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Here you go, grz....(51 posts)

Here you go, grz....speed-chump
Sep 20, 2001 11:43 AM
Hopefully the image comes through. I haven't tried to do that yet...

Free body diagram of a body falling down an incline.

At steady state, all forces must balance.
Fg= force due to gravity = m*g
Fr= resisting force, a composite of wind and rolling
resistance, proportional to velocity
Fn= normal force
- Balance forces in x direction:
(1) Fr*cos(theta) = Fn*sin(theta)
- Balance forces in y direction:
(2) Fg = Fr*sin(theta) + Fn*cos(theta)

From eq. (2), if Fg increases (larger mass) either Fr, Fn,
or both must increase to maintain balanced forces in the y direction.

From eq. (1), if Fn increases, Fr must increase to maintain balanced forces in the x direction.

Therefore, if Fg increases, Fr must increase to maintain balanced forces.

For a given body, the value of Fr can only be increase by an increase in the body's velocity.

Therefore, at steady state, larger masses fall at greater terminal velocities, all other things being equal.
WHOA!cioccman
Sep 20, 2001 11:47 AM
I don't believe you.

Just kidding!
Or, if you prefer...speed-chump
Sep 20, 2001 12:43 PM
We can approach it from the energy balance direction, so maybe I won't seem so "ignorant" to you....

Given: Two bodies of identical drag coefficient, falling at
constant, terminal velocity. Mass of body A = 1kg, mass of
body B= .1kg. We will analyze the energy balance as these
bodies fall through 1 meter.

Body 1, 1 meter above reference plane:
Total Energy = KE + PE
Total Energy = 1/2*1kg*velocity^2 + 1kg*1meter*9.81m/s^2
At reference plane:
The overall KINETIC energy remains the same, because
the body is still travelling at the same speed. The potential
energy has decreased by 9.81 Joules, with respect to the
reference plane. Therefore, 9.81 Joules has been dissipated
by wind resistance.
Body 2, 1 meter above reference plane:
Energy = 1/2*1kg*velocity^2 + .1kg*1meter*9.81m/s^2
At reference plane:
The overall kinetic energy remains the same, because
the body is still travelling at the same speed. The potential
energy has decreased by .981 Joules, with respect to the
reference plane. Therefore, .981 Joules has been dissipated
by wind resistance.

Still with me?

Energy dissipated by wind resistance = drag coef * velocity * distance

If both objects have the same drag coeeficient, and fall through the
same distance, yet one loses 10x as much energy, we reach the startling conclusion that ONE OF THE BODIES MUST HAVE BEEN
TRAVELLING FASTER!

Or am I just stupid?
You people make my head hurt... n/mWannabe
Sep 20, 2001 12:57 PM
n/m
Whos the chump now? Nothing's new here, GRZY is always wrong.a$$ monkey
Sep 20, 2001 1:15 PM
darnit!
You Blew Itgrzy
Sep 21, 2001 8:24 AM
Duh, of course a heavier body has more energy, but it also takes more energy to accelrate it. Your conclusion is a result of incomplete analysis and flawed logic. You have yet to address the inconsistancy of why two different objects fall at the same rate in a vacuum. Something we all know is true. Repeat your analysis for zero resistance and see if you can get it to reconcile. You'll find your error if you do it correctly.

I hate to admit it, but you're being stupid (your words, not mine).
GRZY, you're killing me.speed-chump
Sep 21, 2001 10:13 AM
Here we go. I'm going to force you to adress the real
issue. I will make a series of numbered statements that
prove my point. If you disagree, tell me which number,
and explain why very specifically. No hand waving,
or insults on my intelligence. I expect logic and/or
equations. Don't tell my I'm wrong, PROVE IT. Anything
else is poetry.

1. When an object falls in a vacuum, it is subject to
a "downward" force = m*g. This is the only force
acting on the object.
2. When the same object falls through the atmosphere,
it is subject to the exact same downward force as
in statement (1). This is still the only downward force
on the object.
3. An object under the influence of gravity, and in a
vacuum will accelerate forever, it's velocity
asymtotically approaching the speed of light.
4. An object falling through the atmosphere will not
accelerate forever, but will eventually reach a
"terminal" velocity.
5. The only upward force on the object falling through
the atmosphere is the aerodynamic drag force. Bouyant
forces in air are incredibly small, and may be
neglected.
6. The aerodynamic drag force is independent of the
object's mass. I have proven this and cited sources
elsewhere in this post. I am willing to do it again.
7. Terminal velocity is reached when the aerodynamic
drag force is equal to the "downward" force (m*g)
8. A larger "downward" force, that is, greater mass,
requires a larger drag force for equlibrium.
9. At any given velocity, a hollow steel sphere will
experience the same drag force as a solid steel sphere
of the same size and surface finish. (see #6)
10. The two spheres in #9 are not subject to the same
downward FORCE, because one is more massive. (m*g)
12. If both spheres are to be in equilibrium, the solid
sphere must experience a larger drag force, because
it has a larger downward force. (m*g)
13. For these two spheres, the only way one can experience
a larger drag force is for it to have a higher
velocity.
14. The solid sphere will reach equilibrium at a higher
velocity.

