I know this topic has been beaten to death on this post, but I just wanted to try to explain it in a slightly different manner than everyone else has.

Work is equal to force*distance, if you say that you are going to keep the same torque, the force is constant. If you take the time derivative you end up with dw/dt=f*d/dt. Work with respect to time is power, in this case hp, and force is tq since we are dealing with rotational motion. HP is a function of torque and rotational speed aka rpms. If you look at the equation it is obvious then that d/dt is proportional to dw/dt, showing that if you are creating more power you're speed is greater,assuming a constant torque. HP=torque*rpm/5252 so if you are making torque of say 300ft-lbs at 2k vs. 4k rpms you are making more power. If you are making more power you are going faster because your rate of work is greater. That means that if you want to get down the track faster hp is obviously the number you want to be looking at.

Then why is tq also an important number too? Like someone stated earlier tq is a snapshot of the engine, where hp is the big picture so to speak. tq at low rpms is important when coming off the line because of f=ma, so you will be able to get to the point where you make maximum power faster. Aftermarket cams give the ability to hold your tq output through higher rpm's giving you the ability to make more power, and get down the track faster.

Torque and horsepower are both important because they are related to eachother. If you want to look at how fast you want to go you need to look at hp because that is what that variable represents.

 

Would it be safe to say there are three main factors which "control"
torque:

1. Crank stroke
2. Connecting rod length
3. Number of cylinders

The output torque is multiplied against the cylinder pressure created
on the power stroke.

Horsepower, like "Wattage" is a measurement.

With electricity, you need voltage and current to make power (P= I * V)

The more Wattage indicates a combination of more current, or more
voltage.

Is this a symbolic parallel?

 

Larry, that was great info. Thanks. When I get a few minutes I want to run some of that thru my cheapy simulator.

FWIW, does you sim use thrust at the tire/road surface to figure acceleration?

 

Larry, that was great info. Thanks. When I get a few minutes I want to run some of that thru my cheapy simulator.

FWIW, does you sim use thrust at the tire/road surface to figure acceleration?
===========================================

OldSStroker

this Link is very Quick calculation of Greg Anderson's 6.67 ET

http://www.maxracesoftware.com/HP_Calc_Programs.gif


this Link is another Quick calculation, but with more inputs and accuracy

http://www.maxracesoftware.com/Quick_HP1.gif

and this last Link , i just inputed 6.670 ET and program generated the rest of ET/MPH times
http://www.maxracesoftware.com/HP_Calc.gif

all calculations are baselined off 600 Rpm/Sec accel rate engine Dyno test.... if you use Quarter Pro or Quarter Jr. or any other program.... you might have to multilpy 1.036 to 1.04 times
HP @ 600 rpm/sec to get "Steady-State" non-acceleration HP
to input

example=> 1330 Peak HP @ 600 Rpm/Sec will usually result in
1330 * 1.04 = 1383.2 HP to plug into Quarter Pro version

note=> you might have to plug in as much as 1400+ HP in some programs out there to model ProStock , sort of unrealistic ?

and my 95.0 % percent DriveTrain Efficiency is a "combination" of
the Transmission Eff % times the Differential Eff %

example=> .98 % Trans times .97 % Differential = .9506 or 95.06

Quarter Pro has (2) Inputs , one for Trans and one for Diff

--------------------------------------------------------------------
***FWIW, does you sim use thrust at the tire/road surface to figure acceleration?

yes , or more precisely "Tractive Force" at the Drive-Tire contact patch

i'm not using a Mass Factor , but calculating all rotating objects's rotational inertia and subtracting these losses as RaceCar accelerates down the DragStrip

if you are using Quater Jr. ..it won't model Greg Anderson's Pro Stock record very well without plugging in abnormal HP and Aerodynamic Frontal Area

same but to lesser degree with Quarter Pro version , but will still have to plug in abnormal Aerodynamic and HP curves to model ProStock MPH times correctly

a good book to read or purchase would be
"Fundamentals of Vehicle Dynamics" by Thomas D. Gillespie

look at Chapters 1 pg 10 to pg 19
Chapter 2 (all pages )
Gillespie uses a Mass Factor , its OK but won't get you accurate enough simulation for wide range of Drag Racing applications

