



sactownclimber
Feb 12, 2006, 5:47 PM
Post #26 of 63
(6585 views)
Shortcut
Registered: Jan 1, 2005
Posts: 216

In reply to: In reply to: In reply to: . . . Acceleration is 9.81 meters/second squared. That means that force you'll create falling through the air is less than 900 newtons, so your caribeaner thats rated for 25000 newtons would surely hold you. And that's pretty much all you need to know. Ummm . . . no. My physics is pretty rusty by I know you can't just multiply 90 kg by 9.8 and says that's the force. Ummm . . . yeah. My physics is pretty rusty too, but I do remember that: F= force = ? m= mass = 90kilograms a= acceleration = distance/time squared = 9.81 meters/second squared at sea level Therefore, F=m*a = 90 x 9.81. Thatīs force. it's true that f=ma for a static load, so in that case he's correct, but not in a situation where the climber is falling . . . so fair enough, I rushed to judgement, but I was thinking about a dynamic situation. In my defense, you'll see that a number of other people also corrected woj12_16's post.





sbaclimber
Feb 12, 2006, 6:19 PM
Post #27 of 63
(6585 views)
Shortcut
Registered: Jan 21, 2004
Posts: 3116

In reply to: In reply to: In reply to: . . . Acceleration is 9.81 meters/second squared. That means that force you'll create falling through the air is less than 900 newtons, so your caribeaner thats rated for 25000 newtons would surely hold you. And that's pretty much all you need to know. Ummm . . . no. My physics is pretty rusty by I know you can't just multiply 90 kg by 9.8 and says that's the force. Ummm . . . yeah. My physics is pretty rusty too, but I do remember that: F= force = ? m= mass = 90kilograms a= acceleration = distance/time squared = 9.81 meters/second squared at sea level Therefore, F=m*a = 90 x 9.81. Thatīs force. I am definitely jumping on the 'you have no idea what you are talking about' band wagon!! As curt was hinting at..... 'a', as in acceleration (distance/time^2), is NOT gravity (9.81m/s^2, what you are assuming) when you are talking about catching a falling climber :!: :!: It is the decceleration caused by the rope catching you (or if you will, NEGATIVE acceleration). If you are are 80kg, and travelling at 9.8m/s (you have been falling for 1 second), and the rope catches you.......... (assume it takes 1/2second for you to come to rest): F = 80kg x ()19.62m/s^2 = 1569.6N = ~1.6kN Very rough calculation, please someone correct me if I am wrong! The time to come to rest was a complete guess, I have no idea how long it would take (it would have a lot to do with the rope, I think).





cully_larson
Feb 12, 2006, 6:20 PM
Post #28 of 63
(6585 views)
Shortcut
Registered: Aug 19, 2005
Posts: 28

Where did some of you learn physics. It's almost scary. First, a pound IS a measure of force. It is equal to 1 slug x ft/s^2. If you're on the moon, your pound "weight" will be different than it is on earth (force changes when gravity changes). Your mass is the same on earth as it is on the moon. So, your weight in slugs (or, even better, grams) will remain the same. Second, F=ma will only give you the force applied by gravity. This will work fine if you want to determine how much force you exert when you hang on a rope. However, if you want to compute the force you exert when you fall, you also have to factor in the force of momentum (because of Newton's first law of motion). A force must be exerted to change the speed of a moving body (pun intended). So, it's a little more complicated to compute the force applied to a piece of gear (pro, rope, 'biner, etc.) when something falls on it. First, momentum is equal to mass times velocity (p = mv). The net force you apply when your fall is arrested is equal to change in momentum divided by change in time (F = dp/dt). I'll see if I can do this w/o doing an integral. So, let's say you weigh 150lbs on earth (68 kg), you fall for one second, and your fall is arrested over half a second. I'm not sure if these numbers are correct, but it doesn't really matter. So, you have 1 second to accelerate to a velocity. Your velocity before your fall is arrested is given by vf: vf = vi + at. Your initial velocity is 0, so your final velocity is: 9.8 x 1 = 9.8 meters/second (21 miles/hour). Your initial momentum is actually 0 (your aren't moving) and your final momentum is: 68 x 9.8 = 666.7 kg m/s. So, the total force you apply as your fall is arrested is: 666.7 / .5 = 1333.4 newtons. If we had used the old F=ma equation, we would have come up with 666.7 newtons (which would be our answer if the fall was arrested over 1 second, but I doubt your fall will ever be arrested over the same time period as your are falling). This also shows us why dynamic ropes create less impact force (the more time you have to decelerate (the bigger the denominator in our equation), the less the force). This number probably isn't all that accurate. You have wind resistance. Also, as your fall is arrested, you are still being accelerated by gravity. Also, it would be hard to determine how much force is applied to any one piece of gear during this fall. Also, this is the net force applied in arresting your fall. The max. force applied at any one time during the deceleration of the body would be much less. I hope that helps. Also, I'm sorry if any of the above is incorrect. It's been a long time since I've been in a physics class.





