Got bored doing physics study so I decided to spice things up a bit... hope it is useful for everyone!
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- Dirt Jumping
- Date Posted
- Mar 14, 2009 at 4:33Mar 14, 2009
- Location
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Sydney, Australia
- Trail
- arbitrary
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- imaginary
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179 Comments
- 212 6
lodonnell243
(Mar 14, 2009 at 14:29)
A real physicist would take g to be 9.81, therefore your theory is incorrect. Think of the people who have read this and are now going to get hurt... Tut tut 
- 23 3
my god i thort i'd escaped this once i got home. altho wile i'm here, you should include an air friction coefficient in ur answer and the friction coefficient created by the tyres on the dirt. and 9.8 is a rookie mistake and one more thing just to be annoying. always specify units of angles (degrees or radians) altho i realise it has to be degrees in this case, so i'l let u off 
lecture over hav a nice day kids!
and one more thing just to be annoying. always specify units of angles (degrees or radians) altho i realise it has to be degrees in this case, so i'l let u off lecture over hav a nice day kids!- 12 3
if you was going to do that you would have to include the energy lost with suspention forks ? would you have to use inertia
- 9 2
this is very very very picky
but you'd could take direction into:
if an object is fired at x ms ^(-1)
and fired north, east south and west
it ill travel further when going west
but you'd could take direction into:
if an object is fired at x ms ^(-1)
and fired north, east south and west
it ill travel further when going west
- 6 5
evry one keeps sayin, take into consideration the rebound and preload of the forks for interia, i dont think that you can get an accurate measure ment of a forks, preload and rebound, just beacue each for is different, and no doubt each run the rider takes may be rougher or smoother then the next jump, so it might be differnt measurements each time. but props dude.
- 11 0
but the rebound and the preload influence, as do the air resistance, the temperature, the height, the tires pressure, the mass center, rider position (with variates during the jump).....
But in general cool and useles stuff!!!!
But in general cool and useles stuff!!!!
- 7 2
A real physicist would take g to be 9.81, therefore your theory is incorrect. Think of the people who have read this and are now going to get hurt... Tut tut
Re: that would only make cm's of diffence so he's right
Re: that would only make cm's of diffence
so he's right- 9 0
dude, weight does not affect simple projectile motion; read up on your physics.
and if you want to include wind (which is a force as a function of velocity) it is too hard to integrate nicely
but when you're only going maybe 20 kph it doesnt do a great deal
and if you want to include wind (which is a force as a function of velocity) it is too hard to integrate nicely
but when you're only going maybe 20 kph it doesnt do a great deal
- 3 1
D-Monster:
You are not getting the principles right here. It is the PRINCIPLE he has not used 9.81 as a physicist that I am getting at Doing a physics A-level, I am being picky on the point I would lose marks for that rather than I am being right twat. 
You are not getting the principles right here. It is the PRINCIPLE he has not used 9.81 as a physicist that I am getting at
Doing a physics A-level, I am being picky on the point I would lose marks for that rather than I am being right twat. - 9 1
oh dear lol serious mate u have too much time, just do the jump and if you fail, u try again thats my theory lol
- 16 2
Everybody shut up. I'm failing geometry and you guys are talking about this stuff like you talk about women. My brain is deteriorating from it all. Plus the best and easiest way to know how fast to go is to just guess it and hit it.
- 1 26
im passing geometry and im a year ahead of my grade 
at least im not one of the nerds 2 years ahead in trig/alg2 who might understand this shit
at least im not one of the nerds 2 years ahead in trig/alg2 who might understand this shit - 2 16
just worked it out if you are in the air for 1 second you will of traveled 81.9 miles (130.3 km) therefore if you are in the air for 2.5 seconds and the jump is facing west realisticly you will be jumping 204.75 miles (329.5 km) + the length of the jump. think about that one then!!
- 5 0
Lol, this is what my friend just said who got an offer from cambridge to read Maths: "OMG lol! I had no idea there were so many nerds on pinkbike. I'm pretty sure though that there's an x missing three lines from the bottom (I think it should be v^2sin(theta)cos(theta)x rather than just v^2sin(theta)cos(theta)) and so basically Angus Murray has completely f*cked this up and is going to be the reason that loads of dirtjumpers hurt themselves."
- 6 0
My other "immensely cool" friend who's got an engineering offer from Cambridge has confirmed the above and also says:
Alichapple: You are wrong it makes no difference if you use degrees or radians
Spongosaurus: You are also categorically wrong: "if an object is fired at x ms ^(-1)
and fired north, east south and west
it will travel further when going west" no, simply no. That doesn't happen.
Angusmurray: "he is impressed with your physics but declares that your formula is messy!"
Alichapple: You are wrong it makes no difference if you use degrees or radians
Spongosaurus: You are also categorically wrong: "if an object is fired at x ms ^(-1)
and fired north, east south and west
it will travel further when going west" no, simply no. That doesn't happen.
Angusmurray: "he is impressed with your physics but declares that your formula is messy!"