Regarding this HorsePower -VS- Torque debate ;

i could rewrite my simulation just using only HorsePower values
..without ever using Torque ..and results would come out the same

or you could calculate ET/MPH times again using another method that doesn't need to know engine Torque or Trans Ratios or Differential Ratios ...and still do a very good job of predicting
quarter mile times
sort of like in this Link
http://www.maxracesoftware.com/Quick_HP1.gif

 

Key factors missing include:
4. displacement per cylinder (or just combine 3 & 4 to make it total engine displacement)
5. VE
6. Piston surface area (combustion chamber is pressure, surface area determines the total force on the piston)

There's obviously a bunch more that affect VE and combustion effiency, but you can wrap most of it up into these:
1. displacement
2. VE
3. Piston surface area (combustion chamber is pressure, surface area determines the total force on the piston)
4. Connecting rod length
5. Crank stroke

Quote:

Horsepower, like "Wattage" is a measurement.

With electricity, you need voltage and current to make power (P= I * V)

The more Wattage indicates a combination of more current, or more voltage.

Is this a symbolic parallel?

yep, same idea. In your analagy the Voltage (elctrical potential on a charged particle) is paralleled by force (torque), while current (a measure of charged particles / second) is paralleled by speed (RPM). Wattage and Horsepower are both a measure of power.

The power produced by the engine will determine (via your choice of gearing) how much power you can expend on the rear wheels. Obviously gearing determines how you split this up in terms of rpm and torque, but the idea is to keep forces high as traction will permit.

To go as fast a possible, you DO need to maximize torque to the rear wheels but to do that you need to start with the most hp you can produce at the flywheel.

 

Tractive force = driving wheel torque/rolling radius, right? If you use hp @ rw it just converts to torque to get instantaneous lb-ft.
I suppose if one had GOOD rear wheel numbers that would work also. Using a wheel-mounted dyno(s) would take the traction thing out of it for chassis dyno run.

I'll look for Gillespie's book. I like Milliken and Milliken's Race Car Vehicle Dynamics.

I use an old Performance Trends program, not Quarter Jr.

 

Thanks Steve,

I left out displacement because I figure cylinder pressure would
account for boosted vehicles "effectively" enhancing displacement.

My thought as I typed that list is, using cylinder pressure as a given,
someone could calculate the torque produced using a combination
of displacement, rod length, crank stroke?

This of course assuming perfect combustion, friction, color of the floor mats, etc.

In a perfect world if the cylinder(s) could charge at 100% VE at any
point in the power band, the torque "Curve" would be perfectly
flat from idle to redline...again assuming the floor mats were green!?

The fact that most engines roll off at high RPM has something to
do with pumping losses, friction, intake&exhaust tuninig...and the
list goes on.

So what I wanted to confirm is (let's use a single cylinder motor):

If one power stroke creates 300 ft./lbs. of torque, at 1 RPM, 100% VE ...

At 6000 RPM, I am getting 6000 power strokes creating 300 ft./lbs.
of torque.

Torque hasn't changed because my perfect non friction, 100% VE
motor has the exact cylinder pressure across the tachometer...

but

Horsepower increases because the motor is able to perform 6000
times the work in that minute?

 

Yup. People need to keep in mind, though, chassis dynos correct the torque number for gears. If the Dynojet says you’re putting down 400 lb ft of RWTQ, you’re actually twisting the rear wheels with 4362 lb ft in first, 2919 lb ft in second, etc (M6 with 4.10’s).
Continuing with the line of thinking in your first two sentences instead of going rotational can help people understand. Power is not just a function of torque and rotational speed. It is also a function of linear force and linear velocity. That’s right, Power = Force * Velocity. And Force = Power/Velocity. It really is that simple, folks. For any given velocity, the amount of force you can apply directly depends upon how much power you have.

For example, you have 400 HP and you’re going 100 MPH. The maximum force (and thus, acceleration, going back to a = f / m ) you can apply to the pavement is simply ( 400 * 550 ft * lbs / sec ) / (146.67 ft / sec) = 1500 lbs of force. If the car weighs 3000 lbs you are capable of accelerating at 1500 lbf / (3000/32.17 slugs) = 16.1 fps/s. If you have more power, you can apply more force at that speed no matter what your torque and you can accelerate at a greater rate. If you have less power, you cannot apply that much force no matter how much torque you have and you can’t accelerate as fast.

And that’s all there is to it….

