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# Calculating power at the wheels (newbie question)

## Calculating power at the wheels (newbie question)

(OP)
I'm trying out some ideas for improving drive train efficiencies. For this, I'm trying to model the power necessary at the wheels of a truck.
To validate this, I've entered the specs of a real truck, and the Heavy-vehicle version of the Urban Dynamometer Drive cycle.
My calculation is that the power necessary is dv/dt * the forces to move the truck.
Problem is, I'm getting numbers which are about three times the max output of the engine on the actual truck.

I've considered dividing by pi :), but I would like to know what I'm doing wrong.

Please forgive me if this post is inappropriate. I'm just a big-diesel mechanic toying with some new ideas for hybrid powertrains.
Let me know if I shouldn't have posted here, and I'll ask for the post to be removed.
Thanks

### RE: Calculating power at the wheels (newbie question)

Did you take into account the torque multiplication of the gear box and differential?

### RE: Calculating power at the wheels (newbie question)

(OP)
I'm trying to work out the power required at the wheel. I'm planning to try to specify my powertrain from the power and torque necessary at the wheels.
I'm actually calculating forces to move the truck from drag, rolling resistance, inertia and grade. I'm trying to calculate the power from that.
Power at the wheels and at the engine should be the same, though of course torque is multiplied.
Unless I'm missing something, which I obviously am.

### RE: Calculating power at the wheels (newbie question)

Power is velocity x force, not dV/dt x force.

### RE: Calculating power at the wheels (newbie question)

I think you have mixed your units.

In terms of mechanical energy, one watt is the rate at which work is done when an object is moved at a speed of one meter per second against a force of one newton.

1W = 1Js-1 = 1kgm2s-3 = 1Nms-1

### RE: Calculating power at the wheels (newbie question)

(OP)
Thanks for the help BobM3 and dgallup.

I'm going to have another look at what I'm doing to make sure I've got all of my units in the right place and such.
At first glance it seems OK, but something must be wrong.
I just can't see how 790 kW can be right for a 33 ton truck.

### RE: Calculating power at the wheels (newbie question)

Time for a sanity check by the method of alternate calculation.

Try working backwards from the traction requirement using torques, gearing, and a factor to account for having less than 100% efficiency to arrive at the necessary engine torque.  See what the required HP is at the various engine rpms involved.

Norm

### RE: Calculating power at the wheels (newbie question)

(OP)
Thank you everybody.
I think I shall take a step back and start again at first principles.
I'll try your suggestions and be very careful with any assumptions I'm making.

Thanks for the help.

### RE: Calculating power at the wheels (newbie question)

BobM3 actually already gave you the answer.  You just need to "expand" it a little.

Power = F*v, but F = m*a, and a = dv/dt.  So Power = m*v*a = m*v*(dv/dt).

If you convert units, and express acceleration in G's, velocity in mph, and use vehicle weight in lbs, you should get

HP = w*v*G/375

(I assume you know how to calculate acceleration in G's.)

### RE: Calculating power at the wheels (newbie question)

P.S. Probably should have added that the above is the power needed to accelerate the vehicle, after taking off the various losses.  (That due to rolling resistance, aero drag, driveline losses, etc.)

### RE: Calculating power at the wheels (newbie question)

Almost Clever - Power at the engine and at the wheels is not the same - up to about one third of the engine power (and torque) can disappear on its way to the wheels.

You can roughly calculate the power needed to maintain a steady speed on a flat road - that is, the power needed to overcome air and rolling resistance etc.
Take the vehicle (truck or car or whatever) up to over 60mph, put it out of gear and time how long it takes to slow from 60mph to 50mph. For example assume it takes 15 seconds to slow. This is a change in speed of 15 feet per second (the calculation has to be in fps). If it took 15 secs - this a deceleration rate of 1ft/sec/sec.  The power needed to cause this deceleration can be calculated from F=ma (this is also the power needed to maintain speed).  The weight of the truck 66,000lbs (assumimg US tons) has to be converted to mass by dividing by 32 (gravitational constant). F then equals 66000 divided by 32 times 1 (rate of deceleration) this works out to be about 2063ft lbs (which is the required force). To convert this into power you multiply the force by the average speed of the truck as it decelerates - or about 80fps. This gives about 165000ftlbs/sec. To convert to horsepower divide by 550 (which is the definition of one HP) - this gives an answer of about 300HP - which seems a bit high but is of the correct order (maybe the deceration time would be more than 15secs). This calculation is very approximate but if you use the same method when testing it should show up any changes you make to the truck.
The power needed to climb a grade is a similar calculation. If the road rises one foot after 88ft travelled at 60mph (88fps) the truck has a vertical velocity of 1ft/sec. This is work being done against gravity so this is just the weight of the truck (the gravitational constants cancel out), so the power needed is 66000 divided by 550 which is 120HP - which looks about right. This would have to be added to the power needed to maintain a steady speed - so about 420HP would needed to charge up the grade at about 55mph.
There are also other road test methods that can be used to calculate power required to maintain steady speed.

