Mountain Biking - psi generated from a 60foot drop

just wondering what would be the PSI generated from this.

Could you please rethink the word useage. PSI is for air in a tube or tubeless tire.

Are you thinking of positive/negative g-force?

just wondering what would be the PSI generated from this.

Did a bit of physics and got the following:

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 ;)

Ming

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.

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.

I think it's about 4Gs...

= 3,780 Newton * 0.2248089
= 849.777 Pounds.....

which is a HELL a lot of force on your body and bike.

Ming

Which is the reason 60 ft drops land on a slope...so the force exerted is only partial of that 850lbs.

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!

Making a VERY modest assumption that it takes 0.5 seconds to stop accelerating in the y-direction (no more free fall)


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.
Robert

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



Keith

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.
Robert

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.

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.

Enough to cause a compression fracture of your vertebrae.


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??

Making a VERY modest assumption that it takes 0.5 seconds to stop accelerating in the y-direction (no more free fall)



This is the key key assumption- ie how fast you decelerate.

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 :eek:

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.

So do I (speak from experience).

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.

This sounds like a job for the Mythbusters. Gotta drop Buster!!!!
How true! :D Gotta love that show. ;)

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:

g=v/t
v =gt
d = ½ vt
d = ½ gt2
t= 2d/v
v = 2d/t
t= SQRT(2d/g)

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.

<<Sh-hiv-ver-r>>

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.

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.

<<Sh-hiv-ver-r>>


You sir, need to meet Bender in the land of ridiculously sized drops.

If I had the link I would give it to you.

You sir, need to meet Bender in the land of ridiculously sized drops.

If I had the link I would give it to you.

Yeah but, a 60' to a FLAT????
:eek:

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??

xeno...we need more info. what kind of 60 foot drop are you talking about? A bike dropped on pavement? A rider taking off from a mountain ledge and landing on a 45 degree grassy slope? A stunt rider dropping into a large half-pipe?

Are you hoping to calculate the bending or compressing force on some frame component?

R

How to land a 60ft drop: 500mm of fork travel, and 400mm of rear travel. A bike like that wouldn't be very pedal friendly and its geometry would be screwed up.

EDIT: Im pretty sure your frame would have to be super beefy, as in it weighing around 20lbs.

60 footer off a cliff, landing on a hill. this is all theoretical as of now. my friend wouldnt reveal me any information on this "material" but said its pretty revolutionary, and its main application is not for bikes but can be.

60 footer off a cliff, landing on a hill. this is all theoretical as of now. my friend wouldnt reveal me any information on this "material" but said its pretty revolutionary, and its main application is not for bikes but can be.

There are too many factors that would have to be taken into account, including the angle of the slope, how much forward momentum the rider has, how much travel and stiffness in the suspension, the style of landing (eg. rear wheel first), the rider's skill in handling the landing, the geometry and angles of the frame, etc., etc., so there's no way to make any calculation on this. For example, if the slope was a near match to the rider's trajectory, the landing would be very soft, like a ski jumper's landing.
R

I've seen motorcross bikes land huge jumps (some on flats too) without much problem, so if you have enough and high enough quality of suspension travel, then it is very doable. Better have some good skills though, all the travel in the world isn't going to help you if you dont land with your wheels in line!

Watchman, i dont think the motocross jumps were 60 feet high. Even robbies Knievels jumps probably arent much higher than that, but you also have to take into account the fact that hes moving forward at probably upwards of 50 mph, not easily achievable on a bike, especially one strong enough to take the hit.

60 footer off a cliff, landing on a hill. this is all theoretical as of now. my friend wouldnt reveal me any information on this "material" but said its pretty revolutionary, and its main application is not for bikes but can be.

I hope it's not magnesium.

I hope it's not magnesium.
I hope it's not manganese.

For example, if the slope was a near match to the rider's trajectory, the landing would be very soft, like a ski jumper's landing.


Speed is also a factor. The bigger the jump, the steeper the transition and the runout need to be, which of course means the rider will be going very fast. A bike with the geometry and travel required for a 60 footer is not going to handle well at 80 mph.

OK, I’ve given myself a miniphysics tutorial...

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.

<<Sh-hiv-ver-r>>

Firstly, are your tyres really 8cm deep?
Secondly, the distance shouldn't be multiplied by 2, as you don't add the front and rear wheel travel together do you?
Thirdly, you are treating the suspension (and tyre) compression as if it were a crumple zone with no resistance. In reality, the suspension will have a huge resistance, dampening, the force needed to compress it will incease with the distance it's compressed. It's not simply a straight compression over a distance. The rider would also move a lot on landing, which would essentially increase the distance in which the 'mass' came to a stop. Then the frame woud flex, and various other factors would mean that the value of G you calculated was much much too high...