sbaclimber
Feb 12, 2006, 6:23 PM
Post #29 of 63
(6585 views)
Shortcut
Registered: Jan 21, 2004
Posts: 3116

In reply to: Also, I'm sorry if any of the above is incorrect. It's been a long time since I've been in a physics class. :lol: Well, that makes two of us, but it looks like we were both on the same page (and our numbers aren't too far apart either). PS, oh, and a cubit is the distance from the elbow to the tip of the middle finger on an 'average' person.
In reply to: This natural cubit measures 24 digits or 6 palms or 1― foot. This is about 45 cm or 18 inches.





daithi
Feb 12, 2006, 6:41 PM
Post #30 of 63
(6585 views)
Shortcut
Registered: Jul 6, 2005
Posts: 397

In reply to: If you're on the moon, your pound "weight" will be different than it is on earth (force changes when gravity changes). Your mass is the same on earth as it is on the moon. So, your weight in slugs (or, even better, grams) will remain the same. Crystal clear! :roll: Little gems like this really help to clarify things. Weight is a force measured in Newtons, mass is a proprty of a body measured in kg. The two are related through Newton's 2nd law, F = ma. A slug is a gastropod and a pound is a unit of currency!





curt
Feb 12, 2006, 6:49 PM
Post #31 of 63
(6585 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18271

In reply to: In reply to: If you're on the moon, your pound "weight" will be different than it is on earth (force changes when gravity changes). Your mass is the same on earth as it is on the moon. So, your weight in slugs (or, even better, grams) will remain the same. Crystal clear! :roll: Little gems like this really help to clarify things. Weight is a force measured in Newtons, mass is a proprty of a body measured in kg. The two are related through Newton's 2nd law, F = ma. A slug is a gastropod and a pound is a unit of currency! But waitthere's more. A slug is also a lead projectile fired from a shotgunand describes when you smack someone with your fist. A pound is where stray dogs goand also describes what you do to properly tenderize abalone. :D Curt





daithi
Feb 12, 2006, 6:54 PM
Post #32 of 63
(6585 views)
Shortcut
Registered: Jul 6, 2005
Posts: 397

In reply to: Also, this is the net force applied in arresting your fall. The max. force applied at any one time during the deceleration of the body would be much less. Since you assume a linear deceleration your force of 1333.4 N is a constant and applied over the time interval of 0.5 s. In reality the deceleration force is a function of time.





curt
Feb 12, 2006, 7:03 PM
Post #33 of 63
(6585 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18271

In reply to: In reply to: Also, this is the net force applied in arresting your fall. The max. force applied at any one time during the deceleration of the body would be much less. Since you assume a linear deceleration your force of 1333.4 N is a constant and applied over the time interval of 0.5 s. In reality the deceleration force is a function of time. Or distance. Curt





cully_larson
Feb 12, 2006, 7:07 PM
Post #34 of 63
(6585 views)
Shortcut
Registered: Aug 19, 2005
Posts: 28

