Childish///AMG

Wow, I'm glad that I went to a "Public" high school :p :p

evanswan

Raced an older m5 sometime back. I agree with you. I did a straight freeway from 0 to 100 and it was so even it was funny. He would pull 3 ft when he got to higher revs and then he shifted and it's impossible to be as fast as the auto tran on the amg.

Erik

Long time since I laught so much of any of your posts Mr " I have a Master in EE and plently more....."

Kind of funny how you patronize just about anyone not imidiatly agree to you statements, but when you get some back you dont " appriciate" it.... LOL.... well done Impro.....

Does not look good when a super genious like youselves with an " Master in EE" manage to mix up Newton with Einstein, and thinking the English have the throttle the the left side... LOL... Amazing....


BTW: IMO Reggie was not at all Patronizing you, he was only pointing out a few things, cant see how you got offended by that at all....

In comparisson to what things you have said to and about me, his little neutral comment was more like a love statement



Now, back to point.

You claim with all your expertice, education and brain, that the if the body ( meaning excact the same Cw) of the car is equall the main formual to use is F=ma.

Indicating that if Cw is equall, the only thing important, is the weight to power to weight relation.
Since I do not have a Master in physichs I can not play with this formulas like you, therefor I can not prove my theory.

Meaning in your world I am an idiot ,.... right


So,... according to you a 100 Hp 500 kg C-class MB will have excactly the same performance as a 400 Hp 2000 Kg C-class MB.
a 800 Hp 4000 kg C - class as so on......
Look away from the traction and rolling resistance.

They accelerate equally fast 0-62mph, 0-125mph +++ and have the same topspeed.......That is basically what you are saying.

I would agree if this was done in a vacum with no drag force at all, BUT as you ( probably) know on planet earth we have this thing called drag.


Using your logic that also means a 10 hp 50 kg car should be able to get the topspeed and acceleration of an 400 Hp 2000 kg car, again since the Drag and Power/weight is the same.

One question, since F=ma is the only thing important if the Body of the car is identical ( Cw ), the acceleration will be reduced to half if we double the weight, correct ?,
then using you theory will also the Topspeed be reduced to half.......?

Also according to you, the cars will be identical in highspeed acceleration , they will accelerate equally fast all the way until they do not manage to acelerate anymore, meaning until they reach the same equal topspeed.


Your physic Professor must be very proud , if I where you I would go back to school and ask for my school money back, that was probably a waste of you parents money....

Improviz

I've been waiting for your little celebration post, Erik. Welcome back. I prepared this in advance in anticipation of your crowing...

I've proven you wrong before, Erik, so please spare the cartwheels; in fact, I shall do so again, now.

Yes, Reg was correct: I did forget to divide the NET force by the mass in your hypothetical case. So I went and dug out my physics texts, researched automobile acceleration online some more, and ran some numbers.

Recall that we are discussing the C32 and the M5 at high speed, yes? And your original argument was that the M5, by virtue of its higher power, would outaccelerate the C32 at high speeds. You claimed that physics backed this up, correct?

Yes, correct. You wrote the following:

Here are the physics. Unfortunately, they do not back you up. We'll do a nice little high speed run, from 100-130 mph. To go beyond that would take more than one gear, and as I shall show, the M5 would lose up there anyway...

Specs:
C32: width = 68.0, height = 52.2 => frontal area = 3549.6 in^2 = 2.2900599 m^2; Cd = 0.27; horsepower = 349 , mass = 1636kg

M5: width = 70.9, height = 56.6 = => frontal area = 4012.94 in^2 = 2.5889884 m^2; Cd = 0.31; horsepower = 394, mass = 1818kg

So what would the drag force be for each of these cars at 100 mph and at 130 mph? 1 mile/hour (mph) = 0.44704 meter/second, so 100 mph is 44.7 m/s, while 130 mph is 58.12 m/s; 155 is 69.29 m/s.

Now, the equation for drag force is Fdrag = 0.5CApv^2.

C is drag coefficient

A = frontal area of the car

p = air density, which is about 1.2 kg/m^3 at sea level at normal temps.

So for the C32, we have Fdrag = 0.5*0.27*2.29*1.2*v^2 = 0.371v^2 This gives Fdrag = 741.3N at 100 mph, and 1253.2N at 130 mph.

For the M5, Fdrag = 0.5*0.31*2.59*1.2*v^2 = 0.482v^2 This gives Fdrag = 963.1N at 100 mph, and 1628.2N at 130 mph.

