

03052016, 06:04 PM

#121

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Quote:
Originally Posted by nitrodann
Dont know why you are getting DeKittied, you are right.
Point of this thread is to get a real world graph to prove it.
Dann

It happens. Whoever did it should read more before posting more.
tripple88a posted some graphs that illustrate the relationship between torque/power/acceleration very accurately.
It's funny that some people here are so assertive about being correct/understanding things while proving how little they actually understand.
If you read this thread, there are some hilarious comments in here spoken as facts. My favorite is "engine torque and acceleration have no correlation  Sav".
The best part is, I drove my car earlier today, and when the turbo spooled and torque went from about 100 ft*lbs to over 200 ft*lbs, I actually noticed the acceleration change. Crazy!
EDITED for clarity.



03052016, 06:16 PM

#122

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I think he was trying to show those without much understanding (me?) that horsepower is doing the work.
But yeah its way elementary level for the thread and serves to confuse.
Last edited by nitrodann; 03052016 at 06:36 PM.



03052016, 07:49 PM

#123

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Some arm chair engineering. No “Real” data from some world class engine development shop just a discussion of physics…
The condensed version (with no friction or other losses) is:
HP=T (ftlbf) * RPM / 5252; F(lbf)=m * a(ft/sec^2); T(ftlbf)=F (lbf) * Radius (ft); why gear ratio matters, and the result …
a(ft/sec^2) = (HP * 5252 / RPM) * (GearRatio/TireRadius(ft)) / mass(slug) Or,
a(ft/sec^2) = HP * 5252 * GearRatio_______
RPM * TireRadius(ft) * mass(slug)
And, Kinetic Energy KE = m * v^2 / 2 is not relevant to the acceleration discussion.
Some 100+ posts already on the topic... MOST of what is here has been presented already in one form or another. I’ll try and bring it all together in a single post. (Many previous posts in this thread have little basis in reality… some seriously enthusiastic authors have posts that are comically bad (from an engineering and physics perspective). I’ll try and refrain from flaming – but overall, if you don’t know what you are talking about, stop posting the crap).
Not brief, sorry for the long post, flame away if you want or Grab a beer and take the time if you are interested if not, jump to the next thread… (I really don’t care one way or the other).
______________________________
Part 1) Why Horsepower?
HP (Power) is a linear equation of Torque and RPM.
HP = T * RPM / 5252 (OR: T = HP * 5252 / RPM)
Horsepower (HP) is a common measurement of “Power”
Torque (ft lbf) is a measure of rotational force
There is NO motion (RPM) in Torque, only a force. Torque can be applied to a stationary nut and bolt. Doing so requires NO HP as the speed (RPM) is so slow, HP is effectively zero.
RPM (revolutions/min) is a rotational velocity
5252 is a constant allowing us to use the above units of measure in the equation (ft lbf and RPM)
For ANY race (or other) application the “gearing” will be selected to match the engine’s operational RPM range and power (HP) available to a given vehicle speed.
Consider the following example where we have the same sports racer that can be powered by 3 very different engines ALL WITH THE SAME 100HP output but, a different operational RPM for the peak HP:
Diesel engine – 100 HP Peak at 2626 RPM. Torque at HP peak = 200 ft lbf
Gas engine – 100 HP Peak at 5252 RPM. Torque at HP peak = 100 ft lbf
A sport bike engine – 100 HP Peak at 10504 RPM. Torque at HP peak = 50 ft lbf
Gearing:
Any gear ratio between the engine and wheels does NOT change the power at the wheels. What is changed is RPM and Torque. In an ideal transmission a 2:1 gear reduction will result in engine RPM /2 AND increase the torque by 2x. Example:
The Gas engine above: 100 HP, 5252 RPM, Torque 100 ft lbf.
With a 2:1 gear, the transmission output RPM drops to 2626 RPM, Torque increases to 200 ft lbf.
HP = 200(ft lbf) * 2626 RPM/5252 = 100 HP. As expected.
T(ft lbf) = Force(lb) * radius(ft) = a Linear Force (lbf) applied at some lever arm radius (ft).
An equivalent equation is: Force (lb) = Torque(ft lbf) / Radius(ft)
Applying this second equation, we can determine the force pushing the car down the road for any given axle torque and tire radius.
Under some condition, we might determine we need 400 lbf pushing us down the road at the SAME ROAD SPEED regardless of which of the 3 engines we have in our racer. Using a 1 ft rolling radius tire in all 3 cases:
Force(lbf) = Torque(ft lbf) * GearRatio(n:1) / TireRadius(ft)
The Diesel engine might need a 2:1 overall gear ratio. Torque x gear ratio = 200 ft lbf x 2 / 1 ft = 400 lbf
The gas engine generating HP at a 2x higher RPM would need a 4:1 overall ratio for the same road speed. Torque x gear ratio = 100 ft lbf x 4 / 1 ft = 400 lbf
The sport bike engine with a much higher RPM for its peak HP would need a 8:1 overall ratio at the same road speed. Torque x gear ratio = 50 ft lbf x 8 / 1 ft = 400 lbf
So we have, 3 different 100 HP engines with very different operational RPM ranges and very different Torque outputs. ALL when geared correctly provide the same “road speed” and the same “Force” to the road (400 lbf) for acceleration… (or possibly we need that force to overcome air drag for a land speed car).
