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 10-23-05, 08:46 PM #1 xenochimera Senior Member Thread Starter   Join Date: Sep 2005 Bikes: Posts: 81 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) psi generated from a 60foot drop just wondering what would be the PSI generated from this.
 10-23-05, 08:53 PM #2 Dannihilator User Title     Join Date: Oct 2002 Location: Annandale, New Jersey Bikes: 2008 Trek Portland, 1989 Nishiki Altron Posts: 19,495 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 60 Post(s) 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?
10-23-05, 09:03 PM   #3
sparks_219
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Quote:
 Originally Posted by xenochimera 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

Last edited by sparks_219; 10-23-05 at 09:27 PM.

 10-23-05, 09:22 PM #4 worker4youth Vanned.     Join Date: Jul 2005 Bikes: 2006 Motobecane Le Champ SL, 2006 Mercier Kilo TT, 2004 Gary Fisher Tassajara Posts: 1,244 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) I hope I never have the motivation to try a 60 foot drop. Just me. YMMV.
 10-23-05, 10:24 PM #5 pyroguy_3 Senior Member     Join Date: Oct 2005 Location: Blo-no, IL Bikes: 2005 Specialized Hardrock Sport, 1970's Miyata Liberty ala fixed gear Posts: 556 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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.
 10-24-05, 07:28 PM #6 Prozakk Banned.     Join Date: Oct 2005 Location: Home (don't change this back to "my own personal hell", Mr. Powermad Moderator, and/or computer savvy hacker). Bikes: Posts: 981 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Enough to cause a compression fracture of your vertebrae.
10-24-05, 07:33 PM   #7
sparks_219
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 Originally Posted by pyroguy_3 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.

10-24-05, 09:11 PM   #8
kritter
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 Originally Posted by sparks_219 = 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!

10-24-05, 10:00 PM   #9
cooker
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Quote:
 Originally Posted by sparks_219 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

Last edited by cooker; 10-24-05 at 10:06 PM.

 10-24-05, 10:02 PM #10 NJSurfCowboy Yea, It's that good   Join Date: Oct 2005 Location: South Jersey and Tempe, AZ Bikes: Haro H1 Posts: 22 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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
10-24-05, 10:16 PM   #11
sparks_219
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 Originally Posted by cooker 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.

 10-24-05, 10:42 PM #12 aussiewheeler Senior Member   Join Date: Oct 2005 Bikes: Posts: 59 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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
 10-25-05, 05:54 AM #13 Too Rass Goat Senior Member     Join Date: Sep 2005 Bikes: Posts: 77 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) The pounds per square inch achieved upon landing would be measured in the Cubic Assloads, possibly even Metric Butt-Tons.
10-25-05, 06:29 AM   #14
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 Originally Posted by Prozakk 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.

 10-25-05, 06:59 AM #15 nodnerb Senior Member     Join Date: Jul 2005 Location: Wpg. Manitoba Bikes: Mountain Cycle Rumble. Mostly xt and raceface built. Dirt Jumper 3. Avid BB7s. Posts: 494 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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.
 10-25-05, 07:22 AM #16 Albany-12303 Senior Member     Join Date: Sep 2005 Location: Guilderland NY Bikes: 4 Bikes: A Trek 2300,Old Nishiki lugged frame with sora/Campy wheels, Giant ATX-880 MTB & 2005 Lemond Sarthe Posts: 652 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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.
 10-25-05, 07:59 AM #17 xenochimera Senior Member Thread Starter   Join Date: Sep 2005 Bikes: Posts: 81 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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??
10-25-05, 08:09 AM   #18
Albany-12303
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Quote:
 Originally Posted by sparks_219 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

10-25-05, 10:27 AM   #19
Prozakk
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 Originally Posted by Namenda 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).

 10-25-05, 10:40 AM #20 nodnerb Senior Member     Join Date: Jul 2005 Location: Wpg. Manitoba Bikes: Mountain Cycle Rumble. Mostly xt and raceface built. Dirt Jumper 3. Avid BB7s. Posts: 494 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) This sounds like a job for the Mythbusters. Gotta drop Buster!!!!
 10-25-05, 11:36 AM #21 pyroguy_3 Senior Member     Join Date: Oct 2005 Location: Blo-no, IL Bikes: 2005 Specialized Hardrock Sport, 1970's Miyata Liberty ala fixed gear Posts: 556 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) "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.
10-25-05, 11:51 AM   #22
Drunken Chicken
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 Originally Posted by nodnerb This sounds like a job for the Mythbusters. Gotta drop Buster!!!!
How true! Gotta love that show.

 10-25-05, 12:42 PM #23 cooker Prefers Cicero     Join Date: Jul 2005 Location: Toronto Bikes: 1984 Trek 520; 1990s Peugeot (Canadian-made) rigid mountain bike; 2007 Bike Friday NWT; misc others Posts: 11,990 Mentioned: 59 Post(s) Tagged: 0 Thread(s) Quoted: 2365 Post(s) 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 = ˝ vt d = ˝ 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. <> Last edited by cooker; 10-25-05 at 01:01 PM.
 10-25-05, 01:04 PM #24 Prozakk Banned.     Join Date: Oct 2005 Location: Home (don't change this back to "my own personal hell", Mr. Powermad Moderator, and/or computer savvy hacker). Bikes: Posts: 981 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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.
10-25-05, 01:26 PM   #25
swifferman
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 Originally Posted by cooker 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. <>

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