Physics 101 help (mb 102), m*A*V and hp

[Earendil]

It's been a while since I took physics, and I was never really that great at the math, even though I loved the content and application. My memory is also rather bad.

So one day I decided to stretch my physics muscles and attempt some calculations based on an application I like, my E30.

Here is what I want to do:
Take an acceleration (m/s^2) of my vehicle, and derive the HP requirement for archiving that acceleration based on known values.

My problem: I get a reasonable (i.e. I understand) result using my initial set of values, but nothing else.

Assumptions made:
1. I will find whp, not crank hp. The easiest to find should be whp because it won't take as many calculations, and doesn't involve this vague value called drivetrain loss when approaching the equation from the other end.
2. My measurements are correct.
3. That despite age and potential loss of power, that the power curve of my engine still peaks at 5800rpm.
4. My car is lighter than stock, so a faster than stock acceleration is expected.

First, the equations. Horsepower is a Power, with the metric equivalent being the watt.
Power = Mass*Acceleration*Velocity (m*A*V)

I can supply all those things as they apply to the entire system of energy that is my car. Thus finding the Power should be easy, no?

Mass = 2700lbs or 1224.7kg
Acceleration = 3.92 m/s^2
Velocity = 48.8mph or 21.8 m/s

using m*A*V and the metric units we get a power in watts of 97474.78
Since there are 745.7 watts in 1 hp, we end up with a HP of 140.6
Hey, 140whp looks like a number close enough to verify my work!

But what happens if instead of plugging in the known velocity at the time of the peak acceleration, we instead plugin a velocity of 35mph? Power drops to 119hp. Huh? How does that work (and yes, I can see the equation, I mean conceptually).

Also, I started thinking that this peak acceleration was measured in 2nd gear, not 4th like a dyno. But wait! Should that matter if I'm trying to find whp and not crank hp? After all, I'm trying to figure out how much power the wheels are putting on the road, I don't care if it's an engine and a drivetrain doing it, or gerbils with tiny pedals.

Surely it must matter though, because for the same output I'll get a different measured peak acceleration value depending on the gear I choose. It must be a coincidence then that I chose 2nd gear, and I got an acceleration value that when matched with my mass produced a believable WHP... right?

At this point I'm thinking I must have one or more of the following going on
1. I don't understand the entire system
2. I don't understand the definition of whp, and how it's found
3. I don't understand physics.

Can anyone tell where I've gone wrong in either my math or conceptual thinking?

[billybobjoseph]

I'm not sure where the mass times acceleration time velocity came from. Metric hp is defined as 75 kgf - m / s. The units there are ((kg*m)/s^2 * m ) / s = (kg*m^2)/s. The units from the M*a*v will come out (kg*m^2) / s^3. To calculate horsepower we use engine ( crankshaft ) torque. Here's a pretty good writeup on it http://www.4x4abc.com/jeep101/torque.html
If by whp you mean wheel hp ? there is still a torque involved. The torque output at the wheel is the force from the center considering the distance from there to the pavement as a lever.

[Earendil]

I got m*A*V by doing some math :)

Mass = kg
Acceleration = meters / seconds^2 (m/s^2)

Force = m*A where m = mass(not meters) and A = acceleration (m/s^2)

So Force now has the units (kg*m)/(s^2). Or, a Newton.

Work = Force over distance, or m*A*d. Give d the units of meters, and we have:
(kg*m*m)/s^2

Finally, Power = Work over time, or (m*A*d)/s
But Distance/Time is Velocity. So we have m*A*V,
Or the units (kg*m^2)/s^3

Since watts is our unit of power, we can just look up the units of watts and find it to be (kg*m^2)/s^3

And after proving that to myself again, I see where you went wrong. You divided the entire set of Work in Joules by 's'. I believe that power is Joules "per second" and is thus multiplied.

[billybobjoseph]

Ah ok, the units of mass and weight are getting you. Weight is the product of mass and the acceleration due to gravity ( 9.8 m/s^2 I think ) and the term is in there to make sense of the definition. It's the force required to move 75kgf one meter in one second. I'm using metric hp. The kgf is a obsolete measurement of torque; can use the Newton instead. I see what you are saying about the units, but I'm pretty sure you have to use torque in there. Even if two answers have the same units they aren't necessarily derived from the same thing, we use foot pounds and pound feet in different situations.