In reply to: In reply to: Also, this is the net force applied in arresting your fall. The max. force applied at any one time during the deceleration of the body would be much less. Since you assume a linear deceleration your force of 1333.4 N is a constant and applied over the time interval of 0.5 s. In reality the deceleration force is a function of time. Well, that's the total force applied over the entire 0.5s. At any one time during that deceleration, the force will not be 1333.4 N, even with linear deceleration. Imagine that instead of a rope decelerating me, I have a rocket engine. Let's say it can exert a constant 700 N of force. This will give us linear deceleration. Over a period of time I will slow down and eventually come to a stop (I don't want to take the time to solve the equation for t, so I don't know how long this will take). At any one time, the force exerted on me is 700N. However, using the equation I used before, I will end up with a net force exerted that is larger than 700 N.





daithi
Feb 12, 2006, 7:08 PM
Post #35 of 63
(6585 views)
Shortcut
Registered: Jul 6, 2005
Posts: 397

In reply to: In reply to: In reply to: Also, this is the net force applied in arresting your fall. The max. force applied at any one time during the deceleration of the body would be much less. Since you assume a linear deceleration your force of 1333.4 N is a constant and applied over the time interval of 0.5 s. In reality the deceleration force is a function of time. Or distance. Curt Exactly. When you actually take into account the elasticity of the rope and try and derive a differential equation for the deceleration force, it is written in terms of displacement (and its derivatives) not time. It should be added of course it could be written in terms of time it's just not convenient.





squierbypetzl
Moderator
Feb 12, 2006, 7:22 PM
Post #36 of 63
(6585 views)
Shortcut
Registered: Jul 6, 2005
Posts: 3431

In reply to: In reply to: In reply to: In reply to: . . . Acceleration is 9.81 meters/second squared. That means that force you'll create falling through the air is less than 900 newtons, so your caribeaner thats rated for 25000 newtons would surely hold you. And that's pretty much all you need to know. Ummm . . . no. My physics is pretty rusty by I know you can't just multiply 90 kg by 9.8 and says that's the force. Ummm . . . yeah. My physics is pretty rusty too, but I do remember that: F= force = ? m= mass = 90kilograms a= acceleration = distance/time squared = 9.81 meters/second squared at sea level Therefore, F=m*a = 90 x 9.81. Thatīs force. I am definitely jumping on the 'you have no idea what you are talking about' band wagon!! As curt was hinting at..... 'a', as in acceleration (distance/time^2), is NOT gravity (9.81m/s^2, what you are assuming) when you are talking about catching a falling climber :!: :!: It is the decceleration caused by the rope catching you (or if you will, NEGATIVE acceleration). If you are are 80kg, and travelling at 9.8m/s (you have been falling for 1 second), and the rope catches you.......... (assume it takes 1/2second for you to come to rest): F = 80kg x ()19.62m/s^2 = 1569.6N = ~1.6kN Very rough calculation, please someone correct me if I am wrong! The time to come to rest was a complete guess, I have no idea how long it would take (it would have a lot to do with the rope, I think). Dude, simple, is mass x acceleration = to force? Yes? I never mentioned anything about a falling climber. I was just saying it is force. :roll:





schnoz
Feb 12, 2006, 7:23 PM
Post #37 of 63
(6585 views)
Shortcut
Registered: Jul 19, 2003
Posts: 83