Uh-oh, Erik...looks like that little 13% advantage in Cd that you were telling everyone was "identical" is NOT so identical, and does a very nice job of torpedoing your theory. Also note that although I'm not going past 130 mph, at 155 mph, or 69.29 m/s, the M5's drag force would be 2314N, while the C32's would be 1781N, so we can see that past 130, it gets even worse for the M5.

Now all we need is the force at the wheels. This is calculated by taking the torque (again, to do it over the whole range we'd need to integrate, but I don't have torque curves and even if I did, I'm too lazy) times the product of the gear we're in with the final drive ratio, divided by (radius of the wheel)/2. For the M5, the radius of the 18" rims is 0.4572 meter, while it's 0.4318 meter for the C32's 17" rims, giving the /2 as 0.2286 and 0.2159 respectively

We will be in fourth gear in both cars for this run. The C32 has a 1.00:1 fourth gear through a 3.06:1 rearend, with peak torque of 332 lb-ft = 450Nm. So peak wheel force is 450Nm*3.06/0.2159m = 6338N

But as I pointed out, I cannot use this over the entire run, as it is PEAK...or can I? Let me look...we know that by definition horsepower is defined as torque*rpm/5252, so we can compute the torque at the horsepower peak; we know from the manufacturer the peak torque and the peak torque's rpm, so we can assume a linear drop (not the most accurate, but I would argue it's more accurate than assuming a naturally aspirated engine makes peak torque for the entire run!!) in torque per rpm.

So, let's compute the drop off from peak torque to horsepower peak for each car: M5 makes peak torque of 368 lb-ft at 3800, while at 6600 w/max power of 394 it'smaking 394*5252/6600 = 313 lb-ft. Assuming linear drop-off from peak torque to peak horsepower, it drops at (368-313)/(6600-3800) = 55 lb/ft / 2800 rpm = 0.0196 lb-ft/rpm. (As it turns out, we won't need this, but I left it in here anyway).

Mercedes rates the C32 makes at 332 lb-ft of torque from 3000-4600 rpm. At 5500 rpm w/max power of 349 it's making 349*5252/5500 = 333 lb-ft....which means that they're pretty obviously NOT giving the true peak torque for this motor, which is obviously somewhere north of 332, probably in the 350+ range...but we'll use 332 for this run. But since it's stated max torque is 332, and its torque at its 5500 rpm power peak is 333, the only possible conclusion is that it's making 332 lb-ft OR MORE from 3000 rpm all the way up to its power peak at 5500 rpm, which means it's producing a *minimum* of 6338N from 3000-5500 rpm.

And in our little run, from 100-130 mph, the C32 is spinning at 4260 rpm to 5530 rpm, so we can certainly assume a flat force curve over this range giventhe factory horsepower and underrated torque ratings.

The M5, in contrast, loses torque and thus wheel power steadily from its peak at 3800 rpm. As calculated before, it is spinning at 5100 rpm when it hits fourth at 102 mph. Using the loss figure calculated before, we can see that at this speed (again assuming linear drop off for ease of calculations), the drop in torque should be 1300rpm*(-0.0196 lb-ft/rpm) = -25.5 lb-ft. This would place it at 368-25.5 = 342.52 lb-ft at 102 mph, or 464Nm. At 130 mph, it should be spinning at 6500 rpm, and hey, this is close enough to its horsepower peak (which is at 6600 rpm) to save me another calculation: up above I already calculated that at the 6600 power peak, it's producing 313 lb-ft, which comes out to 424Nm.

Now let's look at the M5 at-wheel force figures: Wheel force at 100 mph is 464Nm*3.87/0.2286m = 7855N. Wheel force at 130 mph is 424Nm*3.87/0.2286m = 7178N.

This gives the following peak force figures for each, along with corresponding calculated acceleration (a) numbers using a = f/m:


FC32 @ 100 mph = 6338N - 741N = 5597N => a = 5597N/1636kg = 3.42m/s^2
FC32 @ 130 mph = 6338N - 1253N = 5085N => a = 5085N/1636kg = 3.11m/s^2


FM5 @ 100 mph = 7855N - 963N = 6892N => a = 6892N/1818kg = 3.79m/s^2
FM5 @ 130 mph = 7178N - 1628N = 5550N => a = 5550N/1818kg = 3.05m/s^2

BUT: remember: I excluded the rolling resistance, which will be higher with the M5 due to its higher mass *and* its larger tires!!! *AND* keep in mind: rolling resistance increases with velocity (though not with the square of velocity as with drag force)!! Iow, if we add in the rolling resistance, the advantage at 100 mph would probably evaporate entirely, and the gap at 130 would be more in the C32's favor.

