I certainly agree that he is being excessively pedantic just for the hell of it (and doesn't have the fundamental disagreement with the others that the others keep wanting to say he has).

I already posted the formula a couple times.
If you don't believe me, then how about NASA?

https://www.grc.nasa.gov/WWW/K-12/airplane/termv.html

Of course it is! Change the weight with all else being equal under normal earth conditions and it will change the rate it accelerates, that is unless the surface area of the object is zero. Check the formula Shimagnolo posted earlier and you can see all the variables that affect the acceleration of an object - one of these is mass.

I think part of the problem is that people are using the terms weight and mass interchangeably...

Just so the latecomers don't need to wade through 6 pages of posts, here is #23 again:
---------------------------------------------------
Vt = sqrt((2 * m * g) / (rho * A * Cd))

where

Vt = terminal velocity,
m = mass of the falling object,
g = acceleration due to gravity,
Cd = drag coefficient,
rho = density of the fluid through which the object is falling
A = projected area of the object.

Do you see that term "m * g"?
That is called w_e_i_g_h_t.
---------------------------------------------------

And if you follow that NASA link I provided a couple posts up, you will see that they substitute the "m * g" term with "W" (weight), when they present the same formula.

I said, channeling that supramax guy
Somedood replied:
It's pretty funny that you say "of course it is," and then proceed to put a very significant caveat. Without that caveat...look again at supramax's statement. Perhaps you'll see how he's been playing with you all...

Yes, if you change the weight but also change the surface area/drag then it'll change the speed as well - hence the all else being equal. The only reason it wouldn't is if there were zero drag. If we lived on the moon then changing the mass of the object won't change it's acceleration, but sadly we live on a planet with an atmosphere.

And therefore the speed of a falling body is not dependent on its weight...per se.

Do you know math?
Here is the formula again (in the NASA form): Vt = sqrt((2 * W) / (rho * A * Cd))
Weight is the numerator of a fraction;
Increase weight -> velocity increases
Decrease weight -> velocity decreases

That is a direct correlation.

It is only a direct correlation if the two objects are of the same shape/external dimensions.

When dealing with air resistance, differences in shape have the potential to cause more of an impact on speed than differences in weight. My weight would increase if I were to carry an open umbrella on my shoulder as I coasted my bike down a hill, but it's quite clear I would go slower without the umbrella with me!

Don't you see the game that guy is playing with you?

False intellectualism. Logic failures. Self-contradiction. Redundancy. Irony. Pig-headedness. Plain stupidity.

Bump.

win

Yes... in a vacuum objects will fall at the same rate.

That fact has no bearing on the topic at hand since we are not riding in a vacuum.

I haven't read all 7 pages, but seriously, how did it get to this? Have people never ridden in pairs before? Regardless of your understanding of physics, surely at some time in your life you've gone down a hill with a fatter or lighter person? Whenever I tour, there's always some big guy that I can't catch down the hills, and by the same token, if I'm chatting to one of the girls as we approach a hill, I just have to say "see you at the bottom", because I'd need to brake to maintain the same speed as her pedalling.

Pfft.

Note: imi is the OP
When a thread goes 7pages, sometimes it's best to just focus on the first post...

Correct

Also Correct, there are more factors than just gravity.(this seems to be the majority of the thread's consternation)

"All conditions being equal" versus "extra girth" are a contradiction, since girth is going to effect air resistance; and I think you are primarily interested in the weight issue right? Please clarify exactly what you meant by this so we can clear up some of the confusion and arguments.
May I suggest modifying the original query to put both riders in recumbents with full aeroshells big enough to hold the larger of the two?

Uhh yeah... I'm not touching this point till the OP clears up some of the hypothetical setup to remove ambiguity.

Apparently you have not seen this, so here it is again, containing all the variables you are talking about:

Vt = sqrt((2 * m * g) / (rho * A * Cd))

where

Vt = terminal velocity,
m = mass of the falling object,
g = acceleration due to gravity,
Cd = drag coefficient,
rho = density of the fluid through which the object is falling
A = projected area of the object.

Do you see that term "m * g"?
That is called w_e_i_g_h_t.

