How much risk?

Originally posted by ~Inshan

If I think I can pull it, and it all calculates out in my head, I will usually go for it. If fear isnt what holds me in check, its definately a huge factor but if i know i can pull something and I believe it in my head i usually can…

This is a good thread…

I've never been one for taking crazy risks, but i think calculated risks are an inevitable part of riding.

I personally don't like doing anything unless I can look at it first and feel confident, totally being able to visualize it the whole way through. AND having a good idea of what could happen if things go wrong, what you can do about it, and what the consequences may be. being a bit older now and having a career to maintain makes me think alot harder about doing stuff that has serious consequences.

The biggest risk I've taken in a long time was the stump gap on fromme, i looked at it and it was super intimidating, but i KNEW i could do it, it was a calculated risk. All went fine on the first try, and then boom, on my second try i flailed it and had to pay the price.
However, getting up and over it and spotting the landing on my first try, knowing i had done it and still being in the air, waiting to touch down was probably one of, if not THE best moment on my bike ever. I still think the risk was worth the reward, but being off the bike for 4 months and permanently damaging my wrist is a big price to pay.
Now that I've seriously injured my right wrist (as an artist it's my source of income and my career depends on it) and had surgery, I think my outook has changed a bit.
I'll take a chance on a line that has fewer variables and that if you try it once and screw it up, you can still try it again, you're not going to the hospital, i won't be rolling the dice on any line that has the potential to do serious harm if something goes wrong unless i'm absolutley sure i'll pull it.

i look at things.. access the line and what i'll have to do to land it or ride it or whatever…

i don't think about getting hurt, if u think about it you'll never progress… ya can't be scared

To start with, I will use the simplest example, doing a step down gap from a flat (0 degree) take-off. Because we have a flat take off, the rate at which you and your bike drops to the ground is the same as simply stepping off the edge of the drop. Your horizontal speed has absolutely no effect on how long it will take you to hit the ground. Therefore, we need to know only two pieces of information:
1) the height of the drop, H
2) the distance of the gap, D

The distance of the gap is the distance from the take-off to the desired landing point in the middle of the transition.

We will assume a 30/40 step down gap, 30 ft down and 40 feet accross. These numbers must be converted to metric, which is approximately 10m down and 14m accross

We must first determine the amount of time we spend airborne. This is done using the equation

H = 1/2 * a * t^2

where H = height of the drop, a = acceleration due to gravity (a known physical constant, -9.8m/s^2) and t = time. Filling in the known information we get

-10 m = 1/2 * -9.8 m/s^2 * t^2

Note: -10 m because we are falling below the start point

Now we need to solve for t, so

(-10 * 2) / -9.8 = t^2 = 2.04

t^2 = 2.04 seconds, so t = 1.43 seconds in the air.

Now that we how long we are in the air for, we can determine the speed needed to cross the 13m gap. Speed or velocity is simply the distance travelled divided by the time taken to travel that distance. As we are in the air for 1.43 seconds and travel 13m the velocity is

13m / 1.43s = 9.09 metres per second, which is about 33 km/h

So to wrap that all together, your velocity is

V = D / [(H * 2) / -9.8]^0.5

Now for the tricky part, a take off that is angled. Let's assume we have a 15 degree angle for our take-off. With the angled take off, there are two components to your velocity, a vertical component (up and down) and a horizontal component (accross). This also means that we must break the time of flight into two seperate times.

If you take a ball and throw it straight up it will reach a peak height and then fall back to the ground. This is similar to taking off at an angle, as our vertical component will let us reach a peak height where we would momentarily stop. This height would be above the lip of the launch and we need to find how long it takes us to reach this max height and how far up we travel. We then need to add this vertical height to the height of the drop. From this total height we can then determine the rest of the time we are airborne.

So to determine the velocity required to clear the gap we need to know how long we are in the air for. To find out how long we are in the air we need to know the vertical component of the take-off velocity. This presents a catch 22, as we cannot determine the velocity required to clear a gap with an angled take off by only knowing the height and distance of the drop and gap. The result is that we must do a couple of estimates to find the correct take-off velocit.

If we look back at the first example with the flat (0 degree) take-off, we needed a velocity of 9.09 m/s to land in the middle of the tranny. We will need more speed on the angled take off to land in the middle of the tranny. So, we simply have to use trial and error to find the right speed. Using trial and error on paper will not result in a missed landing, unlike taking an educated guess for the real thing.