Which I shall now prove: by golly, we want some accuracy here, don't we?? So let's add it in!!

Rolling resistance is defined as ur*n, where ur is a velocity-dependent constant, and n is the normal force = mass*gravitational force. Gravitational force = 9.8m/s^2. For each car we must calculate the rolling resistances with speed:

The coefficient for the constant is related to speed as shown:

fcroll = c0 + c1v + c2v^4

where:
c0 = 0.02

c1 = 0.00002

c2 = 0.000000002

This results in following: 0.029 @ 100 mph, 0.044 at 130, and 0.067 at 155. HOWEVER, this is for a standard width tire. Both the C32 and M5 have wider tires; in particular, the M5's rears are 12% wider than the C32's, and its fronts are 9% wider. But let's be generous to the M5, and assume only a 4% increase in rolling resistance, even though the tire width differential is double this in front and triple it in back (if anyone has width-specific equations, feel free to contribute them). This will also increase the C32's over the "normal" width used to compute the above equation, BUT if we simply normalize with the C32 as the base and scale up the M5's accordingly, the results will be the same from a percentage of difference. So I'll use .030 and 0.046 for the M5 at 100 and 130 respectively....

FrrC32(100mph) = 0.029*1636kg*9.8m/s^2 = 465N
FrrC32(130mph) = 0.044*1636kg*9.8m/s^2 = 705N

FrrM5(100mph) = 0.030*1818kg*9.8m/s^2 = 534N
FrrM5(130mph) = 0.046*1818kg*9.8m/s^2 = 819N

So now the net forces and accelerations (a) become:
FC32 @ 100 mph = 5597N - 465N = 5132N => a = 5132N/1636kg = 3.14m/s^2
FC32 @ 130 mph = 5085N - 705N = 4380N => a = 4380N/1636kg = 2.68m/s^2

FM5 @ 100 mph = 6892N - 534N = 6358N => a = 6358N/1818kg = 3.49m/s^2
FM5 @ 130 mph = 5550N - 819N = 4731N => a = 4731N/1818kg = 2.60m/s^2

Which erased the advantage even more as predicted, but as shown previously even assuming zero roll resistance, at high speeds the M5's acceleration grows progressively worse compared to the C32.

So, what do we see? Well, we see, Erik, that you were wrong: the M5's higher-speed acceleration is *worse*, both because of its torque curve's drop off as speeds increase and its higher drag, while the C32's flat torque curve due to forced induction and lower drag both give it an advantage.

In other words, oh great master of physics, while you were correct in the drag force's increase with the square of velocity, you were incorrect in not comprehending that:

1) higher mass gives higher rolling resistance which also increases with velocity;

2) although it does decrease the drag force component more highly, which helps, higher mass decreases the accelerative force component, which hurts acceleration at all speeds;

3) engines with forced induction have an advantage thanks to their wider, flatter torque band, which increases the work done over time. Again, remember that I used the factory's rated torque and horsepower figures for the C32, which are suspect given that the C32 at its 5500 rpm horsepower peak produces more than its rated peak torque...highly improbable;

4) the M5 has a poorer drag coefficient and larger frontal area, which makes it much more susceptable to your beloved high speed drag.

5) contrary to your claim, the vehicles do NOT have the same drag coefficient; the C32's is 13% better than the M5's, AND has a smaller frontal area.

So who's physics prof is happier now, Erik? I must have missed the part where you actually produced any equations or did any mathematical analysis in this post...can you please point it out for me?

reggid

i think your wheel radius might be wrong! It seems you only accounted for the rim size and haven't included the sidewalls in your radii.

roll rad = 0.5 * rim size + 1 * (width x series)

whether this affects the overall trend significantly

Improviz

I left this out because afaik, the rubber forces one to introduce a constant, and I'm not sure what this was...but if we ignore it (or if I'm remembering incorrectly again), the diameters would be:

C32: 245/40-17 = 24.72" = 0.627888 m => r = 0.314m

M5: 275/35-18 = 25.58" = 0.649732 m => r = 0.325m

This would reduce wheel forces accordingly for both cars....but I'm not sure if you can treat the radius as one piece when it's comprised of different materials.

