Horizontal distance from objects

[guest]

[font size="1" color="#FF0000"]LAST EDITED ON Aug-01-01 AT 08:06 AM (PST)[/font][p]·

Originally this post of mine was an answer to Gaz, but then I thought I could repost it as a new thread because it could come interesting to other fellow jumpers as well. And also it could stimulate discussions, why not.
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.
I hope it helps
Blue Skies :D
Andrea

[crwper]

I get the same answers for the case of the 45-degree jump with a stationary start.

Bear in mind that the 45-degree jump is used to optimize distance when you're landing on the same surface you started on. With a BASE jump, you're trying to optimize your horizontal speed, which means you'll probably put more force into the horizontal component of your launch and less into the vertical component. If you applied the entire force horizontally, you could expect to get at least √2 times the horizontal velocity, or about 4.43 m/s.

Of course there are lots of other variables. I tried the long jump in my office, and it seems pretty easy to jump 2 metres from a stationary start. I'm sure, as you said, this varies a lot depending on the jumper.

Michael

[guest]

owww, my head hurts. Quit it. Go away....

[guest]

Thanks Andrea,
Just what the doctor ordered!

[CF]

Hey Andrea ,

Nice post .
But my geuss is dat the distance from the object is getting lesser then 3.13m/sec after a few sec delay , till zero horizontal speed or then tracking .
What do you think ? and after how many sec is it getting lesser ?

Maybe info from others !

Bye bye ,
CF

[Bryan]

I agree with crwper. Why waste energy in the vertical component of a 45-deg. launch. Donkey kick the s4!t out of the exit straight out and take the longest possible delay(given the conditions).

[crwper]

Hey Andrea,

One more note while it's in my mind... It's easy to over-estimate your horizontal velocity when using the "long jump" method. Rotating about your center of mass, you can gain about 2 meters by this rotation alone (and clearly this is not indicative of your horizontal velocity). What you would need to do is launch forward at a 45-degree angle, and land at the same angle (i.e. you're almost for sure going to fall flat on your face).

The floor is pretty hard, so I'm not going to try it just now, but I'm guessing this will have a negative impact on your estimate of what is a "do-able" long-jump.

We can estimate the acceleration our legs are capable of applying by doing a vertical jump, and measuring the height we reach (without bending the legs). However... Again this isn't going to be so accurate because muscles have a natural "inverse" relationship in the force they are able to apply and the speed with which they can apply it. Jumping straight up, you're fighting gravity and your muscles won't be able to provide the same acceleration they would if you were launching horizontally.

However, since the acceleration should only get smaller, you should at least be able to get a ``conservative'' estimate from the vertical jump method. As a rough estimate, I think I can jump about 1 foot (with my legs straight when I reach the top). That's a quarter second to the apex of the jump. This corresponds with a launch velocity of about 2.45 m/s when I am working against gravity. The number will be higher for a horizontal launch, but it's a bit more of an ``organic'' question to figure out just how much higher the numbers will be.

Michael

[BASE_689]

> I agree with crwper. Why waste energy in the vertical component of a 45-deg. launch. Donkey kick the s4!t out of the exit straight out and take the longest possible delay(given the conditions).
Theoretically true. See later for explanation.

> What you would need to do is launch forward at a 45-degree angle, and land at the same angle (i.e. you're almost for sure going to fall flat on your face).
No: after having launched yourself and having a certain horizontal speed (being “floating” in the air), you keep it (there is actually nothing you can do about that, there is nothing you can do to vary your horizontal speed): you travel horizontally at the same speed, at the same speed until the air resistance starts to significantly show its effects , slowing down you speed, but granted that your speed (in the 3÷4 m/s range) is a low speed, and at that speed the air resistance is proportional to air speed itself, you get a “low” air resistance, so after few seconds your (original) horizontal air speed has diminished by only a little amount.

> The floor is pretty hard
Yes, indeed, in fact I did this “long jump from still” on a grass field!!!!!!!! My 3.13 m/s is quite realistic , conservative and repeatable.
Moreover, I speak for myself obviously, I cannot jump just horizontally (so gaining a real 4.43 m/s horizontal speed) because in this case I just would end “head down”!!! I must exit jumping off and UP at 45° just to remain stable and ending flat and stable.
Plus, the conservativity in the 45° exit stands as follows.
When I jump off and up at 45° after 0.64 s I reach again the level I exited, but when I start going down, I am 2 m away (=OFF) the structure. “Mr. Horizontal push”, even if he got a higher horizontal speed, starts his “climb down” from 0 m off the object, I start my “climbing down” when I am at 2 m away from the object.