I used an example earlier in the thread to illustrate this exact point. It was something along the lines of: assume that you have a parachute with a 1 pound weight attached. Now assume that you have a parachute of the same dimensions with a 1 ton weight attached. I'm not sure if I specified it in my earlier post, but assume that the weights are the same dimension. Anyone with even a cursory understanding of Newtonian phsyics would understand that (1) in a vacuum, these two objects would fall at the same rate, and (2) in an atmosphere, the heavier object would have a higher terminal velocity and would therefore fall faster. Others have provided similar examples.

Our troll's only response to these examples has been to repeat the principle that it is a "scientific fact" that these objects would fall at the same rate. He acknowledges that this scientific fact is proven in a vacuum, but does not acknowledge the effect of an atmosphere. Instead, he repeats his initial mantra regardless of whether it is responsive to the specific examples that are being offered as part of the discussion. You can call that being "excessively pedantic." I call it trolling.

True. And the equations for calculating the relative effects of each of those factors have been provided in this thread at least half a dozen times.

One of our physicists can correct me if I am wrong, but I believe that this is incorrect. Weight is a measure of the amount of force that gravity exerts on a body. If you carry an open umbrella, you are applying another force which is acting on the mass of your body in a direction different from the force of gravity, and therefore changing the net force impacting your acceleration and your velocity, but your weight (i.e., the force exerted on your body by gravity) is not changing.

Of course.

That formula doesn't even taking rolling resistance or the angle of incline into account. How can you expect us to take it seriously?

That is the formula for *freefall*, which is what Troll #1 has been blathering about.
I already posted the modified form for an incline previously.
If you want to see it, go back and find it yourself.

And you are correct that rolling resistance is not in there.
Being on an incline or not, does not change that.

I think you're overthinking this now. If I put on a jacket, I certainly weigh more than without it. If I unzip it and use it as a "parachute" like object to increase my air resistance and slow me down while falling out of an airplane or rolling downhill on my bike, this does not make me weigh any less (by any commonly accepted concept of the word "weight"). I go slower, but I weigh more, than I would without that weight on my body. That's all I was saying: that adding weight in a way that increases air resistance will not result in a faster fall - and therefore, if we want to be totally accurate, we can not hold that falling speed is dependent on weight, (since it is not always so).

We're saying the same thing. I confess that I misread your earlier post

as saying that your weight would decrease as a result of the wind resistance of the open umbrella. Sorry for the confusion.

It is dependant on all of the variables listed in the equation that Shimagnolo posted, among those is weight (mass * gravity). Some of those are pretty constant, such as gravity and the density of air - they do vary by altitude but just slightly. That means you're left with mass, drag coefficient and surface area all of which the speed is dependant on.

If something is independant of a variable, changing it and nothing else will not affect the end result. Falling speed is independant of the color of an object. A skydiver's falling speed is independent of a skydiver's music preference. Falling speed is dependant on mass, drag and surface area. You can also add in there friction from the bikes moving parts (bearings, chain, tires etc.)

Hi, I'm the OP

Thanks for the bounce back xenologer...

What I meant with "all things being equal" was that the only difference between the riders was that one was twice the others weight and would thus also have a greater cross section (bigger guy, I called this girth), but that they were on exactly the same bike, same tire pressure, riding in the same position (say fairly upright), on the same hill, no wind, etc etc...

From your answers (sorry I can't follow the math) would I be correct in concluding that:
1. The heavier rider's extra mass causes him to accelerate faster down the hill solely due to the effect of gravity in an atmosphere?
2. That his assumed larger cross section (which I called girth) would slow him down due to greater air resistance?
3. That his tires have a greater contact area with the road thus increasing friction (heat) thus slowing him down?
4. That the air resistance and friction, however, are less than his greater gravitational acceleration meaning that he would get to the bottom of the hill first?

To the other part of my question... Assume there is another hill after the first, the guys are still coasting, would the lighter guy catch up on the way up the hill or come to a stop further up the hill than the heavier guy (or the other way round)?

Hope this clarifies my question
Thank you for your time and knowledge

As for the coasting uphill part, I honestly don't know. Assuming a higher speed at the bottom of the hill by fatso, I will guess that he would coast a little farther up the hill, but, I'm not as comfortable with this assumption as with the downhill part.

If they are pedaling the uphill side, I suspect fatso will get dropped like a hot potato, assuming comparable fitness levels.