Trekman

Well me and my friend went at it with our cars, we were are a stop light, I shut the ESP and put my foot on the brakes and had my car about 2000 rpm. the light turns green and he get a jump on me by half a car, I feel the car slipping my tire didnt grap so I let my trottle go until I got some grip the hit it again, I started to pull on him by half a car when we shut down at about 80 miles. It was fun have to re do the race again. ohh I forgot we had another car that tried to race with us the same time a Saab 9 5 turbo he was way behind. LOL I think BMW M3 are as fast as our C32 and its all about the driver! I know he will spank me on the turns but my *** will not hurt as much cuz of his stiff suspension.

sunchild

smell the smoke? .... cause someone got burned.

Erik

Hi Impro

BTW, Impressive calculation. Probably correct to, you have probably checked and double check it this time

Maybe your Professor is a little proud after all.

Since you actually behaved nice in this post I shall do to. Thanks.
Not excactly prowd of my behavior, but you honestly asked for it in several other posts.

So shall we both try and act adult from now on and accept that we do not always share the same taste and opinions..

But I belive you have missunderstood me a bit This was NEVER a MB vs BMW disucssion on my part.
Only my remark where I wrote that the M5 will pull in higher speed was.

Offcourse I could be wrong on that, but I still belive the M5 will outrun the C32 at high speed...

Your calculation show otherweise, they actually show that the C32 should be much faster........ and that real life test does not indicate.....

Regarding the drag.....since they has several time been refered to as equal in drag I assumed it to be so. My bad.


BUT

Later it was a discussion about the principle , thats why I specifically used a "phantom" C-Class with variation on weight and power BUT same power to weight ratio..
I still belive what I wrote. No need to repeat....

Have you tried to do you calculation with 2 identical cars, same power to weight, but huge difference in total power ?

Have a nice day.

reggid

The force pushing the car forward acts at the tyre road interface, so you need account for the actual radius and the fact that there are two materials doesn't really matter when dealing with force and torque as its just an equilibrium type problem (rigid body mechanics). If the engine applies a torque which is transmitted to the wheel the road applies an equal an opposite torque by virtue of friction at the tyre road interface. If you include rotational inertia (probably not significant at that speed and difficult to find) and driveline losses the two cars stack up well against each other.

Improviz

Well, I do understand this stuff, although I may not recollect it perfectly at times!

Well, I do confess to getting annoyed, but there was a reason for that: if you look at the calculations I performed, note that even after totalling the forces acting on the vehicle (*including* the drag force ), the equation governing its behavior *still* boils down to (per Newton ) f=ma, so a = f/m, or if you prefer a = fnet/m. So your assertion that mass is less and less of a factor at higher speeds was and is still incorrect; acceleration is inversely proportional to mass irregardless of whether the object in question is in a vacuum or in air.

Air *does* of course change the outcome, but this is *not* caused by any reduction in the effects of mass as velocity increases as you were maintaining; it is through an *increase* in the effects of air! The drag force is independent of vehicle mass; a 1000kg car or a 4000 kg car with a 2m^2 frontal area and a 0.30 Cd will have the same drag force at any given velocity.

So yes, I plead guilty to getting rather annoyed, but I maintained, correctly, Newton's laws hold everywhere, not just in a vacuum, even if I got momentarily confused as to their source and implementation. I confess to having a bit of a temper when I'm challenged on physical laws which I know to be correct, even if the passage of time since college had diminished the knowledge of their calculations somewhat.

Opinions are fine, but again, I was mad at the way I was being treated in what was essentially a factual dispute about Physics. But I'll endeavor to keep a lid on the temper in the future. Oh, and before I forget, likewise: you too have a nice day.

Possibly...but this has nothing to do with mass being less and less of a factor at high speeds.

That's because the position function is a function of velocity at t=0 plus velocity multiplied by time. If you take a limit as t -> 0, the M5's greater launching ability (due to its 12% wider rear tires and better traction from the higher friction coefficient caused by the tires in conjunction with its higher mass) gets a big kick from its initial launch. This advantage will be realized at all speeds, although its effects will be diminished--but will by no means disappear--when 0-xxx speeds are read from any given road test.

I personally believe that magazines should conduct full rolling-start time to distance runs, say from 5mph to the 1/4 mile, as these would be more indicitive of what would happen once the cars are rolling, but unfortunately all we have at best is a small subset of this.

