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Mr. Speed-Chump - One More Time.(15 posts)

Mr. Speed-Chump - One More Time.grzy
Sep 24, 2001 7:59 AM
I posted this in reply to your message titled "See if you can figure this out." late Friday night. I used your numbers and example to make my point - see if it makes sense to you.


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:


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.
Sep 24, 2001 9:34 AM
I'm in agreement with a lot of what you are saying.
I think we've lost track of the original issue between us,
which for me is this -

My thesis statement:
Given two cyclists with identical drag coeeficients and
frontal areas, the more massive cyclist will reach a
greater terminal velocity.

You had stated earlier that you disagreed with this, saying
it violated the physical laws.

Apparently you now agree with that statement, given what
you said above "Sure the heavier mass with proportionally less drag gets to the bottom first."

As best as I can tell, the issue we have now is more
semantics than anything. You want me to say that the
differences in velocity and/or acceleration ARE NOT
dependent on mass, but ARE dependent on the ratio of drag
to mass.

I will not say that, because it is obviously nonsensical.
How can a term that is dependent on D/m not be dependent
on m? If you change m, the value of D/m changes also.

One more example. Go to a mountain, tuck, and hit max
speed. Go home and pour your frame full of molten lead,
let it cool (!) and go do it again. We both know that the
second run will result in a higher max speed. You will say
it is because the ratio of drag to weight is lower. I will
agree, but add that since no external changes were made,
the drag charachteristics are the same. SO... you went
faster "BECAUSE" you were heavier.

You say above that mass does not appear in the drag term.
I agree completely. BUT, you say that the drag force
determines the results of the equation. That is only
partially true. What needs to be considered in the analysis
of a body's motion is the SUM of all forces acting on it.
In the examples we are investigating, the sum of the forces
is the the gravitational force (m*g) - the drag force. This
term is very much dependent on mass.

Will you ever be willing to discuss the situation in which the bodies have reached terminal velocity and are no longer accelerating? I believe it really is the simplest
case to discuss mathematically, as well as the approach
that is most relevant to cycling. You seem to be avoiding
it for some reason.

I know I can't make you specifically adress any of this if
you don't want to, but please read the paragraph beginning
with "one more example" and tell me what you think about
it. If you believe that the acceleration and velocity are
governed by the ratio of drag to mass, and you agree that
there is a conceivable situation in which two objects of
different mass have identical drag charachteristics, then
the unavoidable conclusion is that velocity and
acceleration ARE dependent on mass.

If I have somehow missed your point, please clarify it with
a very specific statement of belief so that we can be sure
of what we are arguing about before we continue arguing.

OK. I promised not to post on this anymore, and I broke
that promise already. I'm sure everybody is sick of seeing
this stuff, so if you would like to continue the discussion
(I'm game) please e-mail me at

I WILL NOT post on this topic any longer. If you would like
to relay our arguments / findings to the board, that's OK
with me, but I am done posting.

Frankly, I'm at my wits end trying to explain this to you.
Good call on not wasting your time.Bas Vanderwahl
Sep 24, 2001 9:54 AM
Its annoying.
Sep 24, 2001 1:18 PM
Hey, Mr. Ludite, you being facetious or hypocritical?

Your sniping is so unbecoming. If you got something to say, well then, say it.
Again...Ballistic CoefficientChris Zeller
Sep 24, 2001 10:27 AM
Again I will point to the ballistic coefficient. Speedchump, the ratio of mass to drag is known as the ballistic coefficient. I don't know what the hubub is all about, you both seem to be saying basically the same thing and what you are both getting at is ballistic coefficient. The higher the ballistic coefficient, the more the object behaves as if it were in a vacum (like ballistic motion) an iron ball falls in air faster than a feather because the ballistic coefficient is higher. The heavier rider descends faster because his ballistic coefficient is higher. You can prove this with free body analysis or energy meathods as you like but you come to the same answer. Demensionlal analysis is usually the easiest way to charachterize a system. The ballistic coefficient tells you everything you need to know about this system.
Again...Ballistic Coefficientgrzy
Sep 24, 2001 1:25 PM
Sure, but speed-chump still thinks that something falls faster b/c it's heavier. It doesn't matter what the facts, data, or proven theories have shown. He can't seperate the two concepts (gravity affects all objects the same regardless of their mass and that things falling through air are geverned by their drag) and is hell bent on being righteous and wrong. What you've got is someone with limited knowledge and the inability to get their head around it, much less understand what the rest of us are trying to say. Most people get past this during their freshman year of physics, but some don't even pass the course.
sounds like you're ballisticET
Sep 24, 2001 6:40 PM
You criticize another poster for sniping and then can't help but insert cheap shots at another. What a classy guy.