BTW, I have never and will never argue with the fact that
objects experience the same acceleration in a vacuum. You
need to get it through your head that I am talking about a
system at equilibrium, and you are not. Your example is
not appropriate, because it is not a system at equilibrium,
and a cyclist at terminal velocity is. Don't come at me
with ANY example that features an accelerating body. That
is not what we are talking about.
or maybe this one....speed-chump
Sep 20, 2001 12:56 PM
Vt = SQRT((2*m*g)/(C*p*A))
Where:
Vt = terminal speed
m = objects mass
g = local acceleration due to gravity
C = drag coeeficient
p = air density
A = frontal area

If frontal area, air density, drag coeeficient, and local
acceleration due to gravity are all the same (in other words - "all other things geing equal") for two bodies, but
the bodies have different masses; once again we reach the
startling conclusion that the heavier body falls at a
higher velocity.

Source:
"Fundamentals of Physics" 4th edition
Halliday, Resnick, and Walker
page 139, eq. (6-19)

Of course, they might be ignorant too.

:)
Thinkgrzy
Sep 21, 2001 8:36 AM
Think about what you're saying. You're confusing yourself. Forget all other things being equal, two different masses fall at the same rate in a vacuum. How can this be if one is more massive? Dang inconsistancies...

You're really working yourself into a lather and demonstrating considerable ignorance of the physical laws. Nothing wrong with Halliday, et al, nor with Tippler or any of the other physics books. Problem is how you can claim them as a source and twist things around in your mind and then think you're right. I've shown things over and over again in the most direct means that the mass cancels out via an energy ballance and you have yet to dispute this. Instead you've taken half concepts and twisted them around in your mind to confirm your beliefs and ignored the inconsistancies. It's a little frustrating trying to have an arguement this way.

In other words you'd be up for the Nobel prize if you were right, but don't quit your day job just yet.
Thinkvanzutas
Sep 21, 2001 9:07 AM
grzy, in a vacum they fall at the same rate. however you are the only one in a vacum. when you do and energy balance out of a vacum there is no way to neglect the energy that the body adds to the air by accelerating said air. In my post earlier and someone elses this is stated. and what I wrote also works in a vacum and when the drag is zero both bodies fall at the same rate. however with increasing drag the heavier body will fall faster because you have to divide the entire equation (not just the terms you want to) by mass, when this is done, if both bodies have the same drag then you will see that the more massive body will accelerate more. I have the FBD and equations to back it up.
You're getting closer!grzy
Sep 21, 2001 9:26 AM
Yes, you're almost there - it's the differences in the drag per pound of mass.

What I've attempted to do is to show that if we remove the entire drag aspect then different masses fall at the same rate. So given that different masses don't cause things to fall at different rates how do you now say that the masses are the cause when drag is added back into the anaylysis? There IS a relationship between mass and drag, sort of. It's just that it's not simple nor is it linear and this is the rub. Larger objects have more drag (for same desinity & shape) but it's relatively less for a given pound of mass. The larger object will have greater energy at a given speed, but less of this energy is required to displace the air and deal with the associated drag. The heavier object descends at a higher rate because less energy is lost, not because it's heavier.

I'm really at a loss of how to better explain this. In fairness it took a longer time to learn.
this word "cause" is the hangupDog
Sep 21, 2001 10:04 AM
I think our entire discussion this has been hung up on the concept of "cause." What "causes" heavier riders to descend faster?

To me, "cause" is the thing that changes when all else is kept the same. The variable. If every physical aspect of a test is kept the same, but one variable is introduced, with a different result, then we attribute "cause" to that variable.

Not sure the word applies in a mathematical world or to the physicist.

Is this the problem?

