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Is body weight equivalent to bicycle weight?
On Fri, 22 Jul 2005 19:43:48 GMT, "Bruce W.1"
wrote: Is body weight equivalent to bicycle weight? In other words, would riding a bicycle that's five pounds lighter be the same as losing five pounds off of your body weight? There's an old Army saying; one pound on your foot (boot weight) is equivalent to five pounds on your back. But I'm not sure what this has to do with anything. opinion: if you're talking about lard around your waist etc. then I'll vote taking it off your body is going to have greater benefits. |
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#12
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Is body weight equivalent to bicycle weight?
Arthur Harris wrote:
...And yes, losing 5 pounds off your body is the same as getting a 5 pound lighter bike (but a lot cheaper!). Yeahbutt... The /bike/ won't gain it back! Meats 'n Cheeses Bill |
#13
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Is body weight equivalent to bicycle weight?
Arthur Harris wrote:
"Eric Hill" wrote: wle wrote: there may be some difference between rotating weight and non-rotating. i e, wheels may matter more because they accelerate differently. The rule of thumb that I've been told goes, "an ounce off the wheels equals a pound off the frame." Oh, it's getting even better! The old saying used to be: An ounce off the wheels equals two ounces off the frame. But it really isn't. For reasonable wheels and tires, and typical accelerations, an ounce is an ounce regardless where it is. And yes, losing 5 pounds off your body is the same as getting a 5 pound lighter bike (but a lot cheaper!). Art Harris Art, I was about to make a post disagreeing with you, but thinking about it, you're more right than you give yourself credit for. A rotating object's resistance to angular acceleration is called it's moment of inertia (I). Whe I=kmrČ with k=some constant dependant on the distribution of the mass of the object m=mass r=radius of the object Since the radii of wheels are pretty set now, you'll have to change the mass distribution to make the wheels any easier to turn. Thanks for making me revisit my freshman physics class! -- Paul M. Hobson Georgia Institute of Technology http://www.underthecouch.org ..:you may want to fix my email address before you send anything:. |
#14
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Is body weight equivalent to bicycle weight?
Ron Ruff wrote:
Bruce W.1 wrote: Is body weight equivalent to bicycle weight? In other words, would riding a bicycle that's five pounds lighter be the same as losing five pounds off of your body weight? The only time when a part of the bike weight is more significant is when accelerating. Then the weight of rims and tires (and a fraction of the spoke weight) is twice the effect of weight elsewhere. I just don't see how this can be so. On what physical principles are you basing this? snip -Ron -- Paul M. Hobson Georgia Institute of Technology http://www.underthecouch.org ..:I go to GT, I like numbers:. |
#15
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Is body weight equivalent to bicycle weight?
On Fri, 22 Jul 2005 19:43:48 GMT, "Bruce W.1"
wrote: Is body weight equivalent to bicycle weight? In other words, would riding a bicycle that's five pounds lighter be the same as losing five pounds off of your body weight? There's an old Army saying; one pound on your foot (boot weight) is equivalent to five pounds on your back. But I'm not sure what this has to do with anything. Thanks for your help. Dear Bruce, The army saying illustrates the principle. When we walk, weight on our back moves along at a steady pace, much like our bicycle frames doing the same thing when we ride. But weight on our feet must be swung forward by our legs, somewhat like the wheels on our bikes, which must spin as well as move forward. With every step, the foot hits the ground, comes to a halt, trails behind and provides a forward push, and is then whipped forward (and upward) again at roughly twice the walker's speed for the next step. Add a pound or two on the end of your hind leg and the repeated violent acceleration and raising becomes much more difficult. Weight added to the hands would be just as obnoxious, since the hands are constantly waving back and forth during walking and running. (If you're not sure about the odd business of the foot coming to a halt with every step, even at Olympic sprinting speeds, step through a puddle and then look back at your wet footprints. They show that your foot never slipped when it was touching the ground, just as the wet track left by your rubber tire will show that your tire was not slipping when it touched the pavement at 20 mph. If the feet and the tire weren't slipping against the ground, then they must have been equally motionless.) Rotating mass like the wheel (and the crank) adds an extra effort. Consider a bicycle sliding across the usual imaginary frictionless physics pond with its wheels locked. Everything--including the wheels--is going 20 mph across the pond and therefore required X amount of force to reach that speed from a standing start. To spin the wheels up to 20 mph would require additional force. But this extra force required for rotating isn't much. Flip a bike upside down, crank one pedal in high gear, and you can spin the heavier rear wheel up to 30 mph in a few moments with a couple of quick heaves of one feeble arm. The other arm would serve to spin the front wheel up to the same speed, if we could attach a crank and chain to it. Another way to judge the effort involved is to remember that no rider can spin a real bike up from 0 to 30 mph in under five seconds on the flats. Now take this already minor effort and reduce it to just the difference between a light and and a heavy wheelset. It could still be the difference between winning and losing in a sprint, where photo finishes can be required: http://www.velonews.com/images/details/8436.11841.f.jpg But it makes scarcely any difference to the rest of us. Much of the claimed difference in feel between lighter and heavier equipment is simply expectation--when something costs a thousand dollars more, we tend to feel a difference, whether it's there or not. Two common scenarios are often raised with the minor difference between light and heavy wheelsets and lighter and heavier bikes. First, the same force accelerates a larger mass more slowly, so the heavier bike should be slower to reach the same cruising speed on the level (and even slower if the extra mass must also be spun up like a wheel or crank). But the masses involved are trivial to start with. Typically, riders are talking about pound or two less weight, which sounds impressive on a sub-20 pound bicycle because the weight is being reduced 5-10%. However, the mass being accelerated includes a roughly 150 pound engine--heavy bike and rider together will weigh 170 pounds, while the 2-pound lighter bike will weigh 168 pounds, only 1.2% less. On the flat, that means only that the rider will accelerate to the same cruising speed 1.2% more slowly--and acceleration to cruising speed is a very small part of a typical ride. After that, we mostly waver at a snail's pace, accelerating and decelerating slightly. Some people want to argue that the constant tiny accelerations must add up, but this really isn't the case--the constant tiny accelerations are pretty much balanced by the constant tiny decelerations, where the heavier bike slows down more slowly than the lighter bike that sped up more quickly. (Any long initial acceleration does tend to be unbalanced because normally we throw away most of our hard-earned momentum by touching our brakes. But that initial acceleration was still only at a 1.2% disadvantage for about 30 seconds.) So the acceleration difference on the flats just isn't going to show up for typical small weight savings on ordinary rides. The second scenario is the dreadful task of hauling the extra weight uphill. Again, the problem is that people think of the extra weight in terms of the bike, not the total mass that must be pedalled up the hill--the difference is a 168 pound mass versus a 170 pound mass. Here's a calculator for speed: http://www.kreuzotter.de/english/espeed.htm Use hands-on-tops, 150-lb rider, 20-lb bike, 7% slope, and 10 mph. It predicts 287 watts are needed. Change to an 18-lb bike, blank the watts, re-calculate, and the prediction is that 284 watts are needed. Change to a 15-lb bike, blank the watts, do it again for 10 mph, and prediction is 279 watts. (Remember, after the initial acceleration to cruising speed, it doesn't matter whether the weight was lost from the rotating wheels or from the linear-motion-only frame.) Not only is the slow-down effect of the extra 2 pounds small (you're just pushing a slightly heavier weight up the same surprisingly gentle 7% slope), but it's somewhat balanced by the speed gained when you descend: http://www.kreuzotter.de/english/espeed.htm Put the same 150-lb rider on a 20-lb bike, put his hands on the drops, set cadence and watts to 0, and roll him down a -7% slope. The prediction is 39.5 mph. Less mass fighting the same wind drag lowers terminal velocity, so it drops to 39.3 mph for an 18-lb bike. And to 38.9 mph for a 15-lb bike. So when you roll back down the mountain, you gain back some of what you lost raising that extra pound or two. In short, typical bicycle weight differences matter only for accelerations, not cruising speeds on the flats. And the differences are so small that they don't matter much even if magnified by rotation. http://www.kreuzotter.de/english/espeed.htm Use hands-on-tops, set the power to 200 watts, the rider to 150 lbs and the bike to 20 lbs, and send him up a 7% grade for 10 miles: http://www.kreuzotter.de/english/espeed.htm The prediction is 83 minutes, 20.0 seconds at 7.2 mph. Reduce the bike weight by 2 pounds, and the time for a 10-mile 7% climb drops to 82 minutes, 11.5 seconds, a full 68.5 seconds faster at 7.3 mph. A 68.5 second advantage in an hour and twenty minutes matters to a racer, but the rest of us aren't likely to notice it. At 7.3 mph, it works out to about a 730-foot lead after ten miles uphill. For fun, recalculate with a 1 mph headwind--the 2-lb lighter bike is right back to 83 minutes, 20.0 seconds. Carl Fogel |
#16
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Is body weight equivalent to bicycle weight?
Paul Hobson writes:
A rotating object's resistance to angular acceleration is called it's moment of inertia (I). Whe I=kmrČ with k=some constant dependant on the distribution of the mass of the object m=mass r=radius of the object Since the radii of wheels are pretty set now, you'll have to change the mass distribution to make the wheels any easier to turn. True, but you need to continue the analysis a bit further. Increasing the angular velocity (w) requires a torque (T). The torque is applied by a force at the axle, the lever arm is r. So the additional force that the rider must apply is f = (I*w')/r = I*(v'/r)/r = I*v'/r^2 = k*m*v' The wheel radius completely drops out. The force required to accelerate the wheel is m*v', so the total force is ftot = (1+k)*m*v' If all the mass of the wheel is at the rim (worst case), then k=1 and we get the factor of 2 previously mentioned. Note that small wheels don't accelerate any faster unless they weigh less. Joe |
#17
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Is body weight equivalent to bicycle weight?
Paul Hobson wrote:
Ron Ruff wrote: The only time when a part of the bike weight is more significant is when accelerating. Then the weight of rims and tires (and a fraction of the spoke weight) is twice the effect of weight elsewhere. I just don't see how this can be so. On what physical principles are you basing this? The wheels must be accelerated rotationally, as well as linearly (along with the rest of the bike/rider). The extra force required to "spin up" the tire and rim is the same as the force required to accelerate it linearly, since the radius times the angular speed is equal to the linear speed. So for acceleration, the rim and tire are double the importance of weight elsewhere. Extra energy is needed to "spin up" all rotating parts on a bike, but angular speed x radius x mass of other parts (like hubs and pedals) is small compared to the rims and tires. -Ron |
#18
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Is body weight equivalent to bicycle weight?
"Ron Ruff" wrote:
Paul Hobson wrote: Ron Ruff wrote: The only time when a part of the bike weight is more significant is when accelerating. Then the weight of rims and tires (and a fraction of the spoke weight) is twice the effect of weight elsewhere. I just don't see how this can be so. On what physical principles are you basing this? The wheels must be accelerated rotationally, as well as linearly (along with the rest of the bike/rider). The extra force required to "spin up" the tire and rim is the same as the force required to accelerate it linearly, since the radius times the angular speed is equal to the linear speed. So for acceleration, the rim and tire are double the importance of weight elsewhere. Thing is, since you can give a loose wheel a spin with a finger and easily get it up to "sprinting speed", it's pretty clear that this "acceleration tax" is minimal. Not only that, but unless you come to a stop, you really don't pay the "tax" again, since any energy you put into the wheel via acceleration is there to prevent DEceleration (think of it as a flywheel and it becomes more obvious). This is certainly the case when you're JRA. Mark Hickey Habanero Cycles http://www.habcycles.com Home of the $695 ti frame |
#19
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Is body weight equivalent to bicycle weight?
On Fri, 22 Jul 2005 19:56:21 +0000, Eric Hill wrote:
wle wrote: there may be some difference between rotating weight and non-rotating. i e, wheels may matter more because they accelerate differently. The rule of thumb that I've been told goes, "an ounce off the wheels equals a pound off the frame." Yeah, I remember that, too. It's nonsense, but I do remember being told it. -- David L. Johnson __o | Do not worry about your difficulties in mathematics, I can _`\(,_ | assure you that mine are all greater. -- A. Einstein (_)/ (_) | |
#20
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Is body weight equivalent to bicycle weight?
Joe Riel wrote:
Paul Hobson writes: A rotating object's resistance to angular acceleration is called it's moment of inertia (I). Whe I=kmrČ with k=some constant dependant on the distribution of the mass of the object m=mass r=radius of the object Since the radii of wheels are pretty set now, you'll have to change the mass distribution to make the wheels any easier to turn. True, but you need to continue the analysis a bit further. Increasing the angular velocity (w) requires a torque (T). The torque is applied by a force at the axle, the lever arm is r. So the additional force that the rider must apply is f = (I*w')/r = I*(v'/r)/r = I*v'/r^2 = k*m*v' The wheel radius completely drops out. The force required to accelerate the wheel is m*v', so the total force is ftot = (1+k)*m*v' If all the mass of the wheel is at the rim (worst case), then k=1 and we get the factor of 2 previously mentioned. Note that small wheels don't accelerate any faster unless they weigh less. Joe I'm 98% sure you just took me to school...but I'm gonna have to think about it tomorrow morning on my weekly ride to the bank. -- Paul M. Hobson Georgia Institute of Technology http://www.underthecouch.org ..:you may want to fix my email address before you send anything:. |
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