Platform Clearance

[dmcoco84 4]

If you have an external platform of 6 feet (give or take 2 inches), how far above it would you need to climb to jump and clear it?

Coco

[SkiD_PL8 0]

I can't really speak for the BASE environment since I have only jumped span's. That being said I can hop 6 feet on the ground. I would think on a BASE exit, I am guessing from an "A" that 5 or so feet would be sufficent depending on how athletic you are. Stand on the bed of a truck or your porch or something and hop off and see how far you go without trying too hard then judge it from there.

Greenie in training.

[cesslon 0]

I was jumping off a small cliff into water the other day (non BASE)
and exit's didn't allow a run up, I would need about 8 foot height to clear 6 foot
and also would most likely have gone head down with another 10 feet of falling lol

[SkiD_PL8 0]

Yep, like I said it will be different for everyone. I have no problem jumping what seems to be about 6 feet on the ground without running. That is on a nice solid floor though, I am sure if the launch point was a bit sketchy it would be shorter, and the rig would probably shorten that number some as well. There isn't a one size fits all answer but I would say if you can climb high enough up a tower to jump you can probably clear a 6 foot platform from 10 feet above it without having to try very hard.

Greenie in training.

[Faber 0]

you cant jump from the platform itself?
could you mail me a pic if you have one.
On my daylight A to 1 side i need to clear a platform aswell,im not sure how far out from the exit point it get out,but sometimes you find yourself folding your legs up under you as you think your about ti hit it
Look at the attached pics(2 different jumps).
Headfuck is just after exit
the other is just to show the platform

im not sure how far out it goes estimate its placed aprox 80ft below exit point,im no way near it as i pass it(seen on video and stills)

Stay safe
Stefan Faber

[dmcoco84 4]

Yea I jump from the platform every time. I was just curious as to how far up I'd need to go in order to clear it. I can probably add another 200 feet to the jump by climbing past it. In the works of determine how high I can climb and stay out of the heat. At the moment it looks like the antenna is on the very top and nothing is below it so I might be able to go 250 more feet up. I can clear it very easily from that high but I was just curious.

Coco

[rendezvous 0]

If you ignore the effect of wind on your exit velocity a general approximation would be HV^2 = 19.6 where 'H' is the exit height in meters (m) above the platform and V is what you think would be your exit velocity straight out towards the horizon in m/s. Of course it's been quite a while since I did Physics so this might not be totally accurate.

[dmcoco84 4]

Well every bit helps, thanks for the input!

Coco

[BASE813 0]

taken from an old post on BLiNC from a known Italian:

Quote

.........The horizontal distance from an object after a certain delay depends only from your "velocity at exit", and so it depends really so much on each jumper. But some time ago I made few calculations making a guess. I supposed the following.
Suppose to have a jumper that performs a "long jump from still" to his "maximum length possible at the given exit velocity" (I know in physics it has another specific word, but in this moment I am missing the English word for it), that is, the jumper jumps with a trajectory of 45° on horizontal plan, and let us suppose this jumper, with his jump to his "maximum length possible", reaches a jump of 2 m, that is, final position on horizontal plan of his feet is 2 m from position at exit. A "long jump from still" of 2 m is doable. Provided what above, let us have V has "velocity at exit" and Vh as "horizontal velocity at exit", and having √2 = 1.41 and g = 9.8 m/s², we have the following results:
V²
— = 2 m after t = 0.64 s · · · V = √ (2 · g) = 4.427 m/s
g

Vh = (√ 2 / 2) · V = √ g = 3.13 m/s

With the above results, we can now write the distance, x (t), "flown" horizontally after 0.5 s, 0.64 s, 1 s, 1.5 s, 2 s, 2.5 s, 3 s, 3.5 s, 4 s, 4.5 s, 5 s, etc. of freefall:

x (0.5) = 1.6 m
x (0.64) = 2 m
x (1) = 3.1 m
x (1.5) = 4.7 m
x (2) = 6.3 m
x (2.5) = 7.8 m
x (3) = 9.4 m
x (3.5) = 11.0 m
x (4) = 12.5 m
x (4.5) = 14.1 m
x (5) = 15.7 m