Doug
Grzy agrees on the cause, sort of, he mostly disagrees thatbill
Sep 21, 2001 10:31 AM
it is the force of gravity on the different masses that makes the difference, even assuming that, though the larger dude is, well, larger, he's presenting the "same" aero profile on a proportionately larger scale. This assumption was throwing me for a loop, because, although the assumption works in the cyclist model (not only are people basically of the same density but our joints generally allow us to bend and tuck about the same -- we're the same shape, relatively), but I misunderstood it's import. Because volume increases by a factor of three as opposed to surface area, which is squared, and because mass therefore increases faster than frontal area, which is the main source of drag force, this relationship is the key to understanding how mass influences ballistic performance, all other things being "equal," which I now understand why he had placed in quotes and which we all should have paid more attention to. How many times did the guy say that the drag force on the heavier dude is larger, it just is proportionately smaller compared to mass, and I, for one, just thought that he was being punctilious, but he was saying what he meant and meaning what he said?
Now that I think that I understand the concepts, I'm disappointed in myself not to have seen that Grzy actually laid it out a bunch of times in a bunch of different ways.
>>>>>>>>>>>Bless You!grzy
Sep 21, 2001 6:33 PM
Yeah! Thank you so much for your note. It really cuts to the heart of what I've been trying to say but have been almost totally frustrated in my attempts to communicate. That one person could understand what I'm saying, see the logic and objectively agree makes it all worth while. I seriously would've given up on this whole thing if it just don't bug the crap out of me that I couldn't present the material and make my point, knowing that I'm right. It's not that I need to hear that I'm right - I already know this and it isn't about to change - I was trying to pass along some semi-useful knowledge (you can decide how useful) and correct some incorrect statements.

I honestly think the my reply titled "One More Time..." is about the best I can do with Mr. speed-chump. If I can't get through then I'll have to borrow a line from Mr. Spock and say that he's being "highly illogical." Problem is people can't see me raise one eyebrow in cyberspace, but that's OK b/c I can't do it in reality. ;-)

I'm going home - it's 7:30 pm on a Firday night and the wife want's to know if I'm having and office affair or what.
YES!!grzy
Sep 21, 2001 3:29 PM
I've been trying to show it's all about the "cause". The free fall in the vacuum is the classical way that the physicists proved their case. Once this is accepted and fully understood we can then make things more complicated. If we can't get beyojnd this then it's just like running that race in the special olympics....

The unfortunate thing is that there is a good correlation between mass and being first down the hill, but it's not the cause. Cause would be the thing that governs the system behavior every single time. Remove the air resistance and the mass thing falls apart. so it's not the mass causing the difference - it's the air resistance which isn't a direct function of mass. if the air resistance didn't matter or were only a strict funtion of mass then it shouldn't be possible for a skinny guy in a tuck to beat the bigger guy sitting up. But we know this can happen. That would be because the mass doesn't capture the whole story.

Cause applies in both the mathematical worls and the physical world since the math is just the means of modeling and communicating the physical concepts. Most of the time if you can prove it mathematically people will accept that it explains the physical world. That is if you're rigoourus enough in the model. The scientific method then challenges people to disprove the model. If it survives then people accept the model. If the model fails then the model is discounted and the guys go back to the lab and try to figure out what they missed.
See if you can figure this out.speed-chump
Sep 21, 2001 10:42 AM
Given:
Two bodies falling through the air at 100m/s
Mass Body 1 = 10 kg
Mass Body 2 = 1 kg
Drag coefficient Body 1 = .5
Drag coefficient Body 2 = .5
Frontal Area Body 1 = .001m^2
Frontal Area Body 2 = .001m^2
Air Density = 1.2kg.m^3

Question - What is the net force on both bodies?
Body 1:
Fnet= downward gravitational force- Upward drag force
= 10*9.81 - 1/2*.5*1.2*.001*100^2
= 95.1 newtons downward
Body 2:
Fnet= downward gravitational force- Upward drag force
= 1*9.81 -1/2*.5*1.2*.001*100^2
= 6.81 newtons downward

Question - What is the acceleration of each body?
Body 1:
a = f/m
a = 95.1 newtons / 10 kg
a = 9.51 m/s^2
Body 2:
a = f/m
a = 6.81 newtons / 1kg
a = 6.81 m/s^2

Why is the heavier body experiencing a higher acceleration,
GRZ ?
They have the exact same size, the same drag coeeficient....

Please adress the equations only.
speed-chump, buddy, how are you going to get two bodies ofbill
Sep 21, 2001 11:22 AM
different masses by a factor of 10 with the same frontal area and the same size and the same drag coefficient in the first instance? I'm not even sure that this is possible, but I'm pretty sure that it has nothing to do with the cyclist model, because it assumes radically different densities for the two objects. Even if we got past that, all you've proved is that the same drag force reduces the NET acceleration of a more massive object less than on a less massive object -- which says nothing about what "causes" the difference, only that there is a relationship between mass and this NET effect on acceleration. Drag doesn't "reduce" gravity's acceleration; drag may alter the way that the body falls through air, but gravity's acceleration remains the same.
Bill...it's easy / And with this I will bow out.speed-chump
Sep 21, 2001 12:05 PM
Make the 2 steel sphere's I mentioned in another post.
Same size and surface finish, one hollow, one solid.