For any given HP and RPM we can calculate the Torque. Using the Torque, gear ratio and tire radius we can calculate the linear force pushing us down the road…
_______________________________
Part 2) Newton’s second laws of motion is F=m * a (Or, a = F/m)
Force (F) equals Mass (m) times Acceleration (a).
 This is a straight line motion equation.
 This simplified equation does NOT include ANY resistance factors. We will ignore all resistance for now (aerodynamics, brake drag, gear efficiency, wheel bearing resistance, tire rolling resistance, ...)
 We will also simplify our car acceleration to ONLY include the mass of the car not the secondary effects of rotational inertia and the Power required to overcome this inertia also (flywheel, gears, driveshaft, axles and CV joints, wheels, tires, brake rotors, …).
 Force is lbf (pounds force)
“Mass” in this equation is a measure of “Inertia” of a physical thing in a linear motion.
 In the case of a car’s motion, the mass is a constant. The change in mass is very small (tank of fuel vs. the total mass of the car) and very slow (1 – 2 hours at racing speeds).
 Inertia also has a rotational equivalent. In the case the force is “Torque” the mass is a rotational Inertia (flywheel effect) and the acceleration is the change in RPM in time. (a discussion for another time).
 Mass is presented here as slug (lbf Sec^2/ft) (a pound mass (lbm), pound force (lbf), gravity and slug are all related – Google slug(mass) or see Wikipedia)
“Acceleration” is a measure of the “Rate of Change of Velocity”
 Acceleration is NOT a constant velocity but the change in velocity in a period of time
 Acceleration is ft/sec^2 (feet per second squared)
“Force” for our discussion is in pounds force (lbf)
______________________________
Part 3) Power to Acceleration
As noted in part 1) given HP (Power) and RPM we can determine the Torque. Given the Torque at this RPM, the overall Gear Ratio and the Tire Radius, we can determine the force that pushes us down the road.
We determined we can go from HP and RPM to determine Torque:
T(ft Lbf) = HP * 5252 / RPM
If we know the gear ratio and tire diameter, we can determine the Force pushing us down the road from the torque:
F(lbf) = T(ft lbf) * GearRatio / TireRadius(ft)
Putting the above equations together:
F(lbf) = (HP * 5252 / RPM) * (GearRatio / TireRadius(ft))
From Part 2) Newton’s 2nd law provides the acceleration given force and mass. We know the mass of the car (a constant in slugs), the force (from HP & RPM, gearing and tire radius)
Using the acceleration equation we have:
a = F/m
Putting it ALL together:
a(ft/sec^2) = ( (HP * 5252 / RPM) * (GearRatio/TireRadius)) / mass(slug)  OtherFactors
OR,
a(ft/Sec^2) = HP * 5252 * GearRatio______  OtherFactors
RPM * TireRadius * mass(slug)
So we can go from HP & RPM to get acceleration for a specific RPM and the HP at that speed.
Apply the above equation every 100 rpm across the operational range, and you will have some approximation of acceleration. Note that you can also include the effects of changing gears as you accelerate, Just keep updating the Gear Ratio to a new value.
But, its likely very inaccurate due to ALL the “Other Factors” not included in this equation. The Other Factors is ALL the resistance factors (aerodynamic losses, gear efficiency, tire flex, wheel bearing drag, brake drag, …), ALL the other Rotational Inertia factors (flywheel, clutch, gears, shafts, driveshaft, brake disks, wheels, tire inertia, …). The transmission and axle alone are estimated to impact power by about 15% alone…
Want an accurate measure of acceleration, measure it at the track with an accurate time/distance recorder…
___________________________
Part 4) Kinetic Energy is a red herring… (in acceleration)
KE = m * v^2 /2
Kinetic Energy equals the mass times velocity squared divided by 2.
Kinetic Energy is energy stored in a body in motion (a mass at a constant velocity or speed).
From Newton’s 1st law of motion, “a body in motion stays in motion unless acted on by some force”.
There is NO power in this kinetic energy equation, no factor as to how much power (e.g. HP) was used to accelerate the mass to its current constant velocity. So the KE of the body in motion is entirely irrelevant with respect to HP, Torque and acceleration…
Kinetic Energy  Why do we care? Kinetic Energy is very important IF want to stop from higher and higher speeds.
Say for a given mass we have a velocity of 1. So,
KE = m * v^2 = m * 1 ^2
The resulting KE = m * 1
If we increase the velocity by 20% the velocity is now 1.2 and the KE calculation is:
KE = m * v^2 = m * 1.2 ^2 = m * 1.44
OR, 20% more speed results in 44% more KE to slow down with your stock brakes…
Double the speed, KE becomes 4X higher. So, your road Miata brakes that work great at 60 mph have 4x the KE to dissipate as you slow down from 120 mph at the track… Also, as you slow from 120 mph to 60 mph you need to lose ¾ of the KE  NOT ½ as many might expect.
During braking, virtually ALL the Kinetic Energy is lost in brake heat. A higher HP racer reaches a higher speed (and KE) between corners and has to reduce that higher speed (KE) that the lower HP racer does not experience…
Want more HP & higher speed? You might consider improved brakes…
Big Brake kit for the street tuner that never goes above 80 mph is likely a waste of $$.
Flame on Miata Turbo readers….