But anyway, I digress; back to mass and its effects on high speed acceleration:

The key phrase here is "huge difference in total power", not weight. You are trying to show the effects of varying *one* quantity, mass, by varying *two* quantities here, mass and force. But a result obtained by varying two quantities cannot be interpreted as being caused only by the variation of one quantity. Since two vehicles of the same mass, frontal areas, gearing, wheels, etc. will produce the same force at the wheels at any given velocity, the thing which will determine acceleration at high speeds (or low speeds) is how much force there is to push the vehicle forward minus how much force there is resisting its forward movement, divided by its mass of course. Drag force does increase with velocity, but this is independent of mass.

If you want to see how mass *only* affects a vehicle's acceleration, you must vary *only* mass. So let's use the hypothetical Benz case again, except in this case rather than calculating a bunch of engine stuff, I'll simply keep engine force constant and vary mass. In this way the direct effects of varying mass with respect to high velocity acceleration in air can be determined. I will use the same calculated drag forces for the C32 from before at 100, 130, and 150 mph.

So let's say that Car A has mass of 2000 kg, Car B has mass of 4000 kg, and Car C has 6000 kg. All other things are equal. Drag forces were calculated previously at 741N at 100 mph, 1253N at 130 mph, and 1781N at 155 mph. Let us further assume for simplicity's sake that these vehicles produce 6000N at the wheels at 100, 5000N at 130, and 4000N at 150, although using the following equations you can simply plug and chug for any values you like; the end results will be the same provided the wheel forces are equal. I will also not be computing the rolling resistance as I am pressed for time, but bear in mind that with the heavier cars as I have shown it will be significantly higher, thus slowing them down even more at higher speeds.

Acceleration at 100 mph:
a(Car A) = (6000N - 741N)/2000kg = 2.63 m/s^2
a(Car B) = (6000N - 741N)/4000kg = 1.31 m/s^2
a(Car C) = (6000N - 741N)/6000kg = 0.88 m/s^2

Acceleration at 130 mph:
a(Car A) = (5000N - 1253N)/2000kg = 1.87 m/s^2
a(Car B) = (5000N - 1253N)/4000kg = 0.94 m/s^2
a(Car C) = (5000N - 1253N)/6000kg = 0.62 m/s^2

Acceleration at 150 mph:
a(Car A) = (4000N - 1781N)/2000kg = 1.14 m/s^2
a(Car B) = (4000N - 1781N)/4000kg = 0.57 m/s^2
a(Car C) = (4000N - 1781N)/6000kg = 0.38 m/s^2

As can be clearly seen: by varying *only* mass, acceleration is still the fastest in the car with the highest force-to-mass ratio. All other things being equal, this will be true at all velocities.

Also note that the relationships of the three accelerations does not vary; at 100 mph, Car B is accelerating at ~50% the rate of Car A, and Car C is accelerating at ~50% the rate of Car B and ~33% the rate of car A. This relationship holds true at both 130 mph and 150 mph, and would hold true at 100,000 mph.

Therefore, mass is shown to be equally important to acceleration at all speeds, all other factors being equal, as acceleration is inversely proportional to mass per Newton's second law.

Improviz

Yes, but what I was saying is that the torque will not be transmitted through the tires with the same efficiency as through alloy; due to the higher flex of the rubber, there will be a reduction of torque across that final few inches, as it will not transmit the torque as efficiently as the alloy, at least if I'm remembering my dynamics correctly...

In any case, this would be relatively equal for both vehicles, and thus should not affect the outcome of the calculations in either vehicle's favor....similarly, while using the overall diameter would result in a reduction of forces for both vehicles, the end result would be the same.

But the main issue is the effects of mass on high speed acceleration, all other things being equal, and clearly the inverse relationship of mass to acceleration still holds true regardless of velocity, per my last post.

SoulBladeZA

Caaaaaan you feeeeeeel the love toniiiiiiight

:p

Josh K

oh my god, what the hell did I stumble upon? It's Prof. Frink and Comic Book Guy arguing about the atomic weight of Bolognium. Everyone needs to take a couple of shots and nail a hot chick.

Childish///AMG

Dude, not all of them are like "Us" errr, you I don't drink, but I am definately nailing a hot chick

Improviz

Hmm...hate to be the one who breaks this to you, but a knowledge of physics and drinking shots + nailing hot chicks are not mutually exclusive.

And who the hell is Prof. Frink anyway?

Josh K

No one said they were mutually exclusive ( ); this thread broke down from a lost race story/request for rematch advice to a physics pissing contest. I was suggesting some recreational activities as an alternative to pointless bickering and hair splitting. Sure I could always jump in and throw some "mathematical blows" of my own, but I'd rather do something productive vs. try and dazzle forum members with my physics powers, or bowstaff hunting skills.

I'll let you go ahead and do the "last word" thing...

reggid

The tyre flex consumes power yes, but usually thats accounted for in the transmission losses since pure gears don't give the 15-25% driveline losses part of this is from tyre flex causing heat losses amongst other things. Since you didn't use transmission losses in your calcs i guess the reduced rolling radius you used does the same thing in the end.

Improviz

Oh, how about simply letting me express my opinion instead? Pointless? To you perhaps, but to me it was amusing, and a nice revisit to physics. Productive? I thought so, maybe you didn't, but it was my time invested in writing it, not yours, so thanks for the concern, but...I enjoy writing, I enjoy math, and I enjoy debate, and to me, well, enjoyable is productive in its own way.

And your projections aside, I wasn't trying to dazzle anyone, I was trying to bring some facts to the table, which is what I normally do in a debate. "Dazzling" has nothing to do with it; it's a kind of mental sport, you see...like chess, another one of my hobbies, which I suppose is also unproductive.

Have a good evening.

Improviz

I assumed roughly the same loss from each transmission, since there's no manufacturer that I'm aware of who will disclose this info voluntarily. And yes, there were probably at least a half dozen other second and third order variables which could also have been included, but I'm not *that* interested in making a point!

Anyway, I believe the analyses were sound and accomplished their intended goal.

mrankovic

Impro-man... Nice to see you being humble... Man you are burying poor old Erik...

mrankovic

Josh - welcome to an Impro-man thread!!

Impro-man - you seriously need to apply that mental energy to something like card counting at casinos or something.....

Improviz

Oh, I do that too...j/k!

Erik

I know, I agree in this. Never meant anything else. Inspite of the temperature in the post we probably mean the same thing, you explain with thery and I try with logic.

You know what they say, ....... a theoretical guy he knows why, but NOT how, the practical know how.... but not why....

Not putting either of us in any cathegory only, but its a difference.

Nothing you have proved or said with physics does change my views, only your numbers on the M5 and the C32 did supprise me a bit but that was not a something I had put much prestige into.
And I did not seriously belive you did not KNOW the affect of the total drag nor did I belive you did not know the difference of Newton and Einstein, just as you probably did not belive I meant that weight does not count when speed increased or.

So lets close the book on this one.

Improviz

Perhaps you didn't mean what you said, then, when you stated that mass is less and less important to acceleration as velocity increases, in multiple posts.

Well, as a Design Engineer, I have ample experience with both theoretical and practical, so it would be difficult to stick me in either one.

Again, just so we are clear: my debate has been with your contention, made in this post , that weight was "less and less important with increasing speed". I stated, accurately, that this was false per the equation f=ma, and I also stated, accurately, that it holds at high speeds as well as low speeds. You chose to argue this point, and you lost.

And make no mistake: this is the point you were arguing. You accused me of not understanding your post in inthis post, where you stated:

False. Acceleration is inversely proportional to mass, at all speeds, per Newton's second law, proven above.

You further argued that f=ma was not the correct equation to be using, in this post:

While I may have mistakenly attributed (later corrected) the authorship of the f=ma equation, and did, prior to looking at my old physics books, forget to divide the drag force by the mass, I was certainly correct in my assertion that this equation still holds, at high speeds, low speeds, wherever.

And I'll say this: when I was wrong about the authorship of the equation, I admitted it. When I forgot to divide the drag force by the mass, I admitted it, *and* ran detailed calculations. These calculations showed that I was still correct in my assertion that whether in air or not, f=ma still holds. Period. And yes, it is the correct equation to use. Period.

I have yet to see you admit that your assertions were wrong. But they were, and the equations above prove this beyond any shadow of a doubt. So please: trying to equate my mixing up the name of the physicist who authored one of the most important laws of physics or making a mathematical error with your arguing the law itself was incorrect due to drag force and publicly ridiculing my supposed lack of understanding of physics is a pretty weak attempt at a glossover.