You are exactly right when you say that all I've proven
is that the NET accelerations are different. That's all
I'm TRYING to prove.

The key is that there is no ultimate "gravity's
acceleration". A body's acceleration depends on the SUM
of the forces acting on it, and it's mass.

Not all objects under the influence of gravity fall with
the same acceleration when drag is present. This is
completely obvious. Drop a feather.

You are also completely correct in saying that that drag
does not "reduce" the force due to gravity which tends to
accelerate the body. But the force due to gravity is only
ONE of the forces acting on the body. The body's
acceleration is dependant on the sum of ALL the forces
acting on the body. So drag does not "reduce" gravity's
contribution to the force that causes acceleration, but it
does reduce the OVERALL force that causes acceleration.

As far as this model being irrelevant - yeah, probably so.
It's irrelevant to cyclist per say, but it is not irrelevant
to understanding the phenomenon. Also, the factor of 10
was arbitrarily chosen. You could have a factor very close
to 1, and the phenomenon would still occur, it would just be smaller in magnitude. I've let this go far beyond
cycling in an attempt to prove my point to grz.

SO... I'm finished. I will post nothing further on the
topic. E-mail me if you like. Grz - if you're ever in
south Louisiana, post and let me know.I'll buy a case, and
we can sit down in front of a chalkboard and argue with each other until we either agree or pass out. Discussions
of these matters in this forum are impossibly fragmented.

Bill and others (you know who you are) - Thanks for
maintaining a civil tone. Grz - thanks for qualifying
your insults by saying "with all due respect". I let a few
insults slip in myself; for that I apologize.

OK, I'm logging off and heading out for a weekend of
mountain biking. Ya'll enjoy your weekend.
Well, maybe now my "wiffle cyclist" model has some currency.bill
Sep 21, 2001 1:35 PM
If we catapulted a wiffle cyclist and a hardball cyclist (same size, shape, surface texture, different mass) through the air in a horizontal trajectory, begining at the same velocity, they're not both going to continue forever, right? Let's make it easier. Remove gravity, but still do it in air. So, they'll continue horizontally until the air resistance stops their horizontal movement. Now, which is going to travel forward farther? You're darn right that the hardball cyclist is going to go farther, and it's because of the kinetic energy of the greater mass, and it has nothing to do with gravity.
One more time....grzy
Sep 21, 2001 5:52 PM
You made a fundamental conceptual error which illustrates the problem we're having. It's all tied up in the selection of the drag force - something that is hard to get an intuitive feel for.

First let's review what you've done. There is the accelration due to gravity which is how you get the weight of the objects. An object weighs the same wheather it's moving or not since acceleration is contstant. In fact we are all under the acceleration due to gravity while sitting here at our computers. However the resistive force of the ground ballances us out so we don't move. Now back to our masses (we're up to our "masses", no?) The key is in your choosen drag force which is the cause of our problems and my point. The drag force is what opposes the weight (force due to gravity) and accounts for the different acclerations. Since the drag force isn't linear (i.e. Fdrag1 = 10 x Fdrag2) the heavier mass has realitively less drag and will experience the higher acceleration as you've calculated. Look very carefully at your equations and realize that the mass term doesn't show up in the drag expression. If you agree with this then I've made my point. Tthe mass doesn't determine the drag force which is what controls the results of the equations. This is what I've been saying all along. Remove the drag and the acclerations are the same. Introduce the drag which is not dependant on the weight/mass. Pick various arbitrary drag forces and the equations can be manipulated in any manner.

Maybe it would help if you turned things around and used a drag force proportional to the masses let's take the smaller one, 3 N., as our reference, for simplicity. This means that the larger mass should have a drag force of 30 N. since it's ten times more massive. Now when you use your equations and do the math

Fnet1= 98.1 N. - 30 N. = 68.1 N.

Fnet2 = 9.81 N - 3 N. = 6.81 N.

Now back to F=ma, and solving for a=F/m

a1=68.1N./10 kg = 6.81 m/s^2

a2 =6.81N/1 kg = 6.81 m/s^2

Wait - how can this be? The key is that both masses now have proportionally identical drag with respect to their masses. They will reach their destination at the same time and rate. Again, this is my point.

Sure the heavier mass with proportionally less drag gets to the bottom first, we ALL agree on this, the point is that it's due to the differences in the drag forces not their weight. This is exactly what I've been saying for two days. The term to describe the mass and speed of descending is a spurious correlation.

