What does Power-to-Weight Ratio Really Mean, Anyway?

"With-a-G is back to not having anything written after his username" (with-a-g)
03/06/2014 at 11:58 • Filed to: None

All the talk about the !!!error: Indecipherable SUB-paragraph formatting!!! and its !!!error: Indecipherable SUB-paragraph formatting!!! got me thinking. First of all, the mixed unit systems in "BHP" and "kg" grates on me for aesthetic reasons, and the fact that the number is "1" is a happy accident but ultimately meaningless. Even deeper than that, however, is the fact that dimensional analysis reveals that it is not immediately obvious how to massage this metric into a dimensionless number that can be used as a figure of merit for automotive performance. Airplanes have a thrust, T, (dimensions of force), and a weight, W , (mass * g , also dimensions of force), which makes it easy to determine if the craft is "ballistic," T/W >1 or not, and how much so.

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Horsepower is, in dimensional terms, power , which is energy per unit time . Reducing to SI base units, 1hp = 745.7 Watts = 745.7 kg*m^2/s^3. Dividing by the "weight" (actually, mass ) of 1 kg gives the quantity 745.7 m^2/s^3. Whatever that quantity is, the Koenigsegg One:1 has it .

So what is that quantity? Well, its units of m^2/s^3 could be interpreted as a squared velocity per unit time, but that has no immediate applicability. How about a velocity times an acceleration? 745.7 (m/s)(m/s^2). What is another relevant physical quantity we can bring into play that will help unravel this? How about the acceleration due to gravity, 9.8 m/s^2?

If we take this power/weight ratio and divide by g =9.8 m/s^2, we get 76.1 m/s, or 170 mph. Ok, this is a velocity , and a fast one, but it's not the top speed of the One:1, or anything like that, so what is it?

It is the speed at which the One:1 could theoretically still spin its tires under acceleration if their coefficient of friction was 1, marking the speed at which the car goes from friction-limited to power-limited. This is, of course, in the absence of aerodynamic effects such as downforce that would prevent wheelspin even at lower speeds than this. It also neglects the number of drive wheels and the F/R weight distribution, but I think it still has a notional value.

So, by the same token, my Miata , at 115BHP and 1050kg, has a theoretical wheelspin speed of 18.6 mph . A late model Viper at ~500 HP and ~1500 kg has a theoretical wheelspin speed of 56.7 mph , meaning that at near-highway speeds, you can likely still chirp the tires more than a little on a hearty downshift into your power band. (Any confirmation on this from Viper drivers?)

To repeat, the Koenigsegg One:1 has a theoretical wheelspin speed of 170 mph (in the absence of downforce).

Some handy conversion factors: If you have a car's power and weight in HP and kg, multiply that ratio by 170 to get its wheelspin speed in mph. If you have HP and pounds, multiply that ratio by 375 to get it.

DISCUSSION (18)

Mikeado > With-a-G is back to not having anything written after his username
03/06/2014 at 12:02

I won't pretend I followed all that (coffee required), but power/weight is really just used as a more rounded comparison reference than just power or weight. Also, technically, the One:1's power is measured in metric horsepower (PS, or sometimes CV in mainland Europe), but that's just me being pedantic. 1PS is about 0.986bhp.

JR1 > With-a-G is back to not having anything written after his username
03/06/2014 at 12:05

Great article wheelspin at 170mph would make lesser men pee their pants. Me though I would not pee my pants, I'd shit my pants.

With-a-G is back to not having anything written after his username > JR1
03/06/2014 at 12:09

You indeed are a great man.

Nonster > With-a-G is back to not having anything written after his username
03/06/2014 at 12:09

I think you're getting bogged down trying to place a real meaning to the number. Power to weight ratio isn't trying to be a dimensionless number, its simply normalizing a vehicles power output to its weight for the purpose of comparison to other vehicles

GhostZ > With-a-G is back to not having anything written after his username
03/06/2014 at 12:16

This article is beautiful.

Thank you.

With-a-G is back to not having anything written after his username > Nonster
03/06/2014 at 12:23

I understand that as given, the power/weight ratio doesn't usually pretend to be more than that. However, dimensional analysis often does give additional physical insights, and I think that is the case here.

One of my favorite applications of this was in my Halliday & Resnick university physics book, where they analyzed that "miles per gallon" has dimensions of inverse area, described how this area is the cross-sectional area of a trough of gasoline that a moving car would consume, and concluded that since it is the chemical bonds in gasoline that are being broken as the energy source, that gasoline is stronger than steel.

Chteelers > With-a-G is back to not having anything written after his username
03/06/2014 at 13:16

As an engineer, my most sincerest Thank You. I can't stand mis-matching dimensions presented as some kind of fact.

JR1 > With-a-G is back to not having anything written after his username
03/06/2014 at 14:23

Thank you sir

You can tell a Finn but you can't tell him much > With-a-G is back to not having anything written after his username
03/06/2014 at 16:20

Are you an engineer? Because that makes perfect sense to me.

With-a-G is back to not having anything written after his username > You can tell a Finn but you can't tell him much
03/06/2014 at 16:50

Kinda. Physicist by training, but what I do every day is closer to engineering. Glad it makes sense. I worried that the caveats (\mu = 1, no consideration of drive wheels or weight distribution, and no aero) would kill the usefulness of this description in the eyes of engineers. As a physicist, I can assume a spherical car and still be all right.

