*a)) / gTravel_Time, t = (Vo * sin(a)) / gMax_Height, h = (Vo^2 * sin^2(a)) / 2*gA typical running-jump: Q=6, Vo=(6*3/3) = 6m/s, a=45.
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d = (Vo ^2 * sin(2*a)) / g
d = (6^2) * sin(2*45) / 9.8
d = (36 * 1) / 9.8
d = 3.67m running-jump distance.
h = (Vo^2 * sin^2(a)) / 2*g
h = ((6 ^ 2) * sin^2(45)) / 2 * 9.8
h = (36 * .5) / 19.6
h = .918m max height of jump
t = (Vo * sin(a)) / g
t = (6 * sin(45)) / 9.8
t = (6 * .707) / 9.8
t = .433 seconds from jump start to finish
d = 3.67m running-jump distance.h = .918m max height of jumpt = .433 seconds from jump start to finishSame running-jump as a Dwarf: Q=6, Vo=(6*2/3) = 4m/s, a=45.
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d = (Vo ^2 * sin(2*a)) / g
d = (4^2) * sin(2*45) / 9.8
d = (16 * 1) / 9.8
d = 1.63m running-jump distance.
h = (Vo^2 * sin^2(a)) / 2*g
h = ((4 ^ 2) * sin^2(45)) / 2 * 9.8
h = (16 * .5) / 19.6
h = .408m max height of jump
t = (Vo * sin(a)) / g
t = (4 * sin(45)) / 9.8
t = (4 * .707) / 9.8
t = .289 seconds from jump start to finish
d = 1.63m running-jump distance.h = .408m max height of jumpt = .289 seconds from jump start to finishCyberSkates in action: Q=6(10), Vo=(Q*6)/3= 20m/s, a=30.
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d = (Vo ^2 * sin(2*a)) / g
d = (20^2) * sin(2*30) / 9.8
d = (400 * .866) / 9.8
d = 35.3m running-jump distance.
h = (Vo^2 * sin^2(a)) / 2*g
h = ((20 ^ 2) * sin^2(30)) / 2 * 9.8
h = (400 * .25) / 19.6
h = 5.1m max height of jump
t = (Vo * sin(a)) / g
t = (20 * sin(30)) / 9.8
t = (20 * .5) / 9.8
t = 1.02 seconds from jump start to finish
d = 35.3m running-jump distance.h = 5.1m max height of jumpt = 1.02 seconds from jump start to finishIt's harder to represent a standing jump using Quickness, I'd be more inclinded to use Strength, but either way: S=6, Vo=(S/3)=2m/s, a=90.
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d = (Vo ^2 * sin(2*a)) / g
d = (2^2) * sin(2*90) / 9.8
d = (4 * 0) / 9.8
d = 0m running-jump distance.
h = (Vo^2 * sin^2(a)) / 2*g
h = ((2 ^ 2) * sin^2(90)) / 2 * 9.8
h = (4 * 1) / 19.6
h = .204m max height of jump
t = (Vo * sin(a)) / g
t = (2 * sin(90)) / 9.8
t = (2 * 1) / 9.8
t = .204 seconds from jump start to finish
d = 0m running-jump distance.h = .204m max height of jumpt = .204 seconds from jump start to finishProblem with using a real-life approach would be making the jump wouldn't really be a test of skill as much as landing the jump. Since Vo for game purposes will equal Vt when landing I would think the TN should be based on the landing speed and conditions of the ground more that how high you jump.
Just some thoughts on a boring day-before-the-holiday at work.