Another way to look at it is to pull down your Halliday physics book and look at the equations of motion, specifically under constant acceleration:

2aX=Vf^2-V0^2

So, it seems that the mass is absent from our equation and with out air resistance the final velocities would be the same. You plug in an intial velocity (let's use V0 = 100m/s as in your example. "a' is the acceleration due to gravity "g" which is everywhere the same (I know you have trouble with this). Now it doesn't matter what the mass is the final velocity (Vf) is determined soely by how much further they have left to fall X. Realizing two things: one the velocity V0 = 100 m/s is only true for an instant and that this works without air resistance.

It's really not worth continuing to argue this. I'm wasting my time and so are you. Frankly I'm at my wits end trying to explain it to you. Probably the toughest thing is your preconcieved notion gets in the way of you looking at things with an open and objective mind. You have the right answer, the bigger guy gets down first, but for the wrong reason. This is all I'm trying to say.
we lawyers can really stir up the engineers, can't we? (nm)Dog
Sep 20, 2001 1:15 PM
We engineers are easily stirred!speed-chump
Sep 20, 2001 1:43 PM
I know none of this really matters in a cosmic sense, but my poor little feelings get hurt when somebody calls me "ignorant" when my high school level physics book clearly shows that they are wrong. It's the arrogance of the thing I find so annoying.

I suppose I'm an example of the social retardation that many claim is prevalent amongst engineers to worry about it, but oh well.

Anyway, a good stirring is pretty fun. These threads today have been much more entertaining than what's going on at work.
I sympathize, a near-scientist of sortsDog
Sep 20, 2001 1:51 PM
I was a biology major in college, until discovering how easy philosphy was. Chemistry, physics, botony, zoology, anatomy, calculus, statistics, were very interesting - but hard. Had to study and do homework. Not fun.

I remember the concepts, just not how to get there or to prove them. Frustrating.

If I had a do-over I'd be an engineer, too.

Doug
Arrogance?grzy
Sep 21, 2001 8:37 AM
I guess it would depend upon your point of view.
grz, my thoughtsET
Sep 20, 2001 2:04 PM
I think the main problem is you're not doing a good job of explaining yourself. I understand the vacuum case, so please don't repeat it. Just give your explanation for the specific cases below; your careful wording in these cases would clear things up a lot:

1. A tandem bike with two riders has nearly the same drag (in fact, slightly greater) than a single rider, yet clearly descends faster.

2. Take one human rider weighing 175 and an iron robot shaped exactly the same but weighing 500 lbs. Who will reach the bottom faster?

3. You can go to analyticycling.com and click on wind on rider at right, then advanced input, set power to 0 for both of them (coasting) and keep everything the same for the two riders (in particular, frontal drag; you can even account for what will be relatively insignificant differences such as rolling resistance if you like), except input different weights (e.g. 65 vs 80 kg). Whether they start at 0 mph or some other identical speed (typical for two riders beginning a hill at the same time) and descend, e.g. down a 10% grade for a nontrivial distance, the heavier rider gets there first. Explain. Not another case, but this case.
I thought we settled this down belowChris Zeller
Sep 20, 2001 2:13 PM
This analysis (by speedchump) is correct (either through force ballance or energy meathods) but I thought we concluded this below. None of this is any different from what GRZY and I have said below.

As for who is really faster? Fat guys don't pedal as hard or as fast as lean guys. I'd bet on the guy pedaling the hardest and with the best control of his bike.
I agree - Grzy is getting a bad rap, here. I've read all thisbill
Sep 20, 2001 2:49 PM
stuff through and through (will someone tell my clients that this is more important, and, oh, btw, pay me anyway?), and Grzy never says that heavier riders don't get to the bottom faster, or at least when you use some other assumptions that are generally true. He agrees with that. He just says that, by focusing on mass as the most important variable is wrong or, at best, misleading. The focus on mass obscures that the real variable is the drag coefficient, which is proportional to mass but is not always DETERMINED by mass, because you can have objects of the same mass but with different drag coefficients. Drag also is proportional and inversely proportional to other variables don't ask me which, but these include speed, surface area, density of air, and stuff. Definitely. Definitely stuff.
You all are proving the same thing. It's somewhat a matter of defining terms, too, because there does not appear to be consistent agreement on what is a drag coefficient, which has screwed up us mere mortals maybe beyond repair. What I'm getting out of this is that we are mixing up drag as a negative force on the system and the drag coefficient, which is an abstraction that is a function of -- that stuff. Which Grzy actually explains, sort. I think that he could have said it a different way, clarifying that of objects the same size, shape, and velocity, but with different mass, the force of drag may be equal, but the drag coefficient for the more massive object is less. Actually, he does say this, but it's hard to catch in context.
I think Grzy went overboard in trying to convince everyone that mass is irrelevant. It is not irrelevant. It can be relevant in determining the drag coefficient. It just theoretically doesn' t have to be relevant in determining the drag coefficient, and it's not relevant in terms of the force of gravity on the object. I think that I understand this, god help my practice if I don't.
Grzy, you may not want my help. But I hope you remember this the next time you are tempted to DRINK MY BLOOD, you cranky engineer, you.
I think he left to go kick his dog. nmNot a weight weenie
Sep 20, 2001 2:56 PM
No, work. Remember work? ;-) (nm)grzy
Sep 21, 2001 9:05 AM
I know I should drop it, but I'll press on...speed-chump
Sep 20, 2001 3:26 PM
Drag coefficient is a funtion of the object's geometry and surface finish.