Oi. How many threads are there on rc.com about this? Weight and mass are not one and the same. An object always has the same mass, regarldess of it's acceleration. An object does not always have the same weight. To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Now, let's use an example of a shotgun slug (since someone decided to bring slugs into this). If I removed a slug from the round and tossed it to you, it wouldn't hurt. It wouldn't pass through your hand or do any other damage to you. If I throw it at you really hard, it might sting a bit. Now, if I shoot at at you and you try and catch it, it's going to hurt. A lot. Why? The slug still has the same mass in all three examples. The outcome is different in all three because, as stated before, F = ma. That is, Force is equal to mass times acceleration. A bullet is constantly accelerating as it leaves the barrel of the gun, it's just not accelerating at a constant rate, until it comes to rest by striking something. How does this translate into the climbing world? I believe the best article to read regarding that (from my searching for reference data and double checking some figures) is one published by PMI. It's located at http://www.safeclimbing.org/..._Climbers_Beware.pdf. It's also an excellent article explaining how ropes are tested and why. As an aside, did anyone else know that PMI stood for Pigeon Mountain Industries? No wonder they go by PMI all the time!





sbaclimber
Feb 12, 2006, 7:33 PM
Post #38 of 63
(6585 views)
Shortcut
Registered: Jan 21, 2004
Posts: 3116

In reply to: Dude, simple, is mass x acceleration = to force? Yes? I never mentioned anything about a falling climber. I was just saying it is force. :roll: Okay, fair enough you did not mention a falling climber. woj12_16, on the other hand, did, and you quoted him :!: I will leave it up to you to decide who my comment was aimed at :wink: Edit: actually, let me clarify: woj12_16 said something stupid > someone said NO, that was wrong > you said, YES. I thought that mean that you were agreeing with woj12_16. My missunderstanding apparently :roll:





daithi
Feb 12, 2006, 7:33 PM
Post #39 of 63
(6585 views)
Shortcut
Registered: Jul 6, 2005
Posts: 397

In reply to: Let's say it can exert a constant 700 N of force. You have given the answer of the force exerted on the body at any given time before any equations are needed! It will be a constant of 700 N.
In reply to: However, using the equation I used before, I will end up with a net force exerted that is larger than 700 N Are you confusing force with momentum? Momentum is the integral of force wrt time. Plot the velocity, acceleration, force and momentum wrt time for this problem. The momentum isn't constant but the force is.





daithi
Feb 12, 2006, 7:37 PM
Post #40 of 63
(6585 views)
Shortcut
Registered: Jul 6, 2005
Posts: 397

In reply to: To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Do they just not teach you about the metric system over there? Please stop this. It is making me cry! :(





cully_larson
Feb 12, 2006, 7:38 PM
Post #41 of 63
(6586 views)
Shortcut
Registered: Aug 19, 2005
Posts: 28

In the interest of science, there are a few things wrong with what you just posted.
In reply to: To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Mass is measured in grams, not Newtons. So, if something has a mass of 98 kg on the earth, it has a mass of 98kg on the moon, or in space, or wherever. It's the 980 newtons of FORCE that changes.
In reply to: Now, let's use an example of a shotgun slug (since someone decided to bring slugs into this). If I removed a slug from the round and tossed it to you, it wouldn't hurt. It wouldn't pass through your hand or do any other damage to you. If I throw it at you really hard, it might sting a bit. Now, if I shoot at at you and you try and catch it, it's going to hurt. A lot. Why? The slug still has the same mass in all three examples. The outcome is different in all three because, as stated before, F = ma. That is, Force is equal to mass times acceleration. A bullet is constantly accelerating as it leaves the barrel of the gun, it's just not accelerating at a constant rate, until it comes to rest by striking something. A bullet, at least in a vacuum, does not accelerate in the x direction after it leaves the barrel. If not in a vacuum, it accelerates negatively (or decelerates) due to wind resistence. When it comes to a stop by striking something, it does not exert a force of F=ma upon that object. It actually exerts a force equal to the change in momentum over time caused by the striking of the object. If it were true that it exerts a force of F=ma, then it wouldn't matter how far a climber falls before her fall is arrested, she will still exert the same force. This isn't the case.





schnoz
Feb 12, 2006, 7:43 PM
Post #42 of 63
(6586 views)
Shortcut
Registered: Jul 19, 2003
Posts: 83