There are too many factors that would have to be taken into account, including the angle of the slope, how much forward momentum the rider has, how much travel and stiffness in the suspension, the style of landing (eg. rear wheel first), the rider's skill in handling the landing, the geometry and angles of the frame, etc., etc., so there's no way to make any calculation on this. For example, if the slope was a near match to the rider's trajectory, the landing would be very soft, like a ski jumper's landing.
R

If the rider landed rear wheel first he would be toast, thrown over the bars at 60mph wouldn't be a prettysight. If they supposedly landed with the wheels perfectly parallel to the ground, they would have a chance at landing it, but a slim chance of steering to safety. Look at Josh Bender, he couldn't even land the 54 foot jah drop.

Stick to a 40 footer :p

I hope it's not manganese.

I hope it's not margarine.
:D :D :D :D :D :D

like said before....TRAJECTORY is the key for the landing.....to bring back up the motocross comment, the landings are designed to slowly decellerate the rider in the Y direction (up and down). conversely with a drop commonly encountered on a trail, the decelleration is quick, also called an impulse force.


to address the issue on where most of the force is taken on the bike: its between the rims/hubs and the bottom bracket. the rims/hubs for obvious reasons but the bottom bracket seems like it would take most of the force because thats where all the riders weight is transferred. if you just took a bike a flung it off the cliff and it actually landed like you would like to if you were on it, not much would happen because there wouldnt be enough force (only weight of bike) acting on the bike but now hop on the bike and do it again this time all the riders weight is being transferred through that BB throughout the bike minus what little weight is transferred through the handlebars

I hope it's not manganese.

I hope it's not a Huffy.

Firstly, are your tyres really 8cm deep?
Secondly, the distance shouldn't be multiplied by 2, as you don't add the front and rear wheel travel together do you?
Thirdly, you are treating the suspension (and tyre) compression as if it were a crumple zone with no resistance. In reality, the suspension will have a huge resistance, dampening, the force needed to compress it will incease with the distance it's compressed. It's not simply a straight compression over a distance. The rider would also move a lot on landing, which would essentially increase the distance in which the 'mass' came to a stop. Then the frame woud flex, and various other factors would mean that the value of G you calculated was much much too high...

Thanks, mp
Of course, my calculations were very hypothetical...no-one's going to drop 60 feet onto the flat (on purpose).
I did make the point that "g forces" would be lessened by bike collapse, and of course they would also be lessened by body movement, as your chest imploded against the stem (ouch).
For tire clearance I actually meant to estimate 4 cm, but I put down 8 cm by mistake.
I agree that deceleration would not be constant...hey, it was a rough calculation, after all. But I don't imply that the "crumple zone" has no resistance...obvously it has resistance or it wouldn't produce deceleration.
The double distance in the formula is not due to adding the front and back suspension...it is a mathematical artifact: distance =time X average speed. The average speed over a period of constant acceleration or deceleration is (v1+v2)/2. But when you're accelerating to or from a stop, one of those v's is equal to 0, so the average speed is the other v divided by 2. When you plug that into these equations, the '2' (or its inverse) gets introduced.

Cheers
Robert

i got a bit more info on this "material"

its called buckypaper
its 250 times stronger than steel, 10 times lighter than steel, and harder than diamond. currently being tested by FSU.

Buckypaper - some info:
http://www.physorg.com/news7435.html

I wonder how much that stuff would retail for.

I wouldn't be surprised if it turns out to be toxic like asbestos. If strands of that stuff break off and fly, look out - airborne nanoparticles can be a major health hazard.
R

Thanks, mp
Of course, my calculations were very hypothetical...no-one's going to drop 60 feet onto the flat (on purpose).
I did make the point that "g forces" would be lessened by bike collapse, and of course they would also be lessened by body movement, as your chest imploded against the stem (ouch).
For tire clearance I actually meant to estimate 4 cm, but I put down 8 cm by mistake.
I agree that deceleration would not be constant...hey, it was a rough calculation, after all. But I don't imply that the "crumple zone" has no resistance...obvously it has resistance or it wouldn't produce deceleration.
The double distance in the formula is not due to adding the front and back suspension...it is a mathematical artifact: distance =time X average speed. The average speed over a period of constant acceleration or deceleration is (v1+v2)/2. But when you're accelerating to or from a stop, one of those v's is equal to 0, so the average speed is the other v divided by 2. When you plug that into these equations, the '2' (or its inverse) gets introduced.

Cheers
Robert

Ah yeah, you're right about the 2 x distance thing, my mistake, sorry!
I know you're calculation was only rough, but I think it's hugely hugely inaccurate due to it's roughness. 80G would completely **** you, whether you're bike fell to bits or not! I know landing a 60ft drop to flat would be harsh no matter what, but it would be nowhere near 80G.
Cheers

Ah yeah, you're right about the 2 x distance thing, my mistake, sorry!
I know you're calculation was only rough, but I think it's hugely hugely inaccurate due to it's roughness. 80G would completely **** you, whether you're bike fell to bits or not! I know landing a 60ft drop to flat would be harsh no matter what, but it would be nowhere near 80G.
Cheers

When Fernado Alonso smashed his F1 racer into the wall from 150 MPH, he suffered a 130G impact. So 80G from a 60 feet free fall might be somewhat realistic. I think 40 G is more like it though.

Ming