You can tell a Finn but you can't tell him much > With-a-G is back to not having anything written after his username
03/06/2014 at 17:55

The spherical car only works if you assume it's in a vacuum though.

With-a-G is back to not having anything written after his username > You can tell a Finn but you can't tell him much
03/06/2014 at 18:40

It shouldn't surprise you that "no aero" is synonymous.

LTIROCKS > With-a-G is back to not having anything written after his username
03/06/2014 at 19:11

What it means in real terms is perhaps best illustrated by this mathematical model, as derived by Patrick Hale (the founder of RSA, which has marketed Quarter Junior and similar drag racing software for quite some time).

Peak "as installed" engine horsepower ~ = (1/4 mile trap speed/234)^ * Race Weight

Where "race weight" = total vehicle weight (vehicle, plus driver, plus equipment, plus all fluids) at launch)

Unlike ET, Trap speed is largely uninfluenced by variables such as track conditions, tires, and, gearing.

The calculated power will come out a little high for vehicles that have very large frontal areas, very poor drag coefficients, and very huge tires on all 4 wheels (e.g. a "monster truck) because the model was derived as drag strips and for the cars that generally race on them.

And the model will seemingly (but not truly) yield "wrong" results in cases where stated/published power figures are BS (e.g. every American car engine that was rated under the old "gross" system prior to the 1972 model year).

With-a-G is back to not having anything written after his username > LTIROCKS
03/06/2014 at 19:17

Not the approach I'm after. I'm aware that you can derive approximate quarter mile times and trap speeds from vehicle parameters, and vise-versa. I'm talking about the general physical ramifications and meaning of the quantity itself. Vehicle dynamics are different when you are friction-limited vs. power-limited, and the crossover point is relevant for a driver.

samssun > Mikeado
03/06/2014 at 19:33

And there's nothing worse than blind unit conversion. The Veyron came out rated for a nominal 1001hp, actually producing 1010-1030 just from rounding error. The 1001 was just to indicate >1000, but ignorant writers converted PS to HP and wrote (986hp) everywhere.

Same thing in news articles with rough estimates: "The crash was heard 3-4 miles away (4.83-6.45km)!" Take a ballpark figure and convert it to a level of precision that never existed, vs. just writing (5-6km) like a human being.

LTIROCKS > With-a-G is back to not having anything written after his username
03/07/2014 at 15:35

How about this one:

Doubling the quantity of horsepower that's available to accelerate an automobile traveling at any given speed will precisely double the rate of acceleration at that same speed.

The meaning of the quantity itself is defined by the ability to lift 33,000 pounds one foot in one minute. That ability = 1 HP.

Work = Force * Distance (e.g., 33,000 Pounds force * 1 Foot)

Power = Work/Time (e.g., 33,000 Pounds force * 1 Foot/1 minute) = 1 Horsepower

And as I mentioned before, trap speed (unlike ET) is rarely friction limited because it varies very little, relative to drivewheel traction.

There's also this to consider:

HP = T * RPM/5252

Drive axle torque = Drivewheel horsepower * 5252/Drive Axle RPM

And...

Tractive Force (which actually accelerates the car) = Drive Axle Torque/Tire Radius

Actual driveline loss (not to be confused with dyno load loss) is only a couple percent.So for purposes of simplicity, engine horsepower can substituted for "drivewheel horsepower" in the drive axle torque equation above.

I'm a 50 year old mechanical engineer and this is the simplest and most practical way I can explain this...

With-a-G is back to not having anything written after his username > LTIROCKS
03/07/2014 at 15:40

That's not true, though, because the relationship between kinetic energy and velocity is quadratic. Doubling the energy transfer into kinetic energy will only increase velocity by the square root of 2 (1.414...) as much.

LTIROCKS > With-a-G is back to not having anything written after his username
03/07/2014 at 17:51

Of course it's true.

Rethink the term "horsepower that's available to accelerate the car*."

Doubling the horsepower that's available to accelerate a car traveling at any given speed will double the tractive force that accelerates the car at that same speed, which in turn doubles the rate of acceleration.

Since Mass is fixed, the rate of acceleration must double because F = MA

* Horsepower available to accelerate the car: The quantity of horsepower that's available solely for acceleration, which is exclusionary of aerodynamic drag and rolling friction.

And I don't understand why your bringing energy into it, since POWER = FORCE * VELOCITY per basic physics:

http://hyperphysics.phy-astr.gsu.edu/hbase/pow.html

http://www.allpar.com/eek/hp-vs-torq…

Here is my mathematical proof:

POWER = FORCE * VELOCITY
FORCE = MASS * ACCELERATION
Therefore, POWER = MASS * ACCELERATION * VELOCITY
Hence, ACCELERATION = POWER/(MASS * VELOCITY)
Thus,ACCELERATION = POWER/ [(WEIGHT/32.2 FT/Sec^2) * VELOCITY]
Where:ACCELERATION= FT/SEC^2

POWER = FT-LB/SEC

MASS = LB/32.2 FT/SEC^2

VELOCITY = FT/SEC

PROOF: FT/SEC^2 = (FT-LB/SEC)/[(LB/32.3 FT/SEC^2)] * FT/SEC

This is the simpler rendition:
POWER = FORCE * VELOCITY
FORCE = POWER/VELOCITY
Hence, doubling the amount of power available to accelerate the vehicle at any given velocity precisely doubles the force of acceleration ("G" force), as well as the acceleration rate.
In English units:
Power (hp) = Force (lb) * Velocity (MPH) / 374