In reply to: In reply to: To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Do they just not teach you about the metric system over there? Please stop this. It is making me cry! :( I fail to see what's wrong with what you quoted me on. Just how is it that 100 kg * 9.8 isn't equal to 980? Nor do I see where the second part of what you quoted me is wrong on either.





daithi
Feb 12, 2006, 7:44 PM
Post #43 of 63
(6586 views)
Shortcut
Registered: Jul 6, 2005
Posts: 397

In reply to: In reply to: In reply to: To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Do they just not teach you about the metric system over there? Please stop this. It is making me cry! :( I fail to see what's wrong with what you quoted me on. Just how is it that 100 kg * 9.8 isn't equal to 980? Nor do I see where the second part of what you quoted me is wrong on either. For fuck's sake, I even highlighted the incorrect bits in italics for you! :roll:





schnoz
Feb 12, 2006, 7:48 PM
Post #44 of 63
(6586 views)
Shortcut
Registered: Jul 19, 2003
Posts: 83

In reply to: In reply to: To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Mass is measured in grams, not Newtons. So, if something has a mass of 98 kg on the earth, it has a mass of 98kg on the moon, or in space, or wherever. It's the 980 newtons of FORCE that changes. It's been quite some time since high school physics. You're quite right, I had that part backwards.
In reply to: In reply to: Now, let's use an example of a shotgun slug (since someone decided to bring slugs into this). If I removed a slug from the round and tossed it to you, it wouldn't hurt. It wouldn't pass through your hand or do any other damage to you. If I throw it at you really hard, it might sting a bit. Now, if I shoot at at you and you try and catch it, it's going to hurt. A lot. Why? The slug still has the same mass in all three examples. The outcome is different in all three because, as stated before, F = ma. That is, Force is equal to mass times acceleration. A bullet is constantly accelerating as it leaves the barrel of the gun, it's just not accelerating at a constant rate, until it comes to rest by striking something. A bullet, at least in a vacuum, does not accelerate in the x direction after it leaves the barrel. If not in a vacuum, it accelerates negatively (or decelerates) due to wind resistence. When it comes to a stop by striking something, it does not exert a force of F=ma upon that object. It actually exerts a force equal to the change in momentum over time caused by the striking of the object. If it were true that it exerts a force of F=ma, then it wouldn't matter how far a climber falls before her fall is arrested, she will still exert the same force. This isn't the case. I had a much longer post composed to address wind resistance, etc. However, I deleted most of it. I was trying to relate a bullet to rock climbing and a rock climber never reached terminal velocity while roped up. Not with a standard 60m rope, anyway. As for length of the fall, that doesn't matter to a climber either. It's the fall factor that matters. It's due to the FF mattering over the length of the fall that I deleted the whole bit about wind resitance and trying to go into forces a bit more. In retrospect, I should have just not said anything. There's already a whack of threads explaining what a Newton is already on rc.com. It probably would have been better to just link to them :)





curt
Feb 12, 2006, 8:22 PM
Post #45 of 63
(6586 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18271

In reply to: In reply to: To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Do they just not teach you about the metric system over there? Please stop this. It is making me cry! :( No, they do. It's just that there are retards everywhere. Curt





curt
Feb 12, 2006, 8:34 PM
Post #46 of 63
(6586 views)
Shortcut
Registered: Aug 26, 2002
Posts: 18271