> from the object is getting lesser then 3.13m/sec after a few sec delay , till zero horizontal speed or then tracking . What do you think ? and after how many sec is it getting lesser ?
Good question. Given air speed of 3÷4 m/s, I would say that after 3” or 4” your original air speed has diminished only by (perhaps) 10%, so well within approximations/differences between jumper and jumper.
But the main point remains the following: while your horizontal speed remains (nearly) unchanged up to 3÷4 seconds of freefall, why the need of knowing what would happen nextn (later in the freefall)?!?!?!?!?
Simply because if it is a “jump and throw” jump, doing 1÷1.5” of delay and you are open after overall 3” and you have the above distances from the object, so we know it all (well, nearly).
If it is (at least) a subterminal jump, I hope that after 3”÷4” of freefall you are NOT still in box but I hope you started (at latest within the end of the 2nd second of freefall) a good track!!!!!!! So the effect of air drag on your initial horizontal speed can be neglected and overwhelmed by far by the (horizontal) speed that you gain with your tracking!!!!! Why would you remain in box after the 3rd second of freefall?!?!?!?!? It would be a nonsense!!!!
Just my 0.02€.
Stay Safe Out There
Blue Skies and Soft Walls
BASE #689 :)
e-mail: base_689@yahoo.com

[crwper]

>> What you would need to do is launch forward at a 45-degree
>>angle, and land at the same angle (i.e. you're almost for sure
>>going to fall flat on your face).

>No: after having launched yourself and having a certain
>horizontal speed (being “floating” in the air),
>you keep it (there is actually nothing you can do about that,
>there is nothing you can do to vary your horizontal speed):
>you travel horizontally at the same speed, at the same speed
>until the air resistance starts to significantly show its
>effects , slowing down you speed, but granted that your speed
>(in the 3÷4 m/s range) is a low speed, and at that speed the
>air resistance is proportional to air speed itself, you get a
>“low” air resistance, so after few seconds your
>(original) horizontal air speed has diminished by only a
>little amount.

I was referring not to air resistance, but to the rotational motion of your body. When you did your long-jump, you no doubt took off from the ground with your legs pushing at something like 45 degrees in the forward direction. When you landed, I think it's likely you landed with your legs about 45 degrees in the backward direction (so you could absorb the landing shock). This change in attitude is due to rotational motion, and gives you about another one or two metres of distance, as well (from where your feet take off to where they land, which I'm assuming is the measurement you took).

A more accurate measurement would come from measuring the position of your centre of mass when you launched (when your feet leave the ground) and when you landed (when they first touch again) -- this will eliminate the distance due to rotation. You could do this with video.

Michael

[CF]

So ,
If my maeserments are correct it schould be +- like this ( i did every half sec - 20% of 3m/sec

0.5 = 1.5m from object
0.65 = 1.9m " " ( same level where you exit from )
1 = 3m " "
1.5 = 5.4m " "
2 = 7.2m " "
2.5 = 8.4m " "
3 = 9m " "
3.5 = 9.3m " " ( this is +- - the last 10% )
4 = 9.3m " " ( from this point it stays the same , or tracking )

Lett us know if it is wrong ,
Cu all ,
Ps : Base 689 , Where are you from ?

[BASE_689]

Now I understand what you mean.

> When you did your long-jump, you no doubt took off from the ground with your legs pushing at something like 45 degrees in the forward direction.
Yes, this is true.

