Drive Power to pull trailer 2

[damnaged]

Anybody care to assist me in understanding this problem i have?

What determines the drive power to pull a certain weight on trailers or wheels? As i know that the weight on the wheels shall subject a certain frictional force and therefore to counter the force you shall need sufficient pull/drive force.

Question: What is this pull force? As normally truck engines are rated in horsepower, i feel that what determines the pull force capacity would be the torque. But how do i convert the torque of an engine into force?

If my assumption above is wrong, i stand clear to be corrected. Thank you.

 

[GregLocock]

On flat ground the drag force required by a trailer is Crr1*m*g

Crr1 is about 0.12 g is 9.81 in Sensible unIts

The power required to supply that force is P=F*v, where v is also in Sensible unIts.

Engine torque is related to power at the drive wheels by T=P*.8/(ERPM/60*2*pi)

where ERPM is the engine RPM



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[JOJOMO]

Greg, what's Crr1?

 

[JOJOMO]

Coeff of rolling resistance, maybe?

 

[GregLocock]

Oops yes Crr1 is coeffcicient of rolling resistance, and I got it wrong, it is 0.012

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[EdDanzer]

You are correct torque does drive the wheels of a vehicle. Power determines how fast something can move. All engines have a torque curve relative to rpm. Using the total gear ratio, and torque you get theoretical drawbar pull at a specific engine rpm.

 

[crysta1c1ear]

http://www.motorauthority.com/news/suvs/touareg-tdi-tugs-a-boeing-747/

 

[Fabrico]

Don't forget the Cow (wind resistance).

 

[damnaged]

Thanks guys for the help!

Greg Locock: Where did you get the formula? (Just curious) fantastic! trust i can use it in my engineering calculations and such?

So from what i can understand, can i say drag force:-
drag force,P = T*(ERPM/60*2*pi) / 0.8 (final output in N?)

Please anybody clarify. the source of the formula would also help. thanks in advance.

 

[MikeHalloran]

When you get done with all that, don't forget to include the contribution of grade.



Mike Halloran
Pembroke Pines, FL, USA

 

[damnaged]

contribution of grade? sorry i don't quite get you.. care to explain more? sorry and thanks

 

[MikeHalloran]

The equations you have so far cover the condition where you are pulling the trailer on a perfectly flat, perfectly level surface. The closest approximation to that is in a metrology lab. Runways are usually almost flat, but these days it can't be easy to get clearance for alternative uses.

Real roads are not flat, even across their width.
To compute the power you need to pull a real trailer on a real road, you have to know the contribution of rolling resistance, now you do, _and_ you have to add the power required to pull the trailer up the steepest hill you plan to encounter at a reasonable speed, which starts out as a trig problem. The latter power is probably much larger than the former, even for a gentle grade.



Mike Halloran
Pembroke Pines, FL, USA

 

[JRLAKE]

With a grade you are now pulling in a plane against some portion of the gravitational force, depending on the degree of the grade. The normal forces exerted on the surface by the tires are also affected (reduced).

Unless you are pointed downhill of course.

 

[JRLAKE]

Mike beat me to it. I thought I was the only one that would be responding on a Friday night!

 

[MikeHalloran]

I'm between jobs, again, still. Entertainment budget is zero.

This "place" keeps me sane. Okay, less crazy.







Mike Halloran
Pembroke Pines, FL, USA

 

[damnaged]

ok understood. thanks guys.. now that you mention it, i think the contribution of gradient should be included in the equation to determine what is the pull bar required, am i right?

i am currently looking at how to determine the pull bar capacity of a tractor or truck base on engine capacity.

Thanks again. Btw, anybody care to confirm the above equations and give out more details?

 

[crysta1c1ear]

I am still uncertain what you want to calculate.

If a tractor is pulling a plough and it catches on some rocks then there will be a shock on the pull bar depending on the mass of the tractor and its speed.

This might be irrelevant to your calculations. What troubles me is that I don't know. I still haven't got enough out of your questions to know if it's irrelevant or not.

===

So from what i can understand, can i say drag force:-
drag force,P = T*(ERPM/60*2*pi) / 0.8 (final output in N?)

Please anybody clarify. the source of the formula would also help. thanks in advance.

NO!

If Greg said P was power, then P is power! So it's not drag force. Greg gave you a formula for force too.
To clarify, he said P is power at the wheels and T is torque at the engine.

The source of the formula?

60: there are 60 seconds in a minute. So to get a rate of rotation, you need to divide revolutions per minute by 60.

2*pi: the Greeks messed up maths by using the diameter instead of the radius as a measure of the size of circles. 2*pi is a basic maths constant, the number of radians in a complete rotation, and is really worthy of its own name, but that's life.

Power is torque times angular velocity.

0.8: An estimate of how much engine power makes it to the wheels.

 

[damnaged]

crysta1c1ear, thanks a lot for clearing that out, boy what a mistake.. thanks again.

Actually i want to look at the capacity of the engine in "force" perspectives. Basically forward force must exceed force backwards to enable something to move. In my case, trailer weight and friction (rolling resistance) shall be my backward forces. So now i need to determine what is my forward forces. This should be the force which is produced from the engine torque in my truck.

I'll try to clear this out again. From the engine torque, i can determine the power to the wheels based on equation given (can anybody tell me the final output unit, would assist me to understand better). Then from that i need to divide P with v to get drag force, F(i don't quite understand what is v,but power equals force times velocity right? so i'll assume v is velocity). Am i right?