Me and my friend have been looking around for a sick place to build a nice backcountry 30 footer. Does anybody have any idea how to calculate the speed and trajectory necessary to clear a step over given distance and weight? Prob not lol but worth a try. Way too pumped for ski season. This is probably a really gay advanced equation which isn't worth it but idk
is it a 30 foot wide 200 foot deep crevasse your planning on jumping? if not don't worry about it...its 30 feet.
when building any sort of jump for a snow sport there are simply too many variables to actually calculate what speed you will need/where you will need to start the in run from to get to the landing.
So yeah just go for it.
distance x gnar + trick = steeze
just go for it, you've probably hit 30 footers before, now it's just not in the park.
This is the part in films where the rider gives a snowball a good toss from the lip, subtracts a few fucks so he doesn't have anymore fucks to give, then mans up and sends it. It's really a simple calculation if you ask me.
it's possible to do it with some simple physics knowledge but being safe and calculated does not get you far in skiing, especially for such a relatively low risk jump
Eyeball it and the send it. I know math goes into consideration for big park jumps but unless you bc booter is huge I think you just need to nut up. Toss a couple snowballs and then send it, and remember, too big is better than too little.
would you have a way of measuring your speed even if you calculated it?
Download-My-Appswould you have a way of measuring your speed even if you calculated it?
I actually have a friend with a radar gun. But I guess I'm probably being too technical about it because it wouldn't be an easy thing to figure out with maths lol
Just make sure to always bring your protractor and tape measure in the BC
yungsteezeI actually have a friend with a radar gun. But I guess I'm probably being too technical about it because it wouldn't be an easy thing to figure out with maths lol
It would be fairly basic math for purely speed and the jump itself. But then you would need someway for measure wind and temperature changes and would have to modify them each jump.
Most jumps I've built have no gap to clear and basically endless landing. It makes it more difficult to screw it up.
Northvansebdistance x gnar + trick = steezejust go for it, you've probably hit 30 footers before, now it's just not in the park.
I agree.
also, if your gapping something, throw a snowball 3 times across it, if 2/3 snowballs make it, so will you.
NinetyFourThis is the part in films where the rider gives a snowball a good toss from the lip, subtracts a few fucks so he doesn't have anymore fucks to give, then mans up and sends it. It's really a simple calculation if you ask me.
Spot on good chap.
It's just something you have to get a feel for. The hardest part when you build your first jumps isn't figuring out how fast you need to go, but how to build a wedge that isn't super awkward and terrible to hit.
9.8 m/s/s...
It's not like you are trying to reach escape velocity, so no, it's not rocket science...
some equations you can use
to find the distance in the vertical direction you’ll fall:
y = y0 + V0 sin(θ)t - 1/2gt^2
Solve for time.
Plug ‘t’ into: x = V0cos(θ)t
to find the horizontal distance you’ll travel.
Make adjustments to your initial velocity and ramp angle using:
x = (V02/g)sin(2θ)
Calculate the length of your inrun.
mgh + ∫μkmgcos(θ)dx = 1/2mv^2
dx is how the angle of your inrun is changing with respect to x (so good luck calculating that)
you’ll also need to figure out the coefficient of friction between your skis and the snow
but you'd do better to just ask yourself where "would big air dave drop?" and then side step up 50 more feet from there.
Skibumsmithsome equations you can useto find the distance in the vertical direction you’ll fall:
y = y0 + V0 sin(θ)t - 1/2gt^2
Solve for time.
Plug ‘t’ into: x = V0cos(θ)t
to find the horizontal distance you’ll travel.
Make adjustments to your initial velocity and ramp angle using:
x = (V02/g)sin(2θ)
Calculate the length of your inrun.
mgh + ∫μkmgcos(θ)dx = 1/2mv^2
dx is how the angle of your inrun is changing with respect to x (so good luck calculating that)
you’ll also need to figure out the coefficient of friction between your skis and the snow
but you'd do better to just ask yourself where "would big air dave drop?" and then side step up 50 more feet from there.
I recromend taking your devilish math and leave newschoolers.
No sarcasm here....just trying to help:
Build your wedge, and try to eyeball the trajectory that it will send you.
Then, try to estimate the high point of your flight (should be in the middle between takeoff and landing). Throwing a snowball can help too.
Then estimate where you want to start your in-run. This start point MUST be higher on snow than the high point of your trajectory.
Speed check it once, see if speed seems reasonable.
Huck.
Land.
Ride away clean.
Afterbang.
Claim.
Hope this helps.
your worried about 30 feet? take you panties off and build that sucka to 90 feet
TWoodsIt's just something you have to get a feel for. The hardest part when you build your first jumps isn't figuring out how fast you need to go, but how to build a wedge that isn't super awkward and terrible to hit.
This is going to be far more important than speed.
Remember to make the wedge a hell of a lot longer than you think as well as an increasing angle of takeoff, not just a perfect triangle.