Drag Force is a function of drag coefficient, air density, frontal area, and velocity.

As far as the statement:
"objects the same size, shape, and velocity, but with different mass, the force of drag may be equal, but the drag coefficient for the more massive object is less"

That is incorrect. Neither drag force nor coefficient is dependant on the object's mass in any way. If two objects have the same size and shape, are travelling at the same velocity, and experience the same drag force, then the two objects have the same drag coefficient BY DEFINITION.

What IS dependant on the object's mass is the force acting on the rider that propels him/her downhill.

My beef with grz is that I specifically asked him to agree or disagree with the statement that - given two cyclists with identical drag coefficients, but different masses, the more massive cyclist will reach a greater terminal velocity.

He specifically disagreed. My three ridiculously nerdy posts above support my statement, and disprove his position regarding it. The whole reason I asked him to agree or disagree with my very specifically worded statement was that I wanted to pin this argument down and settle it.

I have presented three very simple conceptual proofs above. Another fellow presented yet another proof below that is much more sophisticated, but is not well suited to the layman, as it involves a touch of the Calculus. Grz can feel free to argue with me about the first two proofs, which are my own. To argue with the third, he will have to contact the PhD's who wrote my physics book. If he wants to get into arguing calculus, we'll have to start a discussion off of this board, because I don't have an "Integral" button on my keyboard. I'll be more than happy to engage him in a battle of the faxes on a purely mathematical level.

I'm pursuing this because I don't enjoy being called "ignorant" in my field.

Plus it's kinda fun.
I think I'm going to cry.bill
Sep 21, 2001 6:03 AM
I thought that I was frustratingly close to understanding this, and now I realize that I am only frustrated (or, as my daughter likes to say "frusterated"). One of the engineer dudes posted a formula down below (don't ask me where) that described ballistic performance, and I distinctly remember that mass factored into the equation as inversely proportional to the coefficient.
I confess that your drawings came up so outsized on my screen that I couldn't make heads or tails of it, but I guess what you're saying is that, assuming that gravity is the only force driving the cyclists downhill, the effects of air resistance and the phenomenon of terminal velocity sort of negate or moot (to explain the difference between the two objects in air only -- I know that inertia still matters) the countervailing force of inertia in the free-falling, vacuum-packed model, so that the larger force of gravity on the more massive object causes the more massive cyclist to go faster. Right? Is this any different from where I was, oh, about three billion posts ago, where I said that more massive objects with greater potential energy and then greater kinetic energy are better at pushing aside air? And, it doesn't matter where the force comes from, right? If we threw the cyclists through the air, the heavier one would require more oomph to get up there, but if they started at the same velocity, the heavier one would land farther away (like, a hardball cyclist versus a wiffle cyclist)?
It's OK man...let it all out.vanzutas
Sep 21, 2001 7:59 AM
Bill, I think you are reading way too many different theories and trying to put them together, your attempts at describing what you understand definatly have multiple theories jumbled. I tried to explain it below but I am an engineer so I can't write. Speed chump and dog are the coherent ones (and maybe myself,but maybe I only make sense to me). If you simply forget everything and start over. ignore all other posts especially grzy. then read what those two wrote it may make sense. oh and maybe the bulistic dude but that was a theorie in its own so don't try to add that one to any others.
It's OK, Bill.speed-chump
Sep 21, 2001 8:03 AM
What I'm saying is really much more simple than I am making
it seem. The key is to forget about "the countervailing
force of inertia". You can forget about it, because inertial forces only exist when an object is accelerating. When you have reached your terminal speed on a hill, you are no longer accelerating, so there are no inertial forces.
AT STEADY STATE, MASS HAS NO EFFECT ON RESISTANCE. It is
not a matter of effects negating or cancelling each other
out. It is simply a fact; a definition.

What all this boils down to is the simple fact that, once
you have finished accelerating and reach a constant speed,
your MASS has nothing to do with the resistance impeding
your forward progress.

What your mass does have an effect on is the effective force
that propels you down the hill. More massive = more force.