[Earendil]

I'm not sure where the mass times acceleration time velocity came from. Metric hp is defined as 75 kgf - m / s. The units there are ((kg*m)/s^2 * m ) / s = (kg*m^2)/s. The units from the M*a*v will come out (kg*m^2) / s^3. To calculate horsepower we use engine ( crankshaft ) torque. Here's a pretty good writeup on it http://www.4x4abc.com/jeep101/torque.html
If by whp you mean wheel hp ? there is still a torque involved. The torque output at the wheel is the force from the center considering the distance from there to the pavement as a lever.

And thanks for the link. I see I made a few poor assumptions about the way the would works :)

So the peak torque RPM will result in the peak torque to the road (for a particular gear). Which means that my measured peak acceleration did not occur at 5800rpm, but 4300rpm which would result in a different velocity when the peak acceleration was reached.

Time to do some more calculations.

I also believe I understand (conceptually) how velocity can change and the "power" at the wheels can change even though the car is accelerating the same amount. I think...

[Sagaris]

Earendil, I actually just did the exact same thing a couple weeks ago. I created the platform for an iphone/zune HD/droid based dynamometer application. It works and is quite accurate in comparison with aDyno and Dynolicious. I can give you my excel spreadsheet with some of the goodies in it if you want.

I have observed that the peak acceleration occurs at the peak torque output.

The engine produces torque, the torque is multiplied through the transmission gear ratios (all under-driven gears), which is multiplied through the differential and is converted to a force through the tire (torque happens at the drive axle, becomes a force when it is transmitted through the radius of the tire). The force at the tire is the force you can use (assuming Newton's 2nd law in its crude form F=m*a) to determine acceleration and so forth. It is just a matter of re-arranging the equation to do so. You can even get nitty-gritty and subtract the aero-drag force that the car is trying to overcome for even greater accuracy. The equation I derived for my little project is as follows:

I personally prefer Metric units, but am using English in this post because HP is an English unit and will spare me from digging up/deriving a few conversion factors .

Calculating engine torque from acceleration rate:

Engine Torque = mass*accel.*(1/R.Trans)*(1/R.Diff)*T.Rad EQUATION 1

then use Engine Torque in the usual horsepower formula:

HP= Engine Torque * (RPM/5252) EQUATION 2

this will give you wheel-horsepower since the actual acceleration of the car is being measured with an accelerometer and all frictional losses have already been overcome.

To incorporate aero-drag, use the following equation to determine the drag force:



where rho= air density in lbm/ft^3 or other appropriate units, you are looking for an answer in lbf. since the equation gives you the force required by the car to displace air. (If you need help with units or dimensional analysis I would love to help.)
u= velocity (ft/sec)

Cd for E30's is typically between 0.32 and 0.36 depending on bumper style tire width, spoilers, and other things.

Cross sectional area would be in ft^2 or whatever units match properly.

Now, you have determined a force from the aero-drag equation. Use this force with F=m*a

The car must produce this amount of force just to keep the car going at constant velocity. For the car to accelerate, it must provide additional force.
To convert Force into HP you must convert it to torque by re-arranging Eqn. 1 and 2.

For a drag force of 50 lbf. you must work backwards. Imagine the process of the engine transmitting power to the road in reverse: Follow this text with Equation 1. Begin at the point where the rubber meets the road, think of 50 pounds of force acting on a lever (the radius of the tire) which generates a torque at the center (the drive axle), the torque applied through the axle enters the differential, but since this is working in reverse, torque is not multiplied, it is divided (which is why Equation 1 has one OVER the diff ratio). The torque exits the differential and is transmitted through the driveshaft and into the transmission (again torque is not multiplied here because it is working backwards and Equation 1 uses one OVER the trans. ratio of the gear you are in). The torque exits the transmission , through the clutch, and is back at the engine. If you work through Eqn. 1 , replacing mass*accel with the 50 lbf force (remember Force=mass*accel) then you will get the engine torque required to overcome aero drag. Plug this torque into Eqn. 2 and get HP.

Now you have the power requirement to overcome aero-drag. Now you wanted to know what HP was required to obtain a certain acceleration. My car accelerates at a maximum rate of 0.46g and registers in the neighborhood of 160 whp. Say I wanted to find out what kind of HP and TQ i need to upgrade my car to in order to accelerate at 0.50g, I would use 0.50g in Eqn. 1 (0.50g = 0.50 * 32.2 (ft/sec^2)).



-----------------------------------------------------
Engine Torque = ft*lbf
mass=total vehicle mass (lbm)
accel = (ft/sec)^2
R.trans= transmission ratio that acceleration is obtained in (ex. if the ratio is 2.83:1 use 2.83)
R.diff = differential gear ratio (3.73 etc)
T.rad = radius of the tire (ft) tire deformation is far too complex to account for so it is ignored.

I hope this made sense.

[billybobjoseph]

Yeah horsepower is a concept where the power of one horse was equated to the ability to move 33,000 lbs on a cart with 1 foot diameter ( or radius ?) wheels one foot in one minute. But your car doesn't have the same size wheels. Pretty sure it's radius.

[Wh33lhop]

Basically you need to understand the difference between power and torque. Torque is a twisting force directly related to acceleration; power is the ability to accelerate at speed. If you apply 100whp at 20mph you will be accelerating at the same rate regardless of gearing.

Edit: I sat on this page for like 45 minutes before replying. Oh well. :p

[billybobjoseph]

Right but the power to move forward is developed by a torque; the spinning of your drive axle(s) driving your wheel. The distance from the center to the ground is like a lever.

[Wh33lhop]

Yeah horsepower is a concept where the power of one horse was equated to the ability to move 33,000 lbs on a cart with 1 foot diameter ( or radius ?) wheels one foot in one minute. But your car doesn't have the same size wheels. Pretty sure it's radius.

The whole point of horsepower is that the diameter of the wheel does not matter. It is 33000 pounds-feet per minute.

[Sagaris]

to add to that ^ for anyone who cares for further insight:
Horsepower is energy/time

I can ride my bike at 10 mph and get somewhere slow or I can ride my bike at 20 mph and get to the same place twice as fast. Either way I am still burning the same amount of calories (ignoring aero drag and so forth), it is just a matter of the rate at which I am burning them.

So by this logic if an excavator can lift a 4000 lb boulder in 10 seconds, then I should be able to do the same task (same amount of work) in a much longer time.

This is where torque comes into play, even though I may be determined to lift a boulder and am willing to spend the next 10 years trying, it will never happen without adequate torque/force

Work-energy theorem:
Energy=the capacity to perform work
Work=force*displacement (units of energy such as Joules which are N*m, or ft*lbf, note this is NOT torque, it is energy or work)
Power=work / time or energy/time

[einstein57]

To calculate required hp

HP = [1/2*m(V2^2 - V1^2)] / delta t

don't forget your unit conversions


As far as torque goes

W = 2*pi*n*T

So T= (HP*60)/(2*pi*5800rpm)

This is assumiming a 1:1 final drive so you'd need to know gear ratios to determine the shaft speed at a given rpm

[Earendil]

The whole point of horsepower is that the diameter of the wheel does not matter.

Yes, that is the entire point of horsepower

Actually, whether you're talking about whp or wheel torque, you don't need to know the wheel radius per say. However, since you do need to know axle RPM and vehicle velocity, you end up with all the information you need to solve for the wheels if you don't have them.

Torque/diameter = Force being applied to the road.
Force/mass = Acceleration, which is the point where I started.

I now have an excel spreadsheet that is uber fun (for a nerd) to play around with. I have the calculations set up for calculation wheel HP and wheel torque starting at either the engine (using by the book numbers, boring) or starting with the mass of the car and recorded acceleration.

If anyone is interested, I'll upload it some time after I polishing it up and make it usuable. Right now you have to enter numbers in seemingly random and unmarked fields to make everything fill in. I'm no excel wiz though.

Sagaris, I swear your post wasn't that long last night. Lol. I just ended up deriving post everything you said after the first 3 paragraphs
But, now I have a firm understanding, and my own excel spreadsheet!
I would still be quite interested in seeing yours though. Maybe we can do a personal show and tell :pimp:

[Wh33lhop]

Yes, that is the entire point of horsepower

The point stands whether you nitpick or not--horsepower does not concern gearing. It concerns velocity. Torque is where gearing comes into play. That is why we care about horsepower more than torque when we want to make a car fast.

[Earendil]

The point stands whether you nitpick or not-
-

I was not nipicking your point, only pointing out your use of "entire point" :)

horsepower does not concern gearing. It concerns velocity. Torque is where gearing comes into play. That is why we care about horsepower more than torque when we want to make a car fast.

That is completely correct.

Because HP=(Torque*RPM)/5252
Every time a new gearing is selected, the torque is multiplied differently and the RPM changes such that HP will stay constant (assuming engine HP is constant).