Practical example: if you exit with the above velocity and if your canopy opens 3 s after exit, you are 9.4 m far from the object.
Of course, provided the modest horizontal speeds involved, we can neglect any influence of jumper's body with respect to air drag; and it does not take into account at all any horizontal speed you could gain tracking, but that (tracking) would only help, still I do not think you can gain any significant tracking effect below 3 s of freefall (while if you start to track immediately after exit you can have a benefit in the sense the track will be as much efficient as possible, as soon as possible...).
If the jumper runs at exit instead of doing a "long jump from still", what you have to do is calculate (=guess?) your horizontal speed at exit and then do your homework getting the horizontal distance "flown" multiplying the horizontal speed times the elapsed time.
Another guess: if the "long jump from still" gives you a horizontal exit velocity of 3.13 m/s and the fastest men on earth can run at 11 m/s (top velocity), as a run of a BASE jump, with your trekking shoes, clothes, weigth of rig, PC handheld, limited distance for running, etc. etc., reasonably you could have a horizontal exit velocity of 4 m/s, 5 m/s, 6 m/s at best, in any case hardly 7 m/s or more than this...
For conversion meters/feet, provided that 1 ft = 0.3048 m/ft and 1 m = 3.2808 ft/m, you can help yourself.

with discussion:
http://www.blincmagazine.com/forum/showthread.php?t=18156

[yuri_base 1]

Yo!

To clear the platform of length L, assuming you're launching with horizontal speed V, you need to jump at least from height

H = 0.5*g*(L/V)^2

above it. Here g is acceleration of gravity, g=32ft/s^2=9.8m/s^2.

L should actually be the sum of platform length and half of your body height.

E.g. for L=9ft (6ft platform + half your height) and V=8ft/s, H=0.5*32*(9/8)^2=20ft.

Give it a good margin of error -- L=15ft (6ft clearance), V=3ft/s (weak poised launch) -- and you get a whopping H=400ft.

Feel the danger now?

Yuri

Android+Wear/iOS/Windows apps:
L/D Vario, Smart Altimeter, Rockdrop Pro, Wingsuit FAP
iOS only: L/D Magic
Windows only: WS Studio

[JaapSuter 0]

I use a very simple formula; If you need math to clear the object, it's too dangerous.

And that's coming from a mathematician.

[yuri_base 1]

Exactly. On an object where exit point is fixed (cliff with outcropping), you can simply look down and your guts will tell you all the math. In this particular case, he asks how high above outcropping he needs to climb to clear it. Math just shows that vertical safety margin is surprisingly sensitive to the horizontal one.

Android+Wear/iOS/Windows apps:
L/D Vario, Smart Altimeter, Rockdrop Pro, Wingsuit FAP
iOS only: L/D Magic
Windows only: WS Studio

[fastpete 0]

Budapest antenna is quite good example for this. From the top i mean.

_____________________________________________
F......ck the Finns !!!
FastPete www.pete.fi email: [email protected]

[yuri_base 1]

Yo again

A good practical way (and most precise one) to estimate the height necessary to clear an obstacle is to do your typical exit from a step onto pavement while wearing rollerblades or onto ice while wearing skates (to simulate virtually frictionless forward motion). Use a stopwatch to measure the time it takes to clear the obstacle length + half your body height + safety margin. Then use freefall chart to find out the corresponding height. (or use the formula H=0.5*g*t^2)

This method will tell YOUR safe height. As my previous example showed, this height varies greatly depending on your launch speed. 100ft may be enough for you, while someone with weak exit may be surprised to find out they need 400ft.

Yuri

Android+Wear/iOS/Windows apps:
L/D Vario, Smart Altimeter, Rockdrop Pro, Wingsuit FAP
iOS only: L/D Magic
Windows only: WS Studio

[434 2]

Fuck what happend to the old good metod with an orange or a small rockdrop, simulating the force of a jump! Just takes a little swing of your arm to see if it pass the ledge! All this formula stuff will kill someone

[TomAiello 25]

In all honesty, if I look at the obstacle, and I don't know I can clear it, I won't jump, regardless of what any amount of math tells me.

-- Tom Aiello

[email protected]
SnakeRiverBASE.com

[dmcoco84 4]

Quote

In this particular case, he asks how high above outcropping he needs to climb to clear it. Math just shows that vertical safety margin is surprisingly sensitive to the horizontal one.

Right! And I know for a fact I can go 200+ feet above it but I was just curious the minimum height above it to clear it safely.

Coco

[peterk 0]

Why don't you just try it? Do dz.com formulas make you feel better at the exit point?

---------------
Peter
BASE - The Ultimate Victimless Crime