- 5 3
well you can tell your friend to shove his diploma up his ass because that is a basic fact and if he doesnt know that then good luck with his diploma
- 0 0
actually, just put an x next to the sin(theta)cos(theta) on the bottom of the fraction under the square root
and its all good
im sorry if anyone got hurt
and its all good
im sorry if anyone got hurt
- 4 0
so the official equation is as follows::::::::
v=sqrt((-4.9*L^2)/((y2-y1)*(cos(theta))^2-cos(theta)*sin(theta)*L))
v=sqrt((-4.9*L^2)/((y2-y1)*(cos(theta))^2-cos(theta)*sin(theta)*L))
- 2 3
All math aside, I just take a few run-ins and hit jumps. No calculations, just balls and a tiny bit of skill. LOL @ your math and the argument that ensued.
i say gddmm- 1 4
i say you take this peice of paper, fold it up very very carefully... (days, maybe even weeks after) Mail it out to the mo fo that made this, and enclose a letter saying "shove this up your ass you gay homosexual f*ggot! i'm giv'en 'er full steeze!!"
thank you kindly,
Aaron Stacey
thank you kindly,
Aaron Stacey
- 1 0
you need to bring statistics into this. you need a mean and about a 95% confidence interval. min being on cold days, maybe you are fat in the off season also, maybe your forks are stiffer in winter, the ground is harder AND max being in the summer, you are lighter maybe and your forks are plusher and the ground is dusty and loose. so once you figure out the formula for a proper jump in average conditions, you need to perfect it and make it foolproof. or don't, youtube might benefit from more crash videos.
- 17 2
does this mean i need a spedometer? i'll need to lock out my suspension, ride stiff armed and still legged. and i'll need my school supplies with me at all times. i'ld rather just huck
- 8 0
except you're not taking into account the suspension of the bike (regardless of how small the fork of a dirt jumper is) which changes the heights/distance the rider goes significantly.
- 8 1
true, but still, most dj forks only have 100mm of travel so i wouldnt call the changes 'significant'.
the equation is based on the motion of a particle, and neglects factors like wind resistance, rotation, the movements of the rider, etc...
but it does give a pretty good indication of how fast you need to go.
- 1 15
i took higher physics and i hated it
so i dropped out half way through now when i look at my friends jotters its like
wow thts gay as heck im glad im out of there
- 2 0
Nice one for trying. Looks pretty good to me. Although you have not taken into consideration the loss of momentum through air resistance and friction as you leave the ramp. There are a number of factors you have chosen to ignor, so your formula is not exact. But as mathematical models go.... good work.
- 3 0
im working on a new model: one which incorporates wind resistance, rolling resistance, maybe fork compression, etc
shouldnt be too hard...
shouldnt be too hard...
- 3 0
Oh my word - this is like an endless circle, it just gets more and more complicated.
then you're probably going to have to take into consideration all sorts of quantum wotzits
then you're probably going to have to take into consideration all sorts of quantum wotzits
- 2 1
alright, i read about... 5 comments and got sick and tired of the bitchin'. Here's the deal. Grow some balls, strap on your full face, and just rock the gap already. who cares how big it is? DJ, DH, freeride... i dont care. if you clear it and land, good on ya. If you dont, get up, dust yourself, and push yourself back to run in or take the chair back up, and hit it again, until you DO clear it. BAM.
- 1 1
I'll be honest here. I flunked in maths in high school. Infact I have always been failing in maths ever since 4th grade.
So I have no clue about this stuff over here.. i just know there are some trignometry formulas here.. but what they prove.. i dont know dude.
But i know if there are 9 casettes at the rear wheel... and 2 chaingrings infront that = 18 speeds!
So I have no clue about this stuff over here.. i just know there are some trignometry formulas here.. but what they prove.. i dont know dude.
But i know if there are 9 casettes at the rear wheel... and 2 chaingrings infront that = 18 speeds!
- 0 2
you would have to take into account rotational energy or inertia because otherwise you could not go forwards on a jump with a vertical takeoff. Also you would have to have an imaginary velocity since you have the root of a negative number in there either that or the length of the gap would have to be complex (just try putting some numbers in and you'll see). also if you were to take the case where the take off and landing are the same height and the takeoff was vertical you would end up dividing by 0 since (y2-y1) would go to 0 as well as sin(theta)cos(theta) which also makes it all go strange.
- 2 0
when you plug the numbers in you get the root of a positive number, whenever the difference between the heights (y2-y1) is less than the tangent of the theta. Also, theta has to be between 0 and 90 degrees. As you say, when theta equals 90 degrees, everything f*cks up, but thats because this particle sized bike rider would just go up and come straight back down, there is no rotational energy involved in this equation. Most jumps, even if they are extremely steep, don't go past 90 degrees, they might be at about 85 or something, but 90 is just silly.
So actually, for all theta between 0 and 90, and as long as the difference in the height is less than a meter (which is a reasonable assumption for most dirt jumps), you will not get a complex velocity.
so yeh
take that
So actually, for all theta between 0 and 90, and as long as the difference in the height is less than a meter (which is a reasonable assumption for most dirt jumps), you will not get a complex velocity.
so yeh
take that
- 0 2
I just stuck some numbers in (l=2.5, y2=2, y1=1.5 and theta=75) and it came out with 6.256983481i which is an imaginary number!?!?!?!