03052016, 07:52 PM

#124

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Posting messed up the final equation... oh well.
here it is again in a text form that can be displayed.
a(ft/Sec^2) = ( (HP * 5252 * GearRatio) / (RPM * TireRadius * mass(slug)) )  OtherFactors



03052016, 08:10 PM

#125

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Yeah, what pat said in post 2.
But thankyou for your time.



03052016, 09:39 PM

#126

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So then on a corvette where youg et crazy high torque in the beginning do you shift at 5k rpm?
This doesnt seem right.



03052016, 09:48 PM

#127

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Quote:
Originally Posted by triple88a
So then on a corvette where youg et crazy high torque in the beginning do you shift at 5k rpm?
This doesnt seem right.

No, you maximize torque at the WHEELS.
Quick made up example. (not calculating actual RPM drops on gear change, numbers assumed and used for illustrative purposes)
Our engine has 300 ft*lbs of torque at 5K, 250 ft*lbs of torque at 7K for discussion.
If your transmission ratios happen to be 2.2:1 and 1.7:1 for 2nd and 3rd gear.
At 7K in 2nd, trans torque multiplication would be 250 ft*lbs *2.2 = 550 ft*lbs out the tail shaft.
Or, you could shift and fall to 5K: 300 ft*lbs * 1.7 = 510 ft*lbs.
So in this case, reving 2nd gear to 7K is faster than shifting into 3rd all the way to 7K RPMs (and even further since 2nd still puts more torque to the wheels at 7K), despite torque falling off some.
This is maximizing torque to the wheels.



03052016, 09:55 PM

#128

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Ok that makes a lot more sense. Forgot the gear ratios multiply the torque.