You reach equilibrium, or "terminal velocity" when the force
propelling you down the hill equals the drag force.

SO, if you are heavy and have a large force propelling you
down the hill, you will not reach equilibrium until you have
a large force resisting you. The magnitude of the force
resisting you is dependent on velocity. Thus, all other
other things being equal, the more massive cyclist will
reach a higher terminal velocity.

Please note that I have gone to great pains to avoid saying
anything about who has the greater acceleration.

Ballistics - I don't remember this post either, but if
you're interested in the details of wind resistance, read on...

From my Physics book, which I cited above,
D = 1/2*Cd*p*A*v^2
where: D = drag force
Cd= drag coeeficient (see below)
p = air density
v = velocity
AND
Cd = Df / 0.5p V2 Ap2
Where:
Df = Drag force
p= Air density
V = air velocity
Ap = Projected area
"Janna, William, S., Introduction to Fluid Mechanics, PWS Publishing Company, Boston, 1993."

Note from the above that mass occurs NOWHERE in in the
equation. So drag force is not dependent on mass. What
can be confusing is that references on Drag Coefficient
mention density and inertia. BUT they are refering to
the density and inertia OF THE SURROUNDING FLUID, not the
object being analyzed.

Here's an interesting utility to play with. Please note
that it fully supports the notion that given two cyclists
with identical drag coef and and size, the heavier will
descend at a higher terminal velocity, and that drag force
is NOT dependent on mass.

http://www.xsystems.co.uk/machinehead/cyclists_power_calculator.html

Enjoy.
Hold on.grzy
Sep 21, 2001 9:09 AM
He trew up the ballistic coefficient which was normalized for mass. Since mass was in the denominator the coefficient gets smaller as the mass gets larger which is PERFECTLY consistent with what I've been trying to say. Fatt boys have relatively less drag and this is what gives them the edge on the descents.
Thanks!grzy
Sep 21, 2001 9:04 AM
I sincerely mean that. Your help IS appreciated.

It is a difficult thing to explain and you've done a good job of highlighting the points I was trying to make. I keep going back to the vacuum thing because it's the most simple way to illustrate my point. It's even better if you get the chance to see it demonstrated. It also illustrates the flaws in all of the arguements proposed and I find it interesting that no body wants to tackle this simple case, but prefer to get in over their heads with half assed analysis and erroneous conclusions. My point is you can't pick and choose when physical laws are going to apply and when they're not. If it helps people to think that the heavier rider will get to the botttom first, this is OK, but my point is it's not b/c of their mass.

It's also kinda funny that the other practicing mechanical/aero engineers agree with me. Ultimately people can believe what ever they want, but that doesn't make it right. No wonder we can have such violent disagreements about things like religion. This probably helps explain why I'm not religious.
Thanks!vanzutas
Sep 21, 2001 9:14 AM
well the problem with arguing religion is that no one can ever be right because these things can not be proven.
however this can be proven. there may be more than one reason a larger rider goes down faster, but I previously have proven that mass is a large factor in that.
We have all looked at it in a vacum to check our work. however it is when you add resistance that the big guy wins, because of his mass.
Thanks!grzy
Sep 21, 2001 9:30 AM
...no the big guy wins because his drag is relatively less for his mass!

We get the same answer, but the critical difference is why. A little guy can win in a tuck if the big guy sits up. All he's got to do get his drag per poind of mass lower than the big guy.
Thanks!vanuztas
Sep 21, 2001 10:51 AM
grzy, what I desribed changes only one variable. the mass not the drag. which yields the answer that mass effects the velocity. read my equations accompanied by a FBD (yesterdays post you never responded to it) and you will see it. if you want to add more variables that is also possible and the larger rider with the same density will have more mass to frontal area ratio.
you are starting to say that mass has an effect which you viamantly denied in the begining of the argument. so I again say that mass is the determining factor.
GRZY changes his mind?speed-chump
Sep 21, 2001 11:03 AM
So, in other words, if two riders have the same drag per
pound of mass, they will go the same speed?

Ok. I'll buy that.

So, if two riders have the same drag, but one is heavier,
he has less drag per pound of mass than the lighter guy, he'll go faster.

GRZ, that is EXACTLY the statement I specifically asked you
to agree or disagree with earlier.

You disagreed, which is what launched me into all of this
nonsense.

Apparently you have now changed your mind.
Never!grzy
Sep 21, 2001 3:17 PM
JFC - you're the one that just went full circle.

"So, if two riders have the same drag, but one is heavier, he has less drag per pound of mass than the lighter guy, he'll go faster."

