Jump trajectory and distance physics formula

PB Forum :: Trail Building
Jump trajectory and distance physics formula
Author Message
flag proamrider
O+
Posted: May 29, 2009 at 11:46 Quote
driftrc wrote:
rrm1266 wrote:
There's no formula that's gonna help you out with this one. There are an incredible amount of variables hitting a jump on your bike. The shirt you're wearing when you jump is gonna affect how far you go (admittedly not by much, but still). The calculations that are getting thrown around here are for simple projectile motion, in perfect conditions, assuming no acceleration or rotation. Unless you build cheese wedge jumps in a vaccuum and plan on landing at 67.5 degress on your rear wheel, you're out of luck in the formula department.

Just plan on hitting it with a little extra speed the first time and don't try to nose in. It might rattle you a bit, but it'll save you some time in the wind tunnel and avoid a painful nose case.

So, admittedly pinpointing the landing will be tough, but you can get in the basic ball park.

The formula allows for any amount of rotation and an assumed acceleration due to gravity of 9.81 m/s2
The formula does not, however, provide for something that is not a point mass
(Meaning that the formula doesn't account for something with a significant radius)
To add this in, a bike has a diameter of roughly a meter? So overall there will be about 1/2 m * sin(67.5) = .46 m for the takeoff and .46 m for the landing that isn't accounted for, so lip to lip should be a little shorter than 5 meters.
(Even if you landed on your rear wheel, the jump to landing distance should be close to 26" shorter than previously stated.)

If you use and object that doesn't compress, then the shape of the jump does not matter, only the angle of the direction that the object launches in. With a bike and a human body that compresses, it is hard to determine the angle of launch.

And wind resistance is not very significant, The uncertainty in speed (No one can say I am going 30.0 km/h) provides a greater problem than the wind.

So, you don't need cheese wedge jumps, rotation isn't a big deal, acceleration is part of the formula, and your only valid point was the wind resistance, which isn't that big of a deal anyway. You're still in the simple physics mode.
The problems are Compression upon takeoff (Will change the effective angle of the jump) and an imprecise speed (faster = farther), although if there is a decent wind, it will blow you around.

i dont know if anyone agrees but why you need a formula to hit a girt jump. seriously.... if you really need one you might as well not bike because using math/physics in biking is pointless. you learn from mistakes.... learn like every pro in this sport has.... by making mistakes and just keep going never stopping... again like i said before dont be a pussy and hit your DJ
flag westislandbiker
Posted: May 29, 2009 at 11:48 Quote
lots of stuff to think about. id say to much. just go for it. otherwise you will need to calculate trajectory from the lip, your exact speed, the wind, loss of energy due to heat and sound which will vary between tires. its just too much.
flag rrm1266
Posted: May 29, 2009 at 12:04 Quote
I was referring to acceleration of the rider off the lip, not due to gravity. How hard you pop off the lip will have a huge affect on your velocity in the air both horizontally and vertically. You could hit the jump at 40, go 2 feet high and make the landing, while someone else might go 30, boost out 6 feet and land in the same place. As for the cheese wedge comment, most jumps are not flat ramps, they curve up to the top of the lip. As you and your bike ride up the lip, you begin to rotate back as the angle of the jump increases. When you pop of a jump you push forward (whether or not you notice you're doing it) to counter this rotation. That's gonna be another factor in the equation, turning rotational energy into forward velocity.

You could probably get in the ball park with these formulas, but when you do you'll realize you've spent too much time doing math and not enough time riding bikes.
flag Danny-P
Posted: May 29, 2009 at 12:20 Quote
I wanna see pics!! and then a video of either success or pain Big Grin Whip
flag MountainFreedom
Posted: May 29, 2009 at 23:24 Quote
this has gone on long enough just go for it....put hockey pads and foam on you if you must!
flag hambone2
Posted: May 31, 2009 at 18:35 Quote
WhatAreYouNewOrSumpin wrote:
It's a f@ckin triangle kidzz!

oh... i get it.... its A2+B2=C2


hahah Sure.... i think the question is how much speed to you need to fly over C2 with out dieing
flag lew-77
Posted: Jun 1, 2009 at 11:05 Quote
well if A=450 and C=654 then at the rotational angle of 45 degrees you should be able to hit it at 45 M/s if youre lucky
flag fooox
Posted: Jun 1, 2009 at 11:12 Quote
Assuming air resistance is negligible

v=sqrt((-4.9*L^2)/((y2-y1)*(cos(theta))^2-cos(theta)*sin(theta)*L))

v = velocity (m/s) at the lip
l = length of gap (m)
theta = angle at lip (degrees)
y1 = height of kicker (m)
y2 = height of landing (m)
flag christer
Posted: Jun 1, 2009 at 11:25 Quote
this has probably already been put up but nm