 

That’s right, Power = Force * Velocity. And Force = Power/Velocity. It really is that simple, folks. For any given velocity, the amount of force you can apply directly depends upon how much power you have.
===================================

Jon A ,

thats exactly the equation i'm using as the basis of this simulation in this Link

http://www.maxracesoftware.com/Quick_HP1.gif

then subtracting Aerodynamic, Rolling Resistance, Total Drivetrain, and Rotational Inertia losses
to calculate Peak HorsePower required

Power = Force * Velocity in Feet per Second

will let you correctly model a 5 HP Jr Dragster all the way to
Top Fuel Dragster record
=========================

Jon A ,
you did a great job of explaining HP -vs- Torque

Steve in Seattle, also basically posting the same thing.

 

I think after doing a lot of research and reading articles such as this, I feel HP is a term used for convenience "place-holder"...rather than having a practical use in propelling a vehicle faster down the track...torque, I feel, is the relevant force for ultimately going "faster" whereas HP is a functioned result of torque applied...mho...

 

Here's a cool exercise to play with:

Make a graph. The Y-axis is your Torque (ft./lbs.) and the X- axis is RPM.

Along the X-axis make a scale of 500 RPM increments (0-6000 RPM)

On the Y-axis, create a scale of 50 ft./lbs. from zero, to 400 ft./lbs.

Start at 1000 RPM and make a mark at 300 ft./lbs.

Repeat this for every 1000 RPM increment.

When you connect the dots, you will have a straight line from 1000 RPM to 6000 RPM
with an output of 300 ft./lbs. at every RPM interval.

You now have a perfect "flat torque curve" ... that everyone tries to
achieve.

Pick your shifts points using the torque "curve".

Use the same graph, but draw a line starting at 300 ft./lbs. and slope it down
from 1000 RPM to 6000 RPM to end up with 150 ft./lbs. at 6000 RPM.

Pick your shifts points using the torque "curve".

Once again, draw a line using 200 ft./lbs. across the band. Instead of
making it straight all the way across, draw in a little hump from 3000 to 4500 RPM.

Pick your shift points using the torque output.

Write back with your answers for each case.

 

That article is not really very informative and is non-factual. Larry knows a lot about some things and this is not one of them he is relating very well as he is prone to saying crazy things to get people's attention. No one is the modern world of racing goes faster with less power. People are in the business of picking up power not losing it from F1 to Pro-Stock. Power has continued to rise for decades while TQ in the same classes has stayed almost the same such as in F1 for example. They are much faster now even with smaller engines but they make more power even they they make very little TQ.

You can not gain TQ at any rpm without gaining power. You can gain power and go quicker and faster though WITHOUT gaining any TQ but by keeping the same TQ going till higher and higher rpms and then using a lower gear or not having to shift up to the higher gear as soon and this maximizing the REAR WHEEL TQ which is totally different from the engine's TQ. The more POWERFUL engine will always have the possibiblity of more RWTQ at any given vehicle wheel speed. I can take an engine with even more power and even less TQ than the two above and will also go faster as it will make more RWTQ when geared right.

 

I did a lot of research since I last posted some weeks ago on the subject of HP vs. Torque...and I just believe that there is a more scientific basis for Torque getting you through the lights faster...relatively speaking. I know high rpm engines that produce plenty of HP with the right HP/weight ratio will be quick, giving isolating a bunch of other variables. I guess when I read that some "full" race application vehicles take up to a year to work everything out (i.e. optium gear ratio (primary and final) and separate shock settings for each gearthat adjust automatically...well...sort of "overkill" for making a case for HP....My $.02...

 

If a power stroke results in "xxx" ft./lbs. of torque at a given RPM, then it
would seem obvious that 6000 power strokes per minute of a certain amount
of torque will yeild more power than 4000 power strokes per minute of the
same amount of torque.

I'll even give you an extra 50 ft./lbs. @ 4000 RPM to my higher RPM motor.

 

As for the article, he makes good points. It's all relative to this HP vs. TQ, because
the RPM windows he's referring to are not ilde speeds, or cruising RPM.

He would rather build a motor that can achieve a strong band of torque where
gearing can be made effective to propel the car.

Unfortunately, shifting an engine at a peak of 5000 RPM on a drag strip is going
to drop the next shift down around 3000 RPM (depending on the ratios).
How many gears will you need to complete the race?
What sort of power is the motor going to make at 3000 RPM?
Where is the maximum VE occuring?

Certainly not down low on a large volume intake /big cube motor.