This statement is correct and it's what I've said from the beginning, but it's NOT because he's heavier, it's because of the drag. Don't you get it? You're so close. The mass isn't the cause, however it is indirectly related to the drag and can vary significantly depending on a bunch of things. It isn't hard to imagine either the heavier or lighter rider being faster or slower depending on how the drag works out.

I'm beginning to think that you don't fully understand the physical laws nor the reasonng that goes with them.

Gimme some time to work on the FBD.
Mater of the obvious and twister of truth until he's right!BS POLICE
Sep 21, 2001 3:28 PM
THis guy is a master. Lots of practice apparently!
isn't that where we started?Dog
Sep 21, 2001 1:09 PM
"it's all about the mass/drag ratio"?
EXACTLY!grzy
Sep 21, 2001 6:18 PM
My original point has been that the big guy gets down the hill first because he has a higher mass to drag ratio, but it's not attributable to his larger mass but rather his relatively lower drag. It sounds like hair splitting and symantics, but it's not. The vacuum free fall is but one classical way to demonstrate that it's not dependant upon the mass. I didn't make this up - it was around long before any of us were born. The analysis from either and energy ballance or straight equations of motion from any physics book reinforce this. The reality of the situation is attributed to the subtle differences in how the drag force behaves and the fact that we don't live in a vacuum. Since drag isn't dependant upon mass the controlling factor for the descent is the drag force alone. Relating the two is best done by a ratio and the statement:

"The mass with the relatively lower drag will get down first."

The rub is in describing what's meant by "relatively." Given that humans have roughly the same density their drag tends to increase with weight, but at a less than linear rate. People see a larger person descending faster and spuriously attribute it to their weight since that is what is obvious to them. Every one knows their weight, no one knows their coefficient of drag. Subscribing to the KISS princple usually gives you the right answer, but it's for the wrong reason.

Trying to explain this via the internet with people that only have a fuzzy grasp of physics and very strong opinions is like trying to push on a rope. I guess this is why I don't teach engineering, but rather practice it.
The Eddy Merckx therory of descending.....DINOSAUR
Sep 20, 2001 2:18 PM
Don't brake lot's.....
Haven't quite mastered this oneChris Zeller
Sep 20, 2001 2:30 PM
G E E K ! ! ! nm :o)DT
Sep 20, 2001 6:11 PM
$64,000 question:pfw2
Sep 20, 2001 9:43 PM
150lb rider and 190lb rider. Hill, ascent: 2 miles 10% grade, descent: 1.5 mile 15% grade. Power output: 150lb guy averages 350watts, 190lb guy averages 400 watts. 3 mile an hour headwind. Both bikes identical (except one has Shimano one has Campy). Both riders had Wheaties for breakfast. Start line is at the base of the ascent, finish line at the bottom of the descent.
Starting together, who gets to the finish line first?
Answers on my desk 1st thing in the morning.

I hope that keeps the engineers amused for a few hours, might even keep some of the Campy/Shimano fanatics busy too.

Teaching aids here: http://www.improbable.com/
not enough infoDog
Sep 21, 2001 5:34 AM
Got to know drag numbers.
D'oh! (nm)pfw2
Sep 21, 2001 8:57 AM
Ok, that's it. I'm officially taking my B- in physics (both sem's!) andbill
Sep 21, 2001 9:56 AM
voting with Grzy. I think that he's right. Regardless of whatever other forces are at work, the effect of gravity on mass always, 100% of the time, results in no greater or lesser acceleration than it does on a different mass. My attempt at understanding the alternative theories with the wiffle cyclist model is hereby withdrawn, because, in that model, not only does the horizontal force of launching said wiffle cyclist fail to remain constant through the trajectory, as does the vertical force of gravity on a descending cyclist, but, while it may say something about ballistics, the model does nothing to change Sr. Galileo's conclusion. I'm not sure of all the ins and outs of it, but there is something wrong with positing two cyclists possessed of the same drag coefficient but of different masses, because that would mean that they are of different densities and shape and stuff. Definitely, definitely stuff. I think.
RE: not enough infopfw2
Sep 21, 2001 1:01 PM
OK, two options: 1) riders have the ability to hold their breath for a very long time and are cycling in a vacuum, both are using identical wheels with Mavic hubs.

Option 2) - much more fun. This weekend everyone should should ride with a partner to a local hill and do the experiment, both up and down so as to test the stored energy theory, swap bikes and do it again. If the bikes are too disimilar at least do the descent part of the test, no pedaling of course. Will prove nothing but think of all the post-fodder for next week.

A few years ago Bike magazine tested tires(by rolling the tires down a hill, NOT by riding them) and microbrew beer at the same time. Maybe we could get a couple of mountain bikers and repeat our downhill velocity test multiple times, 1 beer per run, untill the bikes are no longer rideable. That was (weak) humor flameboys.