Maybe a final poll where we vote who did a better job of explaining it, you or speed chump, would be illuminating. You don't have to accept the results from us ignorant nincompoops, but it may give you an idea of how well you are expressing yourself. FWIW (nothing in your eyes), not very. I think speed chump did a better job than you did. You failed to make clear the distinction between drag as determined by frontal surface area and that coming from an object with greater mass due to gravity when not in a vacuum. Do you claim it's the same thing? Well, then why did you also ignore repeated pleas to explain the case where the frontal surface area is the same but the masses are different? And yes, I agree with speed chump that you backtracked from what you said previously without admitting to it. Sure, you know a lot, but that doesn't mean you explain yourself well, nor that you can't ever be wrong.
Some unbidden crankiness crept in, but I also think that bothbill
Sep 25, 2001 7:14 AM
ends of the debate eventually moved their battle lines a bit to something more tenable. What I got out of this was that it pitted Grzy's umbrage at the suggestion that weight is what matters and his attempts to show that it doesn't against Speed-Chump's somewhat easier to follow but, let's face it, sometimes misleading proof that mass, in this model, with certain assumptions, is decisive without giving Grzy's argument its due. Let's put it this way, I think that I learned more from Grzy's postings, as ably abetted by Mr. Chris Zeller with a crucial assist, about the phenomenon than I did from Speed-Chump. Not to say that I think that Speed-Chump misunderstands as utterly as Grzy at times said but that Speed-Chump has done little to disabuse us peasants from the notion that it is weight that matters as opposed to, I believe that I now understand, the behavior of an object's given mass at a given density and volume, with all that those values and the relationships among them imply, moving through a fluid, the effects of which exist without regard to gravity's effect on mass (i.e., it ain't mass, with density posited as a constant, it is the relationship between mass and volume that is decisive). To illustrate that it isn't gravity's effect on mass, let's suppose that Nolan Ryan were going to bean you with a thrown object on the space station, so that now gravity is not an issue but air is, would you rather be hit with Speed-Chump's 1kg hollow metal sphere or his 10 kg solid metal ball (same volume, different mass)? Okay, now would you rather be hit with a 1 kg solid steel sphere or a 10 kg solid steel sphere (more like the cyclist's, since the density is the same for both, greater mass also means greater volume)? All of them leaving Nolan's hand at the same initial velocity? Intuitively, we know the answers. A little different, since the acceleration of gravity continues to act on cyclists on earth (all equally, thank you very much, so that it actually doesn't matter a whole lot -- you'd still want to be hit with the less massive per unit of volume object), but I think that it goes some distance to show that it is mass vs. volume as the decisive ballistic characteristic and not weight (i.e., gravity's force on the mass) or mass in itself.
Sep 25, 2001 9:42 AM
Try this link below and the related one for air resistance. It's what I've been trying to say.

Being correct and being able to explain it are different things. With apologoies to Dog and others, that's why we have lawyers.
troublesome commentDog
Sep 25, 2001 10:09 AM
I think this is the comment that messed us up, grzy: "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." grzy "Hey, Grzy, larn me physics. And make it stick this time." 9/20/01 9:31am

not sure this is so troublesomeET
Sep 25, 2001 11:59 AM
I may just agree with grzy on this if by "drag coefficient" he means, as I think he does, to include the effects of mass, as opposed to just a measure of resistance to surface area. Let's say one cyclist has a very wide torso and weighs more, another a narrower torso and weighs less, but their net "drag" in descent could be the same, in which case they descend at the same rate.
We're Hopeless.grzy
Sep 25, 2001 1:15 PM
It's like talking nuclear science to artists. Not to put anyone down, but we're at two different places. I didn't/don't communicate as well as I'd like - that's why I'm an engineer, not a lawyer.. I'm sure I made some errors in the way I stated things, but you can't change the laws of nature - and at the end of the day that is waht matters. We start talking in generaltites then move towards specifics, but the problem is that the language means differerent things to different people and logic seems to be optional.

I figure that someone else has done a much better job of explaining it so I passed on what I deem to be a useful link. If you want to take me to task for making some misstatements, then I'm going to need a spin doctor. However, the people with the incorrect misconceptions and muddled understanding are going to need more than that. If I say I'm wrong on one of my statements would anyone else they were wrong on their entire reasoning? I think not - this is the net - it's not like anyone is actually all that accounatable. Give me a physics test and I may not get a perfect score, but I'll certainly score higher than my opponents.

Take a look at the link and carefully work through it and see if your understanding isn't improved and that I'm much closer to being right. Do I get my Special Olympics medal now? ;-)
I agree, this comment was the problemChris Zeller
Sep 25, 2001 2:27 PM
Yep, I think GRZY and Speedchump know what they are talking about, but this statement is dead wrong.

The basis for the most realistic simplified example is that the drag coefficient of a heavier rider and a lighter rider are nearly the same. GRZY stated something to the effect that drag coefficient and frontal area are not linear functions of the mass. This was particularly insightful.

If you look at the ballistic coefficient, the mass term of the heavier rider will be greater while the drag term would be smaller giving the heavier rider a higher ballistic coefficient. The rider with the highest ballistic coefficient will win because his motion more closely approximates ballistic motion (motion in a vacum). Imagine a feather and a feather cast in lead. Both have the same shape and product of drag coefficient times frontal area, but with the higher ballistic coefficient of the lead feather, it will fall faster. So will the heavier rider.

Prove this with energy meathods or kennematics as you like but demensional analysis above is much easier.

Given identical drag constants (area*CD) the heavier rider will descend at a HIGHER rate. Indeed this is what happens.
Sep 25, 2001 2:32 PM
The misleading part is attributing the cause to the weight when it's the drag that makes the deterimination and is the cause. Some would say this is semantics, but it's not to the people that study and understand it. The common over simplifcation is misleading and confusing.

I guess if you're going to say something over and over and over again, there's a decent chance you'll mess it up once or twice. Call it noise and crank up the filters. ;-)
I think we all understandDog
Sep 25, 2001 2:49 PM
I agree. Just at one point, I got (more)(all) confused about your position when it seemed like you were arguing that given the same drag (in air), a heavier and lighter rider would descend equally. I didn't think you meant that, but I think that I and others understood you to say the same that, and that sparked a bigger debate.

You've done a hell of a job enlightening me. I pretty much have a more thorough understanding of something we all know from observation to be true. I appreciate it.