In reply to: In reply to: In reply to: To say that an object with a mass of 980 Newtons weighs 100 kg is true iff that object is at rest on Earth's surface. Another example is to say that an astronaut can step on a scale here on Earth where gravity is 9.8 m/s^2 and weighs 98 kg. That same astronaut with the same mass stepping on the same scale on the Earth's moon where gravity is 1.6 m/s^2 would weigh only 16 kg. Do they just not teach you about the metric system over there? Please stop this. It is making me cry! :( I fail to see what's wrong with what you quoted me on. Just how is it that 100 kg * 9.8 isn't equal to 980? Nor do I see where the second part of what you quoted me is wrong on either. I'll field this one. Nothing "weighs" 98kg versus 16kg, dependent on the acceleration caused by local gravity. Mass is constant irrespective of the acceleration due to gravityand is not a unit of weight at all. Got it yet? Curt





sactownclimber
Feb 12, 2006, 9:43 PM
Post #47 of 63
(6586 views)
Shortcut
Registered: Jan 1, 2005
Posts: 216

In reply to: In reply to: In reply to: In reply to: In reply to: . . . Acceleration is 9.81 meters/second squared. That means that force you'll create falling through the air is less than 900 newtons, so your caribeaner thats rated for 25000 newtons would surely hold you. And that's pretty much all you need to know. Ummm . . . no. My physics is pretty rusty by I know you can't just multiply 90 kg by 9.8 and says that's the force. Ummm . . . yeah. My physics is pretty rusty too, but I do remember that: F= force = ? m= mass = 90kilograms a= acceleration = distance/time squared = 9.81 meters/second squared at sea level Therefore, F=m*a = 90 x 9.81. Thatīs force. I am definitely jumping on the 'you have no idea what you are talking about' band wagon!! As curt was hinting at..... 'a', as in acceleration (distance/time^2), is NOT gravity (9.81m/s^2, what you are assuming) when you are talking about catching a falling climber :!: :!: It is the decceleration caused by the rope catching you (or if you will, NEGATIVE acceleration). If you are are 80kg, and travelling at 9.8m/s (you have been falling for 1 second), and the rope catches you.......... (assume it takes 1/2second for you to come to rest): F = 80kg x ()19.62m/s^2 = 1569.6N = ~1.6kN Very rough calculation, please someone correct me if I am wrong! The time to come to rest was a complete guess, I have no idea how long it would take (it would have a lot to do with the rope, I think). Dude, simple, is mass x acceleration = to force? Yes? I never mentioned anything about a falling climber. I was just saying it is force. :roll: sbaclimbr thanks for the backup, I knew I was right . . . for the record:
In reply to: . . . Acceleration is 9.81 meters/second squared. That means that force you'll create falling through the air is less than 900 newtons, so your caribeaner thats rated for 25000 newtons would surely hold you. And that's pretty much all you need to know. which is NOT correct. To clarify: m*g is the force on an object due to gravity, whether falling or static, and when that falling object does come to a stop (when the rope catches you) m*g is NOT the force felt by carabiner. Edited to more precisely spell out what I was trying to say . . .





daithi
Feb 12, 2006, 10:12 PM
Post #48 of 63
(6586 views)
Shortcut
Registered: Jul 6, 2005
Posts: 397

In reply to: for the record: In reply to: . . . Acceleration is 9.81 meters/second squared. That means that force you'll create falling through the air is less than 900 newtons,.... which is NOT correct. As previously stated, m*g is the force due to an object at rest. Incroyable! Actually the force acting on a body that is free falling is also m*g (ignoring aero drag of course). Unless you are propelling yourself downwards there is no other force acting on you. The high forces that your rope etc. are rated for arise when you attempt to arrest this falling motion. Even by rc.com standards, the number of erroneous statements on this thread beggars believe. What compounds the disbelieve is the simplicity of the subject at hand! Mods please lock this thread before my head explodes! :)





blondgecko
Moderator
Feb 12, 2006, 11:30 PM
Post #49 of 63
(6586 views)
Shortcut
Registered: Jul 2, 2004
Posts: 7666

I have to get off this thread before I perform a little experiment in acceleration by hurling my computer out the window. :evil:





jarin_italy
Feb 13, 2006, 12:16 AM
Post #50 of 63
(6586 views)
Shortcut
Registered: Sep 17, 2003
Posts: 2

1KN=1000N=101.9716KG