So, we will try for a speed of 11 m/s or 40 km/h.
The vertical velocity component (Vy) of 11 m/s from a 15 degree take-off is

Vy = 11 m/s * sine 15 degrees
Vy = 11 m/s * 0.259 = 2.85 m/s

Now to find the time of flight to our max height we use

t = (Vy final - Vy initial) / a
t = (0 - 2.85) / -9.8 = .29 seconds

Now that we know the time to our max height, we can find the vertical distance travelled up.

D up = (Vy * t) + (.5 * a * t^2)
D up = (11 * .29) + (.5 * -9.8 * .29^2)
D up = 3.19 - 0.41 = 2.78

Now we use the equation from the first example to get the second half of our time total (from max height to the ground)
Don't forget to add in the height we just calc'd to the original height of the drop, 10 m + 2.78 m = 12.78 m drop

-12.78 m = 1/2 * -9.8 m/s^2 * t^2
-12.78 = .5 * -9.8 * t^2
-12.78 = -4.9 * t^2
t^2 = -12.78 / -4.9 = 2.61
t = 1.61

Now add this time to the time to reach max height and we get

1.61 + 0.29 = 1.9 seconds in the air.

With the total time in the air we can find the hroizontal distance travelled by multiplying by the horizontal component of our initial velocity of 11 m/s

Vx = 11 m/s * cosine 15 degrees
Vx = 11 * .966 = 10.63

Finally (about frickin time!) we get our horizontal distance travelled which is

10.63 m/s * 1.9 s = 20.2 m

So at 11 m/s (40 km/h) we blow our tranny by about 6m which equals carnage. We know we need to go a little slower.

Vy = 9.6 m/s * sine 15 = 10 * 0.259 = 2.49 m/s

t = (Vy final - Vy initial) / a = (0 - 2.49) / -9.8 = .25seconds

D up = (Vy * t) + (.5 * a * t^2)
D up = (10 * .25) + (.5 * -9.8 * .25^2)
D up = 2.5 - 0.31 = 2.19 m

-12.19 m = 1/2 * -9.8 m/s^2 * t^2
-12.19 = -4.9 * t^2
t^2 = -12.19 / -4.9 = 2.48
t = 1.57

total t = 1.57 + .25 = 1.82

Vx = 9.6 m/s * cosine 15 degrees
Vx = 9.6 * .966 = 9.27

Finally (about frickin time!) we get our horizontal distance travelled which is

9.27 m/s * 1.82 s = 17 m So now we only miss by about 3 metres, which is about 9 feet.

So you can see that if the take off angle is not that steep, the flat take-off equations (which are WAY easier to figure out) will give you a solid idea of how fast to go.

Can anybody tell why math and physics are important?

So to wrap that all together, your road gap take-off velocity is:

time to fall = the square root of [ (Height of drop * 2) / 9.8 ]

Velocity = Distance of the gap divided by the time to fall

I spent an hour typing this out, hopefully somebody will make use of it. If there are any mistakes (shouldn't be) pm me and I will fix it up. Just so you know, the angled take-off is a good example of a first year college level physics course. The fourth year stuff will really make your head spin.

:thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp: :thepimp:

Crazy.

Originally posted by Tha Niggla
**This is a good thread…

Now that I've seriously injured my right wrist (as an artist it's my source of income and my career depends on it) and had surgery, I think my outook has changed a bit……… i won't be rolling the dice on any line that has the potential to do serious harm if something goes wrong unless i'm absolutley sure i'll pull it. **

Listen to the wisdom here.

synchro I have a calculus final next week, if I pay for it would you mind coming down this weekend to Cailifornia and helping me out? Man, that's a bit of work, good job figuring it all out.

The Ito

That is sick that you actually took the time to calculate all of that

you do know that having suspension will sorta screw that formula up for hte one wihta take off .

The amount of vertical velocity that your suspension would add would not be fairly negligible. It might result in a diff of a couple feet on the landing. If you're aiming to lan in the middle of the tranny it won't matter much.

meh…just eye it up and go:D

Originally posted by Vret
meh…just eye it up and go:D

;)

exactly

:werd:

mad props man i will use it for good!:thepimp: