Archive Home >> General(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 )

Hey, Grzy, larn me physics. And make it stick this time.(47 posts)

 Hey, Grzy, larn me physics. And make it stick this time. billSep 19, 2001 7:02 AM I was reviewing the thread on cornering, and you posted some formulas about how mass doesn't matter to downhill velocity. Or at least I think that's what you did. But, it has to matter. Far too often, the fat guy you passed on the way up passes you on the way down (and, far more often, the skinny guy that passed you on the way up you pass on the way down, thinking, I knew those love handles were good for something). I know that in a vaccuum, the bowling ball and the beach ball will fall at the same velocity, but not in air. So, what's the answer and why? You make the point that the mass is greater for the fat guy relative to his frontal exposure, with which I certainly agree (and I think that a fat guy's aerodynamic resistance is going to be the same as a skinny guy's, if not greater) but then you took us (well, me; maybe us, maybe no one) through a formula negating mass as a factor. So, what gives? BTW, I took a semester or two of physics in college, but that was my sophomore year. I really still should be in therapy over my sophomore year.
 simple concept DogSep 19, 2001 7:09 AM all about your mass to (wind) drag ratio. The higher the mass compared to drag, the faster you go. If drag is zero (vacuum), all fall the same. Doug
 Well, that's what I thought, but Grzy seemed to be saying billSep 19, 2001 7:39 AM else. What, I cannot explain.
 speculation about explanation DogSep 19, 2001 7:45 AM I think it got hung up on that "free fall" concept. I think that he was saying that, forgetting air drag, a bike is a free falling object, even if not falling straight down, but even on an incline. It's free falling in the sense that gravity alone is propelling it. But, it is not free falling if there is air drag. I think that's what was being said. In any event, it doesn't matter. We know what happens. Doug
 speculation about explanation SteeveSep 19, 2001 8:25 AM Dog, your above statements are true. The readers should also note that there are other forces that retard acceleration down a hill (such as bearing frictional forces and the rolling frictional forces due to the tires). These frictional forces are directly proportional to the weight of the bike and rider. We should note that weight is a force, mass is not a force. Weight is the force an object exerts due to it's mass and gravity. All frictional forces (including the afore mentioned air resistance/drag) retard acceleration. The acceleration of an object down an incline (without any fricional forces) is the acceleration due to free-fall multiplied by the sine of the hill's angle to the horizon. Examples: The "hill" is straight up, the angle is 90deg, the sine of 90deg is 1, thus the object accelerates at 1 X freefall. The "hill" is 45deg, the sine of 45deg is .707, thus the object accelerates at .707 X freefall. It is interesting to note that often a heavier rider will coast farther on level ground after the hill ends due to his greater kinetic (moving) energy.
 So, what is the relationship of weight to velocity? Because of billSep 19, 2001 8:51 AM the variables you mention, I don't believe that any of them (other than air friction as a proportion of mass) matter to the relative speed of a heavier rider and a lighter rider down an incline. The drag force of tire deflection and hub friction may exist, but, if nothing else, because they would have to exist (as a negative vector against the force of gravity) in direct proportion to the rider's weight, they certainly can't account for the higher resulting velocity of the heavier rider. Even air friction, to the extent that it's different for the two riders of different mass (and I doubt that it is significantly different for the two riders) if anything would be greater for the more massive rider (the bigger dude). I get that gravity exerts a greater force on an object of greater mass. Sure. That these more massive objects then possess a greater kinetic energy, I get. Sure. The other side of the equation, though, is that objects of greater mass resist movement more than objects of lesser mass. That's the whole Galileo thing. So, if you have two objects of different mass, beginning at equal velocity, one having greater potential energy than the other, all other opposing forces being equal to or even greater for the more massive rider, how does the one with greater potential energy get to the greater velocity?
 Nope grzySep 19, 2001 9:24 AM Dog got it right and used far fewer words (of course). See if my note below helps. Otherwise just remember to eat a big breakfast and tuck when descending!
 Ah, Grasshopper grzySep 19, 2001 9:22 AM You're pretty much on the right track. Problem is when people try to simplify the real world down to something that can be analyzed without a wind tunnel. The part that I showed where the mass didn't matter was to counter the half developed arguement by a previous poster who was attempting to show that the heavier (actually more "massive") rider had more Potential Energy (PE = mgh) which is true. However this isn't the reason why said heavier rider would be going faster at the bottom of a hill. To complete the analysis in the simplified approach one uses a technique called an "energy ballance" - that's just a fancy way of saying all of the energy in a system has to be accounted for. The classical first year physics (Newtonian Mechanics) takes all of the Potential Energy and converts it into Kinetic Energy (KE = 1/2mv^2). the poster claimed that since the heavier rider had more PE that they would then be going faster at the bottom of the hill. I showed that while, yes the heavier rider would have more KE at the bottom of the hill, they'd still have the same velocity since it's purely a function of the height of the hill and not the mass which cancels out of both sides. One is left with gh = 1/2v^2. Now enter reality and wind resistance due to the mass and viscosity of air as a fluid. Typically a more massive rider has more frontal area (one aspect of drag), but it's not linearly proportional to his weight. The lighter rider has a smaller frontal area but it's relatively greater with respect to his mass (it helps to think of two spheres of the same density). So the heavier rider is at an advantage on a descent, but it's not due to his weight, but rather his drag. Now to throw a twist in things to bring the whole thing full circle. If two riders of different mass have the *exact* same drag (and no other losses of energy) their velocities at the bottom of the hill will be the same. Unfortunately this isn't often found in the real world since people all have pretty much the same density. So what one should take away is that, yes more massive riders have and advantage down hills, but it's not due to their mass but rather their relatively lower drag for a given mass.
 was that an insult? 4byknSep 19, 2001 10:36 AM Did you say that we are equally dense ;-)
 I think that I understand everything that is being said, but billSep 19, 2001 11:00 AM there is one set of dots that I am not connecting. Given Galileo's principle about heavier objects falling as fast as lighter objects because of the equal and opposite tendency of things at rest to remain at rest (the more massive, the more they will resist moving), why does equal drag, which I understand is going to affect three-dimensional humans of different masses more or less equally, volume increasing by a factor of three with increased dimensions, as opposed to surface area, which is squared, why does drag permit a heavier person to go faster downhill? What is the formula? Is it simply F=MA? If we focus on the force of drag, and compare the way that the equal forces (close enough) of the drag on the heavier object vs. the lighter object, F1=M1A1, F2=M2A2, F1=F2, M1A1=M2A2, M1>M2, therefore A1Right?
 Ah, Grasshopper Daniel H.Sep 19, 2001 11:13 AM Well explained, made perfect sense to me.
 well, I also wonder about this one, Grzy -- billSep 19, 2001 11:33 AM " If two riders of different mass have the *exact* same drag (and no other losses of energy) their velocities at the bottom of the hill will be the same. " Is that right? That would seem to contradict the conclusion that people with greater mass DO seem to reach the bottom of the hill with more velocity. I'm not following why density is anything but a red herring -- I think that people can have all sorts of different frontal exposures for a given mass, which, relative to drag, would be the same thing as having different density (compact little guys with lots of muscle probably ARE more dense than very flabby guys, but the flabby, heavy guys get to the bottom of the hill first, and I'm proof of that. And, what about plain wide guys?). It seems to me that, for our purposes, the only variable that really is changing between heavier and lighter riders is mass. I'll bet that mass can vary by 30% without an appreciable difference in frontal area -- how much more drag does a 210 lb guy raise as opposed to a 140 lb guy really? Relative to the lighter guy, he is 50% heavier! Unless I don't understand what you mean by "*exact* same drag." Can you explain your statement in the context of the conclusion -- heavy guys hit the bottom faster?
 doesn't sound right DogSep 19, 2001 11:45 AM If two riders of unequal mass have the same drag (truly the same), the more massive would reach the bottom sooner. No doubt about it. Drag is not only a function of frontal area. Given the same frontal area, the object that is longer would tend to have less drag, if surface drag is not large. So, given same mass, the longer torso rider, if making his torso parallel to the ground, should have less drag. So, a 5'5" rider might be slower than a 6' rider of the same weight, if the increased height is concentrated in the torso, and they are keeping their backs flat. Very complicated. Much of the shape of the bodies could be offset by the placement of the bodies on the bikes. Some might well have a better aero position, getting smaller and relatively longer. For example, for some unknown reason to me, I routinely outpace much heavier riders down hills, even with straight roads. Given the physics, I must be producing much less drag. So, the drag component cannot be neglected, as it could well offset the mass component. All the theory in the world cannot overcome a simple, real example. Often what we think may be negligible differences could be significant. Doug
 Doug, I'm not sure whether you mean to agree or disagree with my billSep 19, 2001 12:29 PM analysis, but I think (at least one of is in agreement, I am confident) that we are reaching the same conclusion. That is, the vagaries of drag are so complicated that they are too complicated and too subtle, probably, to have any real meaning in this discussion (at this point in the lecture, the professor would say "and the delta approaches zero, so, let's call it zero -- blew me away every time). So what we are left with to explain why the fat guy hits the bottom first? Because we've all seen it, time and time again, through, we have to believe, countless permutations of these subtle variables that you raise. GRZY, WHERE ARE YOU? Are you sticking by the statement that I questioned -- your statement that if the two riders of different mass had the *exact* same drag, they would reach the bottom with the same velociy? Or whatever it is you said exactly? If so, why?
 Agree in general DogSep 19, 2001 12:40 PM I agree that the differences can be too unknowable to tell in advance which rider might be faster down a hill, for example, based purely on mass. But, that is not to say the differences are not important or even subtle. They can be dramatic. There is an agressive rider I ride with routinely. He is about 20 pounds heavier - significant. Yet, I beat him down hills every time, straight or curvy. So, the effect of drag, him relative to me, must be not only significant, but huge enough to overcome the weight advantage he has. So, the general statement that drag being equal, the heavier rider will descend faster, is true. But, the disparity could be, and sometimes is, overcome by a greater decrease in drag. This is not a matter of neglecting aerodynamic drag as "near zero." Just the opposite is true - drag is overwhelmingly important. It's just difficult to ascertain short of testing. Doug
 Oh brother! grzySep 19, 2001 2:08 PM What we all see and know to be true (bigger guys are faster in the descents) isn't caused by their mass, however there is a great correlation between mass and drag since density doesn't change. In a sense you can inderectly attribute it to their mass, but it's really controlled by their drag. Turn it around - what if their drag was zero (i.e. vacuum). Would you agree that they'd both reach the bottom at the exact same time and speed? If so then it's not their weight that accounts for the difference - we didn't change the laws of gravity - just took away the air. Try this one of for size: The square root of three is two, for large values of three.
 As the limit of 3 approaches 4 Chris ZellerSep 20, 2001 6:50 AM And the way I heard it: 1/2 = infinity for large values of 2
 Could be several things Chris ZellerSep 19, 2001 2:08 PM You are right, it is more than frontal area, it is the product of both drag coefficient and frontal area. Drag coefficient takes into account the shape of the body, although when we are talking about cyclists, frontal area is most of the story. If you can reduce this, you have done much to reduce drag. It is possible that you are in a more aerodynamic position than the other havier riders while descending a hill (down in the drops rather than heads up). You could also have more aerodynamic wheels. It could also be that you are pedaling harder than the other rider as well. Potential energy is not the only source of energy to this system.
 Consider the source. grzySep 19, 2001 2:12 PM Agree with your first paragraph. In the example I tried to keep all other things equal and focus on the bigger factors. We're having enough trouble with just gravity and frontal area - more terms and details will just cloud the water. About your "source statement" - the PE (and initial KE) are the only true sources of energy - everything else is a a sink or a loss of energy. Less drag just means a smaller loss, not an additional source.
 Sure, but there is also mechanical energy Chris ZellerSep 20, 2001 6:56 AM Ok, but we are talking about relative energy here when comparing the two riders--essentially subtracting the two energy ballance equations. The source I was refering to was the mechanical energy of pedaling. Perhaps the other rider wasn't pedeling as hard or not at all. Perhaps even hitting the brakes (an additional sink) as I often do while descending a hill.
 well, I also wonder about this one, Grzy -- grzySep 19, 2001 2:02 PM Sure, you can say heavy guys hit the bottom faster (b/c it's true), but it's not b/c of their mass. If you put two riders in a vacuum (beisdes going anerobic) they'd reach the bottom in the same time at the same speed. The key is the relative drag between them. I threw the density in so that one could imagine two solid spheres of the same diameter but with different weights (i.e. densities) - they will fall at the same rate since the drag is the same. Probably the key to understanding is that just b/c you weigh 50% than another guy you don't have 50% more drag. to illustrate the point we all know the difference between riding up high on the flat part of the bar and getting all low in the drops. Your weight didn't change but your speed does.
 I'm a-gonna beat this dead horse one more time. billSep 19, 2001 3:10 PM If you are more aero on a flat, you will go faster. No conceptual cluster-f*ck there. You go faster because the force moving you forward remains the same, but your resistance decreases, so that the vector of the opposing force of drag is less. The resulting net force vector forward is greater and, voila, you go faster. Okay. When you are going downhill, we already have assumed (with all deference to my friend and barrister colleague, Doug the Dog -- do you know that I always have wanted to get a dog and name him Doug? that really is no lie -- who pointed out that aero differences on a hill can exceed whatever this other difference is that remains, in my feeble mind, inadequately described) that drag remains the same for two joes going down the hill, one of whom happens to weigh more than the other. So, in contrast to your last example, where the negative vector has changed, in the hill example, the negative vector is equal and constant for both. Mass is the only difference. And, we have agreed, resulting velocity is different. Tell me why. What is the relationship between mass and velocity, because there has to be one. In a formula, prefereably. If M1
 it IS because of the mass DogSep 19, 2001 3:35 PM The concept of gravity acting equally on all objects in a vacuum has several concepts tied up in that. First, gravity acts in proportion to mass. The greater the mass, the greater the attraction. However, accelleration due to gravity is constant, because, the increased attractive force is offset by the inertia of the object. Greater mass has greater inertia in proporation to mass. So, more massive things have more resistance to accelleration. If not for the greater inertia, more massive things would in fact accellerate (fall) more quickly in a vacuum. So, when falling in air (coasting down a hill), the more massive an object the more attractive force there is due to gravity. However, now, in addition to inertia as a resistive force, you have air drag, too. If two objects have the same air drag (resistive force), then the more massive will accellerate faster due to a stronger pull by gravity. Bottom line, then, increased mass does "cause" a rider to descend faster, keeping drag equal. Make sense to look at it that way? Doug
 Sorry, but... grzySep 19, 2001 4:13 PM That reasoning is totally wrong. You're melding a couple of distinct concepts together and coming up with a new theory.. One that is not supported by the rest of the world. Yes, weight and inertia are larger for a larger mass - no one would argue that. You can't bring the gravitational forces between two bodies into this b/c the earth is a trmemendous number of magnitudes larger than a rider. It's generally assumed that the acceleration due to gravity is everywhere the same on the face of the earth at sea level Think about what you've said: "If two objects have the same air drag (resistive force), then the more massive will accellerate faster due to a stronger pull by gravity." Let's carry the logic a little further. How about we set the drag forces equal AND to zero? We're now back to two different masses falling in a vacuum, which everyone has agreed happens at the same rate. You'd get the same result if you picked some value other than zero. The key is that everyone focuses in on the mass or weight and *assumes* that the drag is roughly the same, but it's not. Ultimately you're free to look at things any way you want, but it doesn't mean that it's correct.
 we can turn it around, though DogSep 19, 2001 7:13 PM I know we all realize what happens, that is, heavier riders generally descend faster. That's reality. We can turn your example around, though. Let's assume that drag really does remain constant. Let's say we have two riders, each wearing an identical packpack. One's is filled with ping pong balls, the other's with lead. They are exactly the same shape. One weighs 1 pound, the other 100 pounds. So, the only variable between then is weight/mass. The heavier one will certainly descend faster, don't you agree? Therefore, greater mass "caused" one to fall more rapidly; nothing else could have caused it, as that was the only variable. True, mathematically, the difference is their mass/drag ratio, but it is certainly true in this case that increased mass caused higher speed. I can't see any way around that. In the real world, differences in drag due solely to mass are fairly neglible, as you have pointed out, or at least compared to differences in mass. Pretty much the same thing. Doug
 Doog, I think I've got it. billSep 20, 2001 6:38 AM All other things being equal, objects with greater mass are more aerodynamic. Makes perfect sense; the object that is more massive, with more kinetic energy (not velocity, at least at the start), pushes aside more air molecules with greater dispatch than a less massive object. So, as Grzy has been trying to tell us, it's not the effect of gravity on mass that makes the more massive object go faster at the bottom of the hill, it's the effect of mass on air molecules.
 Thanks Bill....... Len JSep 20, 2001 7:26 AM I've been trying to figure out a simple way of conceptualizing this. If you are right, this helps me understand. Len
 No Sir grzySep 20, 2001 8:31 AM Actually since you've got two identical riders with identical drag coefficients and the only difference being their weight they will descend at the exact SAME rate. This is what I've been saying all along. the problem is that we all know something intuitive: heavier riders descend faster. Wrapped up in that intuition is the fact that their drag is not equal, but we attribute it to their weight since that's easily measured. Remember: none of us had any trouble agreeing that two different masses in a vacuum will descend at the same rate. The vacuum is just an easy way to ensure that the drag is exactly equal, but your twin riders with backpacks is the same thing. I actually said that the diffferences in drag are NOT neglible between masses and as you yourself said drag accounts for something like 80% of the resistance - even more as speed increases as a squared function. The key is that the heavier rider does have more drag, but it's proportionaly less for the given mass compared to the lighter rider. I fully understand how you're looking at it. You're trying to reconcile what you know intuitively and trying to attribute it to the wrong thing. If you go back to the energy ballance between potential energy and kinetic energy you see that the mass term cancels out of both sides. It's just not there. Change in PE = Change in KE so, mgh = 1/2mv^2 cancel the "m" term from both sides yielding, gh = 1/2v^2 if we have a hill use the sine of the incline angle (a) in degrees thus, g(sin(a))h = 1/2v^2 solving for velocity, v = (2g(sin(a))h)^(1/2) We could start the whole energy ballance thing over again with a term on the RHS for energy lost due to frictional forces (drag, bearings, tire losses, etc.) The key is that the energy lost due to drag is a function of shape, not mass. When we deal with people the density is pretty much the same and we have limitied control over shape. This gives the simplification that mass and shape are *roughly* correlated, but the mass is not the cause. Dunno how else to explain it other than using a chalk board and developing it in great detail and assigning homework ;-). Suffice to say if you want to think of the heavier rider being faster down a hill that *does* work, but it's not the cause.
 I think that the conceptual stumbling block here is the use of billSep 20, 2001 10:17 AM the term "drag." Without following all the formulas, what I'm getting out of this is that your use of the word "drag" may be different from what I, and I think others, were assuming. When you posited what would happen if "drag" were a constant, we poets were getting confused about whether you were saying that the same "drag" would resist two objects of different masses with more or less the same frontal exposure moving through air at the same velocity with the same resulting effect on velocity or whether you were saying that the same "drag" would cause identical loss of energy from the same two objects but with different effects on resulting velocity. What does the "same drag" mean? I know that I, anyway, was thinking that the same frontal exposure would create more or less the same aerodynamic, or maybe more properly, ballistic profile, when I gather that it doesn't. More massive objects of the same size and shape have different ballistic characteristics, which is intuitively correct. To be fair to the rest of us, I'm not sure that you made the distinction very well, perhaps assuming some sophistication that we didn't have (and still may not). You left out steps, the baby ones, undoubtedly. What I think that you are saying is that "drag" is the total aerodynamic effect of air on a moving object. Period. Which is why you say that the same drag gives the same velocity, regardless of mass, and that different drag values mean different resulting velocites. Although the resistive force of the air on objects with the same frontal exposure (simplistic, but can we go with it, please?) is of the same absolute value as an abstract force vector, the RESULT on the two objects is different, because the more massive object is, all things being equal, more aerodynamic, more resistant to the resistant force of air friction (or something like that). The negative force vector may be abstractly the same, but, relative to the mass it is smaller. I think that when you said that this means that the drag is not the same is when all of our little circuits fried.
 Any Engineering Profs out There? GarySep 20, 2001 8:53 AM This problem would make a great extra credit or final exam question in applied mechanics. I think Doug has the right idea with the ping pong balls in the backpack experiment. The point I think has been missed is the effect of friction as a function of mass. Everyone is focused on the effect of wind resistance but what about the effect of mass on the rolling resistance of the wheel bearings and the stickiness of the tires? Think about it this way: The same friction applied to the break pads of otherwise identical bikes, riders and aero profiles will stop the lighter one faster, or slow the heavier one less. Same thing if you put a vehicle on a slight incline at rest, brakes off and begin adding weight - it will eventually begin rolling when the downward force overcomes the friction of the tires and bearings. The weighted backpack experiment should isolate mass as the only varible, whether or not my theory explains it. Had college been this interesting, I might have actually used my degree in engineering!
 I Play One on TV grzySep 20, 2001 9:55 AM You're confusing the issue by adding additional terms in an effort to avoid explaining the fundamentals. To make things simple we'll use identical steel wheels and ABEC 5 precision bearings from the space shuttle transport crawler with no significant differences due to slight changes in rider weight of say 50 lbs. All of the prior analysis remains unchanged. Let's take one rider and send him down the hill a couple times - once sitting up and once in various tuck positons. Why are his times down the hill different? Can't be b/c of his mass or rolling resistance - it didn't change. Now take two vastly different mass riders and make their coefficients of drag exactly equal, by dropping them down a hill in a vacuum. They both get to the bottom at the same time regardless if it's a free fall or and incline plane. To me mechanical and aerospace engineering was this interesting so that's why I continue to use it. It gets a little scary when people start believing lawyers and discounting the engineers. Not that they aren't both smart and competant, but one has more skill with rhetoric and the other is more skilled with the physical laws. Not that someone couldn't be both - now that's a potent combination!
 Thats scary that you still use it and youre still so wrong. a\$\$ monkeySep 20, 2001 1:00 PM shoot!
 Get a bigger stick grzySep 19, 2001 3:58 PM You were doing pretty good until you got to the part about the two Joes going down the hill with the same drag, but different masses (asses?). This point is critical - if you weigh more you have larger drag - all else being equal. The density of humans is roughly the same (but remember: witches float). This next point is the crux: In a vacuum, where there is no air resistance, two objects of different masses will fall at the same rate under the accelration of gravity (or a component if on a hill) with g=32.2 ft/s^2. So now we have two different masses falling at the same rate. Keep everything the same and bring the air reistance back into the picture - now the heavier object falls faster, but it's NOT because of it's weight - we just proved that. The key is that although the drag is typically larger with a more massive object it's not linear, primarily due to volume increasing faster than surface area. The key is that the drag is not the same - in fact one would think that since the more massive object has larger drag maybe it should actually be slower. The devil is in the details and this is one of those situations where approximations, generalities, and hand waving will get you into trouble. Ultimately, if it helps for you to think that the more massive object falls faster you have the right answer, but the reasoning is wrong. At the end of the day all that matters is that we get out and ride, hopefully with friends. Just remember to press your advantages when you get the opportunity. If the other guy is heavier, just make sure he suffers on the climbs.
 Ballistic Coefficent Chris ZellerSep 19, 2001 2:00 PM What GRZY is hinting at is known as the ballistic coefficent wich is basically a measure of a projectile's ability to coast. It is defined as Cb = M/CdA where M is the projectile's mass and CdA is the Drag Form Factor which is proportional to the drag coefficient times frontal area. The more balistic an object will act is dependant upon how areodynamic it is and how massive it is. The more massive and aerodynamic, the more the object will act like an object in a vacum--which is the fastest something will roll down a given hill. This turns out to be much more important than I would have expected. Lance Armstrong's training mannual (and I'm sure this is correct because it comes from wind tunnel data conducted by HED) that an average cyclist at 20 mph on flat expends 80% of his effort for forward motion to negate aerodynamic drag. That is huge, when you consider that drag goes as the square of velocity. A cyclist at 40 mph descending a hill would expend considerably more.
 Yup grzySep 19, 2001 2:20 PM You got the picture, but remember a coefficient is simply a quick way to charactize a system. The balistic coefficient can mislead people into thinking that mass is a critical value, when the energy ballance shows that it's not. In just about anything that gets raced in the speed/time/distance realm drag is a serious consideration. An exception has to be running since the Reynolds Number (oh no, a nother dimensionless ratio) is so low that much more energy is spent just getting all the body parts to move. In swimming the drag is amore important factor since the fluid is much more dense & viscous (higher Reynolds No.) and more energy is spent moving though the fluid than supporting and moving the body parts.
 Ok, but should include drag in energy ballance Chris ZellerSep 20, 2001 6:42 AM Ok, To the point of the original post, mass does fall out of the m g h = 1/2 m v^2 energy balance if that is how you define your system but to add another layer of complexity to get at the reason why heavier riders roll down hills faster than lighter riders you should set up your energy ballance to include energy lost due to drag. More like: mgh = 1/2 m v^2 - 1/2 cd rho a v^2 x if you assume for simplicity some constant speed down the hill where rho is the density of air, area is frontal area, cd is the cefficient of drag, x represents the distance along the road. To be more precise you could express the drag energy as: Integrate[1/2 cd rho a v(x)^2 dx,{start,finish} ] where rho is the density of air, area is frontal area, cd is the cefficient of drag, and v(x) is the velocity as a function of distance along the road and dx reprtesents an incermental change along the road. From this more complete equation you can see that the mass does not cancel and for a given aerodynamic signiture Cd a the more massive rider has the advantage.
 I'm cool. I've got it. Not all the numbers, necessarily, but billSep 20, 2001 7:10 AM I can see the relationships, and that's all I was really after. Thanks.
 Fully agree grzySep 20, 2001 8:40 AM Sure, I fully agree, and while we're at it we'll add *all* of the other frictional forces. Problem is this raises the level of complextiy significantly and we're having major problems with the simplified case. Being rigorous and adding more detail will yield a better model and should prodcue more accurate results, but I didn't want to confuse the issue any more than it already was. Engineers are used to the drill - start of with something simple and as you understand and verify that it's correct you keep adding terms. For example we haven't even discussed the thermal issues and the heat transfer of the rider, tires and various componenets. If we can't get agreement on the basics then increasing the complexity isn't going to help.
 Sorry, grzy, but I got to call you out speed-chumpSep 20, 2001 9:03 AM The statement that two riders of equal mass but identical drag coeeficients will decend at the same rate is incorrect. A 1" diameter hollow plastic sphere and a 1" solid steel sphere, assuming identical surface finishes, have identical drag coeeficients, but the steel sphere obviously attains a higher terminal velocity. The problem here is that everyone is fixated on ACCELLERATION, when the true issue is FORCE BALANCE. As an engineer, I'm sure you are familiar with free-body diagrams, right? I suggest this excercise: Draw a free-body of a cyclist on an incline. The component of force down the incline in the direction of travel = sin(angle of incline) * mass * g This is the force that accellerates the cyclist down the hill. Obviously this force is larger for a more massive cyclist. Now add the component of force resisting the cyclist's accelleration, pointing up the incline. This will be the sum of rolling resistance and wind resistance. Steady state (constant or "terminal" velocity) is reached when these two opposing forces equal each other, that is, there are no unbalanced forces acting on the cyclist. Given two cyclists with equal mass, but different drag coeeficients, the accelerating force is equal (F=sin(angle)*mass*g) But the resisting forces are different, as it is dependent on the drag coeeficient. The rider with the lower drag coeeficient will reach steady state at a higher velocity, and will therefore reach the bottom sooner. All discussion of acceleration and energy ballance is immaterial. Please work out the free-body before responding, and if you find a hole in my analysis, let me know. RESPECTFULLY, Zachary Broussard
 Step Outside ;-) grzySep 20, 2001 9:40 AM Sure the more massive cyclist has more weight and therefore a larger accelrating force, but he's also got more inertia (or lack there of) to over come. When using gravity as the accelrating force it's the same "g" regardless of mass. The resistive force of drag is a function of shape and doesn't have anything to do with mass. The same rider with and unchanged mass can have greatly different results down a hill depending on his shape (i.e. position on the bike). Now, here's the hole in your arguement: How would you account for vastly different objects falling at the same rate in a vacuum. Afterall the more massive object experiences a larger accelerating force (i.e. F = W = mg)? Seems that the resistive forces ARE equal at zero, yet they fall at the same rate. How strange. You start off with the example of two 1" spheres with different masses and identical coefficients free falling at terminal velocity then shift the example to two cyclists with equal mass and different drag coefficients. You've set up an apparent paradox and attributed it all to the same cause. What you're neglecting is the fact that the force of boyancy comes into play with our 1" spheres due to the different densities. To see how this might be possible, lets change the fluid from air to water (both are fluids, no?). It is now possible to have two 1" spheres where one sinks and the other floats. One is moving and the other is at rest: this "must" prove your case, no? The key here is that the densities of the objects are VERY different, but this isn't really possible with humans given that we're mostly water and our bone structures are pretty similar (i.e. no hollow "bird" bones.) Of course we can skew things a bit with different % body fat, but so far we've been thinking about two reasonbly fit cyclists and I've brought the "red herring" of density up several times to qualify my statements. With all due respect the "All discussion of acceleration and energy ballance is immaterial." is just plain ignorant. Since when do you get to pick and choose which physical laws you're going to consider and which you're going to neglect? That's a rhetorical question, you don't have to answer - just think about it. I'd suggest that you take some advice and work out the FBD's yourself - I've already done it many times over. Fact is the other engineers here agree with me. It's just the "poets" I'm having trouble with. So, do you agree that different
 Step Outside ;-) speed-chumpSep 20, 2001 9:53 AM The point I am attempting to prove is that given two cyclists with identical drag coeeficients, but different masses, the more massive cyclist will attain a higher terminal velocity on a given incline. Do you disagree with that statement? If so I will post a neat, legible free-body diagram with all necessary calculations to prove my position. If you agree with that statement, we have nothing further to discuss. Perhaps "immaterial" was a poor choice of words. What I meant was that while the problem CAN be thought of in terms of accelleration and energy balance, it is UNNECESSARY, and seems to cloud the issue. If you find an error in the diagram I post tomorrow, I will be willing to discuss the problem from other directions.
 Put up Ur Dukes ;-b grzySep 20, 2001 10:14 AM No, I don't agree with your statement. It violates the physical laws as we know them. The reason why the more massive cyclist attains the higher terminal velocity is that the drag coefficients are not equal. A more massive cyclist is going to have a larger drag coefficient, there's no practical way around that. Although his drag coefficient is larger it is proportionally less given his weight. the drag doesn't increase linearly with weight. The problem is the use of your conditional "if" statement. Like my dad still says to me, "If a frog had wings he wouldn't bump his ass every time he hopped." While this statement is true, it isn't reality, frogs don't have wings so what's the point? Ultimately you should be able to analyze any given problem from a force/mechanics stand point OR an from an energy stand point and get the same answer everytime. After all why would the choosen method of analysis change reality? Typically the energy ballance is a much quicker way to analyze a complex system when the actuall forces are very hard to model. It's one of the ways you cross check your analysis to see if your force modeling is on track. I've shown the basic energy ballance time and time again, and shown how the mass term cancels out. No one seems to be able to dispute this, yet everyone is reaching for some magic straw to try and cloud the issue. Even the guy with a grasp of aerodynamics is in agreement - he just thought we should add more terms to be more accurate. This is fine, but it doesn't change the basics. I could produce your FBD diagram for you and we'd both be in agreement. The issue isn't going to be the diagram, it's the analysis that follows and understanding how and what changes.
 you can't be right DogSep 20, 2001 10:38 AM Grzy, we all very well know that heavier guys descend faster. At least admit to that. The question is, why? You note that even the heavier rider will have a higher drag, not lower. Therefore, given all else the same, and even assuming HIGHER drag for the heavier rider, it MUST be his higher mass that gets him down the hill faster. With identical drag, the heavier rider will most certainly descend faster (in air). We all know that. We have observed that, and it makes sense. Cannonballs fall faster than softballs (in air). I think that is undeniable. I think your reasoning is getting hung up on trying reason that since all objects fall equally fast in a vacuum, then they will do so in air, too. This is a false assumption. Add up all the forces acting on an object falling in vacuum. You have gravity pulling down, and inertia acting upwards. That's it. But, falling in air introduces an additional force acting "upwards", effectively. This force is not zero, as in a vacuum, and is not proportional to mass, as inertia is. In fact, is could be wholly unrelated to mass. I don't have the tools handy (long forgotten) to pick apart your math. But, your concepts and conclusions defy reason and reality. Heavier objects of the same shape fall faster in air. We all know that. If the only variable is mass, then the mass causes the difference. Now, I can't show this mathematically (once I could), but this "poet" deals in reason and logic, too (not to mention evidence and reality). We've beat this to death; maybe we need an experiment. Doug
 One more thing before I go back to work... speed-chumpSep 20, 2001 10:29 AM Speaking of ignorant... You keep falling back on two bodies falling at the same rate in a vacuum as a specific example of a case in which the resistance force is equal. This model neglects the whole concept of STEADY-STATE, which is really the crux of my argument. In a vacuum, the bodies will accelerate forever, never reaching a constant velocity. On long descents, cyclists DO reach steady state, so any model neglecting that is, for the purposes of this discussion, IMMATERIAL. Please think about the problem at steady state, and draw your free body at steady state. There should be no acceleration involved. It's pretty darned obvious that if given a relatively larger force component DOWN the slope, that is - a more massive rider, you need a relatively larger component UP the slope to balance it, that is - higher velocity or greater drag coeeficient. Again, I'm not making any argument about who has the greater acceleration. I'm arguing about who ends up at a higher velocity WHEN ALL ACCELERATION HAS CEASED. If the tone sounds heated, I apologize. I really do find this sort of discussion great fun. OK, I'll talk to you tomorrow.
 Grzy talks out of his A\$\$ quite often. Hes not as smart as he a\$\$ monkeySep 20, 2001 12:50 PM claims. Dodo.
 be nice, we were having a civil discussion nm DogSep 20, 2001 1:01 PM
 grzy, you are wrong. vanzutasSep 20, 2001 12:56 PM grzy, I will give the equations and the Free body diagram which you have not yet produced. First of all using the energy balance, mass does not cancel because you do have to integrate the drag coefficient on the kinetic energy side of the equation. I will not discuss things in a vacum because I do not live in a vacum grzy does. Drag cannot be neglected, because the two largest forces acting on the rider are gravity and wind resistance. here is the energy balance equation for any point of a falling body. I will neglect the sin of the angle because we are all going down the same hill. PE=Potential energy KE=Kinetic energy DL=energy lost due to drag m=mass of object h=initial hieght g= acceleration due to gravity v=velocity p=(rho)air density Cd=coefficient of drag A=frontal area of object y=vertical position PE=KE-DL mgh=.5mv^2-.5pCd*integral of V^2 from y1 to y2 mass does not cancel, I am sorry. that is how your argument is wrong. now I will show an FBD with the sum of the forces on two separate bodies. they both have the same shape (spherical) and they both are the same size (this is to simplify the problem which you are always trying to do) Both sphere start at the same height (y) with a velocity of 10m/s. therefore the force of drag on the two objects is initially identical. I am going to give this force a value of 10N (10 newtons or 10 kg*m/s^2). I will prove that the acceleration of the more massive object is greater than that of the less massive object. m1=10kg mass of sphere one m2=20kg mass of sphere two v1=v2=10m/s velocity of one and two are equal g=9.8m/s the acceleration due to gravity Fd1=Fd2 the drag force on 1 and 2 (do not argue this point because I am assuming that they are the same size and shape with the same initial velocity) a1=? the acceleration of spere 1 (this is what we are trying to find) a2=? the acceleration of sphere 2 Now look at the bottom and study the free body diagram. My argument is this if both riders start at the same speed the heavier one will accelerate more therefore he will go faster there is no arguing this point. now I say that in reality the heavier rider will be bigger and have more frontal area giving him more drag. however the extra drag in the case of a cyclist does not overcome the increase in acceleration that occurs due to the weight. the magic happens when the mass from the left side of the equation is divided by the right side. see how the effect of drag becomes relatively smaller because of the larger mass of the rider. One poster said that he does not want to sound heated. I on the other hand have every intention of sounding heated. You are constantly ignoring the facts and have not yet given any evidence of proof. A previous point on increased terminal velocity of the heavier rider is also correct. the downward force of the larger rider is greater so the required drag force to stop acceleration is greater for a more massive person. I end this saying that grzy is wrong. I have been prven wrong infinite times in my life and I await your retort to prove me wrong again. I don't see it coming. Adam