just wondering what would be the PSI generated from this.
just wondering what would be the PSI generated from this.
Did a bit of physics and got the following:Originally Posted by xenochimera
Only working the de-acceleration of vertical-direction
60 ft = 18.288 m
1) t = [vo +/- sqrt(vo^2 - 2 * (a) * (x0 - x)] / a
t = [0 + sqrt(0 - 2 * (9.8) * (0-19.288)]/9.8
t = 1.93 seconds
Making a VERY modest assumption that it takes 0.5 seconds to stop accelerating in the y-direction (no more free fall)
a = 18.9 / 0.5 = 37.8 m^2/s
Now the force of the drop for a 100 kg rider and bike (220 lbs) becomes
F= m * a
= 100 kg * 37.8 m^2/s
= 3,780 Newton
Convert Newton to Pound Force
= 3,780 Newton * 0.2248089
= 849.777 Pounds.....
which is a HELL a lot of force on your body and bike.
I hope you have the skills and THE bike before you try a 60 foot drop
Last edited by sparks_219; 10-23-05 at 10:27 PM.
I hope I never have the motivation to try a 60 foot drop. Just me. YMMV.
How many g's is that? I can't cipher a conversion at the moment... I'm sort of in the whole electrical circuit mode of physics... Did the question maybe mean how many psi is exerted on the fork's reserves? However, given the ammount of info given in the question I believe you are correct in your assumptions sparks.
Enough to cause a compression fracture of your vertebrae.
I think it's about 4Gs...Originally Posted by pyroguy_3
Which is the reason 60 ft drops land on a slope...so the force exerted is only partial of that 850lbs.Originally Posted by sparks_219
a 60 ft drop to a flat land would break bike or body or both...that would be like jumping off a 6 story building and landing on the sidewalk...suspension is good, but I dont think that good...but I could be wrong as I dont watch or care about 60 ft drops!
Sorry my physics is rusty so I can't do the calculation, but I assume it would be better to estimate the deceleration distance (the "crumple zone") and then figure out the deceleration time from that. I bet the impact would occur over much less than 0.5 sec, and would therefore involve much more than 4 X total weight.Originally Posted by sparks_219
Last edited by cooker; 10-24-05 at 11:06 PM.
well if you take that ~850 lbs of force and divide it by the footprint of your tires you can get psi....
2.5" wide tire and assume a circular footprint....A= 4.90625 sq in
850lbs/4.90625 sq in = 173.25 psi if landed on one tire at a time
if flat land, divide that by two and get 86.6 psi
that is of course in addition to that pressure already in your tire
That's what I thought too, but I figured i'll make the assumption modest due to absorbsion from the tires and possibly suspension. So 0.5 seconds sounded reasonable to me at the time.Originally Posted by cooker
yeah like someone already stated. its 859lbs of pressure assuming you land on flat land and decelerate almost instantly. if you are rolling (landing on a slope) then alot of that force will just add to your momentum and not to fracturing your back and makeing you bike into toothpics
The pounds per square inch achieved upon landing would be measured in the Cubic Assloads, possibly even Metric Butt-Tons.
Originally Posted by Prozakk
And, if you are fortunate enough NOT to get a fracture, you almost certainly will get one or more disc herniations. And speaking from experience, you don't want that.
I think he may have meant what rise in psi in your tires(rear tire I'd imagine) would you get when landing a 60 foot drop. Which would then lead me to think he also meant 6 feet, not 60. Not sure though.
The rise in PSI in the tires would not be very much - the most that the tire volume will compress is about 10% max or only a few PSI. You are only pushing in the footprint region and the most it will push in is to the rim (any more and the rims will bend).
If we are talking about the pounds of force generated on the drop and not PSI in the tire then SPARKS 219 has the answer.
i mean 60, its for a calculation, my friend at umich said they have some sort of new material, light and durable, that can replace carbon fiber as the lightest source of frames. what would be the PSI generated on the frame??
This is the key key assumption- ie how fast you decelerate.Originally Posted by sparks_219
If you have shocks the size of stilts, there will not be much force. If you have a hardtail with rigid fork, then the force would be thousands of times higher
So do I (speak from experience).Originally Posted by Namenda
This sounds like a job for the Mythbusters. Gotta drop Buster!!!!
"i mean 60, its for a calculation, my friend at umich said they have some sort of new material, light and durable, that can replace carbon fiber as the lightest source of frames. what would be the PSI generated on the frame??"
There is no general psi generated over the entire frame. the majority of the blow would be taken on the welds. the drop outs, where the head tube and the top and bottom tube meet, and probably the chainstays. You would be better off calculating the forces not in psi, metric is where it's at, and also would need some sort of backround in engineering of metals. It would be some decently heavy physics. ask your friend, if they are making this material, then they should have some idea of the stresses it can take.
How true! Gotta love that show.Originally Posted by nodnerb
2005 Ironhorse 7.3
2005 Specialized Hardrock Sport
OK, I’ve given myself a miniphysics tutorial, and here are my very shaky conclusions about a mountain biker dropping 60 feet onto a flat surface.
Acceleration due to gravity = g = 9.8 m/s/s
Assuming starting velocity = 0, and final velocity = v, then here are some general equations we can apply to this situation:
d = ˝ vt
d = ˝ gt2
v = 2d/t
So in our example, for falling 60 feet
d = 60 ft = 0.305 X 60 m = 18.3 m
the time to fall 18.3 m:
t= SQRT(2d/g) =1.93 sec (as sparks_219 calculated)
The impact velocity would be:
v= 2d/t= 36.6m/1.93sec = 19 m/s (about 40 mph)
Now suppose we impact flat ground and thus decelerate quickly. If the bike and rider were of some super material that would stay intact, the deceleration would occur over the distance of tire compression (about 8 cm, or under two inches) and suspension travel (let’s say 6 inches, or 15 cm). Thus the total deceleration distance = 23 cm, or 0.23 m, the distance over which you have to slow from a speed of 19m/s to 0 m/s.
The time it will take to decelerate is:
t=2d/v = 0.46m/19m/s = 0.024 sec
The deceleration rate would have to be equal to v/t, so (19m/s)/0.024s = 791 m/s/s
Which is about 80 g.
Why so huge? You had 60 feet of falling to get up to speed, and less than a foot in which to stop.
Of course in reality, the bike would crush under your body, allowing you some extra travel distance, so the g force on your body would be softened by that, but it would still be a lethal force…plus the broken bike parts would impale you.
Last edited by cooker; 10-25-05 at 02:01 PM.
The 1st 20" bike I got, I jumped a 4' high ramp on pavement & broke the frame in two. Bike lasted 2 hours. The next bike I had to pay for.
Originally Posted by cooker
You sir, need to meet Bender in the land of ridiculously sized drops.
If I had the link I would give it to you.