> When you landed, I think it's likely you landed with your legs about 45 degrees in the backward direction (so you could absorb the landing shock)
No. When I do my long jump from still, I actually do not impress any rotation at all to my body: I leave the ground with my legs/feet trailing back myself at a (more or less) 45° angle and I keep that position "fixed in space" (=with no rotation but just traslation). With such an attitude I would hit the ground feet first, with my feet behind me, ending smashing the ground with feet, knees, pelvis, torso, face. To avoid deteriorating rapidly my day of testing, at the right amount of space and time before hitting the ground I simply lift my legs and feet and I hit the ground with my feet just in front of me, absorbing the impact ending with a squat position. The same way a long jumper (having jumped with a running launch) would end his jump in the sand.
Yes, it is true, to exactly calculate the 2 m long jump from still I should touch down with my feet having my center of mass on same level as I took off. Ending the long jump from still "squatting" down, I understand that I gain a little bit of distance ending the jump with my center of mass few dm's above the ground (in fact, long jumper lift as much as possible their legs/feet to gain as much distance as possible, ending with their buttocks being squeezed (more or less) over their heels). But if I jump the way I described above, let's say, 2.5 m, I could guess that I would have done 2 m calculated centre of mass at launch/center of mass at landing (in a stand up position with feet/legs trailing back).

> this will eliminate the distance due to rotation
See above: it would eliminate not the distance due to rotation but rather the distance due to squatting.
I guess my 2.5 m jump "my way" would be equivalent to the "perfect" 2 m jump.
What do you think then?

Stay safe out there
Blue Skies and Soft Walls
BASE #689 :D
e-mail: base_689@yahoo.com

[BASE_689]

Reply to CF:

No. Given the “table without the effect of speed diminishing because of air drag” being:

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

taking into account the effect of horizontal speed dimishing after the 3rd second of freefall by 10% every successive second of freefall, we would have that after the 3rd second you get a decrease of 10% in your forward speed, meaning that after 3rd second your speed is not any more 3.13 m/s but now is 2.82 m/s (10% less than the previous value of 3.13 m/s), and let’s suppose again that after the 4th second your speed has become 10% less than the value at the 3rd second, getting for the “after 4 s speed” a speed of 2.54 m/s.
So taking into account the diminishing of speed with time (due to air drag) we would get the following “corrected” table:

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) = 10.8 m
x (4) = 12.2 m
x (4.5) = 13.5 m
x (5) = 14.7 m

But, again, the above values (after 3rd second of freefall) are really theoretical values, given that THERE IS NO VALID POINT TO REMAIN IN BOX POSITION AFTER THE 3rd SECOND OF FREEFALL!!!!!!!!!!
If within the 2nd second of freefall you start a good track, from the 3rd second of freefall what “would” happen to your initial horizontal speed (due only to your launch) is completely overwhelmed by the (horizontal) speed you get by means of your tracking.
Just my 0.02€

> BASE 689, where are you from?
BASE #689, aka NIGHT BASE #124, aka NAKED BASE #14, aka ITALY BASE #8, is from Italy!!!!!!!!!!! ;-)


Stay safe out there
Blue Skies and Soft Walls
BASE #689 :D
e-mail: base_689@yahoo.com

[BASE_689]

Very latest news from an article posted on the French BASE Board (see http://fba.base-jump.org/index.php?T...rticle&Lang=fr )
They have been done REAL measurements performed during tests done at Service of Human Physiology of Department of Physical Education of University of Liege (Belgium).
These measurements have been performed using laser beams to start and to stop "stop watches" and so they have been capable of taking real measurements of real horizontal exit speed; the bloke under test was a BASE jumper with his donned "ready-to-jump" rig and jumping from a "stool" 45 cm - 18" high.
They did 2 sets of tests: the first set (20 jumps) was performed by the BASE jumper jumping with normal push off the object while the second set (20 jumps) was performed by the BASE jumper jumping "as his life would depend on the jump".
Results:
the normal jump yielded a horizontal exit speed of 3.5 m/s ± 10%
the strong jump yielded a horizontal exit speed of 5 m/s ± 10%
The article explicitely mentioned that, in case of "strong jump", after 5 s of freefall you would be at 25 m off the object (while being 115 m below).
Again, they say that "after 5 s of freefall the aerodynamic supports become largely sufficient to increase this horizontal speed thanks to the tracking".
The above are results quite in accordance with my theoretical values (my data having been more conservative!) ;-)

Stay safe out there
Blue Skies and Soft Walls
BASE #689 :D
e-mail: base_689@yahoo.com

[Mac]

Andrea, Do you have all your calculations and thoughts about the various things you have hit mathematically in one file? You have ALOT of good information about alot of things that should be collated.

[FlikIt]

....."So, it's a social club.....demented and Sad..but social".....