### RE: Calculating power at the wheels (newbie question)

Hi Everyone

Im doing practically the same thing, im designing a vehicle and trying to calculate the power required by its motor.

To do this i am using the following formula;

Tractive Force = Resistive Forces + Inertial Forces + gravitational forces

Resitive forces = Aero drag + Mechanical losses + Rolling resistance

Inertial forces = Me(dv/dt) (zero for constant velocity)

Me=vehicle effective mass = M + 4.Iw/Rr^2 + Ig.Gfd/Rr^2
M=vehicle mass,
I=inertia
Gfd=final drive ratio

What i did was find the required torque at the wheel by

Torque_max = Tractive force x wheel radius
then
Power = rotational speed x Torque_max*

*torque can be scaled down by efficiency coefficents and other transmission losses if you like.

### RE: Calculating power at the wheels (newbie question)

robin - Yes, you have that right.  In that form, (P = ω * τ), the angular velocity ω is in rad/sec.  Watch your other units, too, of course.

### RE: Calculating power at the wheels (newbie question)

Thanks Blackcobra,

Almost Clever:

I think the main issue for this type of calc is the unknown values for efficiencies, this data is extrememly difficult to get hold of and the only realisitic way to obtain it is by having a chassis dyno.

Even with all this though, at high speeds and with bluff body shapes the overiding resistive force is aero drag.

If anyone has any rough figures for transmission efficiencies i would be interested to see them?

### RE: Calculating power at the wheels (newbie question)

Im having a big issue with my calcs,

In order to accelerate my vehicle from 0-60mpg (0-26m/s) I use F=m(dv/dt), this is giving me a force of around 1000 N at the tyres, which translates as a torque of around 440 Nm which is way high!

My stats are;

M=410
0-26m/s (0-60mph)=10 s

I did F=ma to get force at contact patch (im modeling the vehicle as just a wheel at the moment, i just want to know the torque required by the electric motor).

### RE: Calculating power at the wheels (newbie question)

Maybe the torque seems high because there is normally torque multiplication through the gearbox and diff when accelerating - maybe 3.5:1 in the gearbox (average of 1st. and 2nd. gears) and 3.5:1 in the diff -  very roughly approx. 12:1 overall. Dividing 410 by 12 gives a more reasonable answer. I think most electric cars use gearboxes these days.
410Kg is a very light car.

### RE: Calculating power at the wheels (newbie question)

Yes true, I did think all these things, though the car being light should reduce the required torque.

I wasnt aware electric cars had much in the way of gearing?

My overall aim is to spec an electric motor and its torque curve. In that sense i just need to know the maximum torque and power required and then pick a motor which can provide it and pick and choose between the most preferable torque curve....

### RE: Calculating power at the wheels (newbie question)

(OP)
Thanks for all the input everyone.

I am now convinced that I've got the torque numbers right.
They add up, and they make sense in relation to what the actual machine can do.

However, I'm still mis-calculating the power requirement somehow.
Using the same formula as Robin, I'm trying to calculate power at the wheels. I'm getting answers at about 750 kW, for a truck whose engine can only produce about 300 kW.

I'm not taking any mechanical/efficiency losses into account yet, because I'm still at the wheel. Once I progress back towards the power plant, I'll start doing efficiency calculations as well.

It has just occurred to me that perhaps my drive-cycle data is a bit off.
It seems to have the truck accelerating from 27 mph to 32.5 mph in the space of two seconds. I'm not sure if this is reasonable or not, but it seems rather quick for a 33 ton truck.

### RE: Calculating power at the wheels (newbie question)

If you are calculating torque at the wheels, make sure you are also using angular velocity at the wheels in the expression, P = ω * τ.  (Not sure that's it, but it's something else to be careful about.)

### RE: Calculating power at the wheels (newbie question)

Simply put.

With no losses, the HP seen at the engine flywheel or output, would be the same at the rear axle or where ever the power is sent.

What does happen is torque multiplication through the gearing. This is why an under powered class 8 truck,(all of them are under powered) can pull 80,000 lbs up a steep grade, though not very fast.

As you can see if you multiply the torque, and reduce the rpms the HP will stay the same no matter what gear its in.
That is with no losses of course.

Its the LBs/HP thats important for moving a vehicle up a hill.

Thats why a 1800 lb Geo metro will beat any loaded class 8 truck, the Geo has 55 hp, and say the truck has 600 hp.
32 lb per HP for the Geo, 133 lbs per HP for the truck.
Lets see that means for the truck to do what an under powered Geo metro will do the engine in the truck needs to be 2500 hp and thats to equal a metro. Pretty bad huh??

### RE: Calculating power at the wheels (newbie question)

dicer's elegantly simple computations drive home the importance of driveline efficiency and gear ratio selection for maximizing performance of large trucks.  Use of spiral bevel gears at the differentials boosts efficiency to about 95% to 98% from about 80% to 85% for hypoid gears for that portion of the drivetrain.  Within the transmissions (main or auxiliary) either spur gears (noisy but strong, efficient, and with no axial loading for bearings) or helical (quieter, nearly as efficient, but with significant axial loadings for the bearings), each contact will have an efficiency of 97% to 98%.

As a truck driver (in a former life, so to speak), I was always impressed by the wide variety in the performance capabilities of different trucks due primarily to the selection of gear ratios available--actual engine power always proved to be a secondary factor.  Trucks with transmission ratios that were selected for the purpose of keeping the engine operating within its optimum speed and power range always performed very poorly compared to the trucks that had transmission ratios that were chosen to be progressively closer as road speed increased.  At lower speeds, relatively wide steps were not much of an issue because full engine power and torque were rarely needed for acceleration or hill climbing. It was only as road speed increased significantly that using full engine power really mattered very much.  At low road speeds, as long as the minimum engine speed was well above idle speed, the available torque was always more than sufficient to accelerate the truck because of the great torque multiplication provided by the "lower" transmission ratios.

The trucks that usually performed the best had transmission ratios selected to provide increments of about 5 mph at maximum normal engine speed for each gear change.  When going from near level road to climbing a hill, each drop in gear selection then only lost 5 mph.  Usually, 1, 2, or 3 gear changes would be necessary to climb a hill losing 5, 10, or 15 mph in increments before topping the hill.

With transmissions that were selected to keep engine speed within its optimum speed range, the truck could easily experience a drop of 15 to 20 mph in just one increment at higher road speeds when beginning to climb a hill.  If a second gear change became necessary, the next increment would result in a drop of an additional 10 to 15 mph for a total loss of 30 to 35 mph.  Almost always, after such down-shifts, the road speed was limited by engine rpm red-line and not available power.  At low road speeds, the ratio selections were so close together that it was almost always more practical to skip one or two ratios at at time while accelerating in the "lower" gears.  (I always guessed that this type of transmission was designed, selected, and purchased by people who never had to drive a truck for a living.  I remember these types of transmissions getting lots of positive "press" at the time, and I was never able to make sense of all the "hoopla" for something that performed so very poorly in practice.)

Staying with dicer's nice, clean basic approach and looking very simplistically at a truck with a 600 hp engine, a main transmission, an auxiliary transmission, and a single fixed ratio at the differential, effective available power at different overall ratios can be represented as follows:

In lower ratios where neither transmission is in direct drive:

600 hp at engine
600 hp x .98 x .98 = 576 hp at shaft between main and auxiliary transmissions
576 hp x .98 x .98 = 553 hp at shaft between auxiliary transmission and differential
553 hp x .85 = 470 hp to drive axles with hypoid gearing (130 hp drivetrain losses)
553 hp x .97 = 536 hp to drive axles with spiral bevel gearing (64 hp drivetrain losses)

With one of the transmissions in direct drive:
600 hp x .98 x .98 = 576 hp at shaft between auxiliary transmission and differential
576 hp x .85 = 490 hp to drive axles with hypoid gearing (110 hp drivetrain losses)
576 hp x .97 = 559 hp to drive axles with spiral bevel gearing (41 hp drivetrain losses)

With both transmissions in direct drive ("top gear"):
600 hp x .85 = 510 hp to drive axles with hypoid gearing (90 hp drivetrain losses)
600 hp x .97 = 582 hp to drive axles with spiral bevel gearing (18 hp drivetrain losses)

From this, it is easy to see the distinct efficiency and performance advantages of fewer gear contacts, using the highest efficiency gear contact types, and having the most truly practical selection of gear ratios to serve the actual needs of the truck and the duty for which it is being designed.

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