your max height will equal V²=u²+2as
V= 0 as at the top of the arc you will be traveling at 0m/s upwards
u= your vertical take off speed off the lip [(speed lip is hit)x sin (angle of takeoff)]
a= -9.81
s= height (displacement)

horizontal distance= s=vt
v= horizontal speed= [(speed lip is hit)x cos (angle of takeoff)]
t= ['u' from top equation] divided by 9.8

thats as far as i could take it physics wise im afraid

edit* this will only show you how far away you will land horizontally from the lip and obviously -9.8 is acceleration due to gravity
double edit* don't quote me on any of this
flag crankenstein
Posted: Jun 1, 2009 at 11:43 Quote
the "Paralysis of Analysis" will get you nothin but frustrated. Just hit it and you'll probably learn the answers to your questions the first time...if you survive.
flag konakoolkid
Posted: Aug 3, 2009 at 12:28 Quote
MnM1618 wrote:
bankturn wrote:
I don't necessarily understand the application because how are you going to be mathematically exact about the speed your going into it? Maybe a radar gun I suppose... BUT, cool thread. It's so hard to judge speed jumping, as I've begun to jump I've found that to be the hardest part of the equation. You get the right speed and its easy cheesy fo sheezy.

lol well i have a spedo on m bike and when i place the jump on the traili can minupulate the run in to give me thr right amount of speed ... iff you come in to fast put a small rock garden in befor ppl automaticle slow down a little ... iff you comin in to slow you build a little "a" fram ladder ... then you can gadge the speed therefor makeing the corect jumping distance and of cors you allways by rule of thub make the landing bigger for ppl that ride well so they dont land flat and for not so good riders right

ur friggen smart dude i didnt understand but wowsszers smarty pants over here can jump wid maths its true n official u need maths to get jumps rite n landings i use to guess were to put the landing
flag twingate
Posted: Aug 3, 2009 at 17:56 Quote
proamrider wrote:
driftrc wrote:
rrm1266 wrote:
There's no formula that's gonna help you out with this one. There are an incredible amount of variables hitting a jump on your bike. The shirt you're wearing when you jump is gonna affect how far you go (admittedly not by much, but still). The calculations that are getting thrown around here are for simple projectile motion, in perfect conditions, assuming no acceleration or rotation. Unless you build cheese wedge jumps in a vaccuum and plan on landing at 67.5 degress on your rear wheel, you're out of luck in the formula department.

Just plan on hitting it with a little extra speed the first time and don't try to nose in. It might rattle you a bit, but it'll save you some time in the wind tunnel and avoid a painful nose case.

So, admittedly pinpointing the landing will be tough, but you can get in the basic ball park.

The formula allows for any amount of rotation and an assumed acceleration due to gravity of 9.81 m/s2
The formula does not, however, provide for something that is not a point mass
(Meaning that the formula doesn't account for something with a significant radius)
To add this in, a bike has a diameter of roughly a meter? So overall there will be about 1/2 m * sin(67.5) = .46 m for the takeoff and .46 m for the landing that isn't accounted for, so lip to lip should be a little shorter than 5 meters.
(Even if you landed on your rear wheel, the jump to landing distance should be close to 26" shorter than previously stated.)

If you use and object that doesn't compress, then the shape of the jump does not matter, only the angle of the direction that the object launches in. With a bike and a human body that compresses, it is hard to determine the angle of launch.

And wind resistance is not very significant, The uncertainty in speed (No one can say I am going 30.0 km/h) provides a greater problem than the wind.

So, you don't need cheese wedge jumps, rotation isn't a big deal, acceleration is part of the formula, and your only valid point was the wind resistance, which isn't that big of a deal anyway. You're still in the simple physics mode.
The problems are Compression upon takeoff (Will change the effective angle of the jump) and an imprecise speed (faster = farther), although if there is a decent wind, it will blow you around.

i dont know if anyone agrees but why you need a formula to hit a girt jump. seriously.... if you really need one you might as well not bike because using math/physics in biking is pointless. you learn from mistakes.... learn like every pro in this sport has.... by making mistakes and just keep going never stopping... again like i said before dont be a pussy and hit your DJ

It would be interesting to spend some time on sketching out the math of the algebra involved. To the best of my knowledge no-one has put together a (relatively) complete mathematical model of a bike going over a jump, let alone involved in the "flow" of a complete trail. All the things going on with bike, rider, context and jump is a complex equation.

The math would be useful for a couple of things:
1. Improve safety on standard models for technical trail features. It would make building jumps that are well calibrated to a given trail and its conditions easier.
2. Insurance statistics is based on analytical models. Better understanding of jumps would probably lower insurance premiums for bike facilities as the insurance company would be able to calculate risk better.
flag cougar797
Posted: Sep 17, 2010 at 21:23 Quote
wonder how this ended

 


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