03062016, 08:37 AM

#129

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This whole thread is a bit ridiculous. The OP is a trick question. Strictly speaking, the answer would be that the greatest momentary acceleration would be at peak torque, but that's not actually a useful answer. If all you cared about was maximizing momentary acceleration, you would go get a 16:1 R&P made so you can redline in fifth at 30 MPH.
What's actually relevant is maximizing acceleration at any given wheel speed. Let's assume that 30 MPH is too slow for you, and it's a requirement for you that your car go at least 100 MPH, and get there as fast as possible. Now we have a set speed, and we're looking to maximize forward force (which isn't quite wheel torque since the wheels are a gear ratio). And since power is speed times force, that means we're looking to maximize power.
Power in is power out, so it doesn't matter whether you have a high torque, low revving engine or a low torque, high revving engine. What matters is how much power it makes. That doesn't just mean peak power though, since you don't have a CVT that will hold your engine at a set RPM while the car accelerates. Maximum power between shift points is what it takes. Ideally a broader power band with a wider ratio gearbox is better than a narrower power band with a closer ratio gearbox because shifting costs time, but "more torque" is not the same as "a broader power band."
In the end, acceleration comes down to five factors: power, gearing, mass, friction, traction. A dyno sheet gives you 20% of that. I suppose you could overlay it with a gearing chart and weight them to account for air resistance in higher gears and and pull a number out of your *** to account for tire grip, but in the end it's all just guesswork. You're better off just driving the car and having someone time you. You'll be more accurate measuring power with a stopwatch than measuring acceleration with a dyno.



03062016, 08:44 AM

#130

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"This whole thread is a bit ridiculous. The OP is a trick question. Strictly speaking, the answer would be that the greatest momentary acceleration would be at peak torque, but that's not actually a useful answer. If all you cared about was maximizing momentary acceleration, you would go get a 16:1 R&P made so you can redline in fifth at 30 MPH."
It's a super useful answer. There is a tonne of useful reasons for understanding this relationship.



03062016, 09:22 AM

#131

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Quote:
Originally Posted by Failure
This whole thread is a bit ridiculous.

Most accurate thing I've read so far



03062016, 03:21 PM

#132

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Quote:
Originally Posted by nitrodann
It's a super useful answer. There is a tonne of useful reasons for understanding this relationship.

Read the last paragraph. Your question is the automotive equivalent of "what is the difference between a cat?" The reason there's so much disagreement in this thread is that people are filling in the blanks to figure out what they think you meant, and (mostly) everyone is more or less correct. The disagreement is between the people who are offering the answer to "what is a cat?" and the ones answering "what is the difference between one cat and another?" or "what is the difference between a cat and a person?". Their answers are just too couched in mathematical abstraction to realize that they're not even answers to the same question.



03062016, 05:32 PM

#133

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I don't mind if you missed the point of the thread.
It's my thread and I'm really satisfied with it.



03062016, 08:07 PM

#135

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Quote:
Originally Posted by nitrodann
I don't mind if you missed the point of the thread.
It's my thread and I'm really satisfied with it.

No, I saw the point straight off. I see you shitposting on /o/ often enough that I know you've seen how my monthly "torque > horsepower" threads turn out. The only question in my mind is whether I inspired you to try it here or you didn't understand what was going on and you're genuinely looking for an answer. Your satisfaction implies the former.
Quote:
Originally Posted by triple88a

Nifty. Mind if I add support for shift points to this?



03062016, 09:21 PM

#136

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Someone post a RPM datalog while on a dyno along with the dyno plot. You should then be able to derive ΔRPM/s. See if it makes a graph that looks like the torque curve or the HP curve.



03062016, 09:39 PM

#137

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Quote:
Originally Posted by FatKao
Someone post a RPM datalog while on a dyno along with the dyno plot. You should then be able to derive ΔRPM/s. See if it makes a graph that looks like the torque curve or the HP curve.

You mean like this?
Acceleration curve vs torque curve vs HP curve.
That ground's already been covered. That was about the time the thread went nuts.



03072016, 12:18 AM

#138

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Quote:
Originally Posted by Failure
No, I saw the point straight off. I see you shitposting on /o/ often enough that I know you've seen how my monthly "torque > horsepower" threads turn out. The only question in my mind is whether I inspired you to try it here or you didn't understand what was going on and you're genuinely looking for an answer. Your satisfaction implies the former.

I wanted to see a multitude of explanations regarding wheel torque vs acceleration so I have lots of ways to explain it to people. Ive got visual graphs also, which is nice too.
Dann



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