Let's say you had a 300rwhp & 320rwtq. The max hp was obtained at around 6000rpm but the motor reved freely up to 6800rpm which would translate to about 170mph.

Okay now let's say you built another larger motor with around 370rwhp & 380rwtq with the max hp coming at about 5400rpm but the car only reved easily up to 6200rpm. Would it pull a lower top speed a higher top speed or about the same? Everything except the motor is the same.

If your car ran 170 @ 6800 rpm, it will run 155mph @ 6200 rpm. Increased HP and torque will just make it accelerate quicker. You're gonna need a taller gear to take advantage of the increased power at lower rpms.

BTW, the hp/torque numbers are hypothetical since the car has never been on a dyno.

quote:

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Originally posted by steve'66:
Jas,

If your car ran 170 @ 6800 rpm, it will run 155mph @ 6200 rpm. Increased HP and torque will just make it accelerate quicker. You're gonna need a taller gear to take advantage of the increased power at lower rpms.

SteveW

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This is an issue I've really been thinking about alot lately too. Do I build a 513 cubic incher that I shift at say 6500 or a 460 cubic incher that I shift at say 7000 RPM. It seems that either can make good peak horsepower, but is there a clear advantage in the extra cubic inches? I'm talking striktly in the horsepower and torque department, not durability.

I'm currently building a 513" stroker. My primary concern is for the car to run around 6.00s in the 1/8 and not have to turn the car to hard so it will last. I plan on shifting the car a 6400 according to my figures. I initially thought I would just build the 460. My issue was cost. I believe it is just as cheap to build a 513. I am going with the 6.8 rod, 4.440 bore and 4.14 stroke. If you look at the cost of maintenance in the future I believe the 513 is the better way to, at least in my situation.

Just my .02

Torque = Horsepower * 5252/RPM

HP, Torque, and RPMs are directly related to each other.. If you compare 2 engines with the same amount of HP and one has more Torque, then the higher Torque engine would have to be making more Torque at a lower RPM.. Theoretically, if you gear the car with the higher Torque number, to keep the RPMs in the "Peak Power" level of the engine, then the higher Torque engine will always accelerate faster given the same set of circumstances.

Also consider this: If you compare 2 engines and both have the same amount of Torque and HP, and one has "Peak Power" in a higher RPM range, then you will be able to generate more Power to the rear wheels because of Torque multiplication coming from the ability to use a better gear ratio..

Yes. See the tech tips section for a detailed explanation:

https://mustangsandmore.com/ubb/DanJonesAerodynamics.html

>If you compare 2 engines with the same amount of HP and one has more Torque,
>then the higher Torque engine would have to be making more Torque at a lower
>RPM..

Correct.

>Also consider this: If you compare 2 engines and both have the same amount
>of Torque and HP, and one has "Peak Power" in a higher RPM range, then you
>will be able to generate more Power to the rear wheels because of Torque
>multiplication coming from the ability to use a better gear ratio..

No. Rear wheel torque is what accelerates a vehicle. Horsepower, being the
rate at which torque is produced, is the indicator of how much potential
torque multiplication is available. In other words, horsepower describes
how much engine rpm can be traded for tire torque. It doesn't matter where
in the RPM band the horsepower is made. The word "potential" is important
here. If a car is not geared properly, it will be unable to take full
advantage of the engine's horsepower.

It's fairly basic physics to derive the equations for top speed. A vehicle
will continue to accelerate until the total external resistive forces
(aerodynamic drag and rolling resistance) cancel out the propulsive force
As a first approximation, lets ignore the effects of rolling resistance and
concentrate on aerodynamic drag. The power required to overcome the drag
force can be expressed as:

Preq = D * V

where:

Preq = power required
D = aerodynamic drag
V = velocity

Drag is usually expressed in terms of non-dimensional coefficients:

D = Cd * A * q

where:

Cd = coefficient of draq
A = frontal area
q = dynamic pressure

and:

q = (rho * V**2)/2

where:

rho = air density (a function of temperature and altitude)

Substituting into our original equation yields:

Preq = Cd * A * (rho * V**2)/2 * V
= (Cd * A * rho * V**3)/2

Using typical units:

A in ft**2
V in MPH
Preq in HP

Noting:

1 HP = 500 ft-lb/sec
rho = 2.3769E-03 lb*sec**2/ft**4 (sea level standard day)
5280 feet = 1 mile
1 hour = 3600 seconds

Assuming I've done the math correctly, the equation reduces to:

HPreq = 6.817E-06 * (Vmph**3) * Cd * A

Rearranging yields:

/----------
/ HPreq
Vmax = 52.739 * \ 3 / ----------
\ / Cd * A
\/

Plugging in some numbers for a cars frontal area (say 30 square feet)
and drag coefficient (0.32) and solving for 175 MPH:

HPreq = 6.817E-06 * (175**3) * 0.34 * 30
= 373 HP (rounded to the nearest horsepower)

We've ignored the effects of rolling resistance but it typically scales
with velocity raised to something a fraction over one, so the effect
should be small compared to the aerodynamic drag. As a crude rationality
check, Mad Dog Antenucci competed in the 165 MPH class at the Silver State
run in his Pantera. He was running a 377C that made something on the order
of 400 HP at the rear wheels and was able to average right at 165 MPH,
hitting something like 175 MPH in spots, IIRC. Given the equation above,
that seems to be in the right ballpark. Panteras are fairly slick and Ford
supposedly had one in the wind tunnel along with GT40. I think the
published number for a skinny tired 1971 model was 0.29. Wider tires and
wings increase that a bunch.

HP = Torque x RPM / 5,252

kW = Nm x speed / 9549)

Because the maximum torque for a given displacement is constant, the power per cubic inch is directly proportional to engine speed. Since engine maximum speed is inversely proportional to engine stroke, the shorter the stroke, the more power per cubic inch the engine can develop.

Newton's Second Law of Motion covers that as well and states that the
sum of the external forces acting on a body is equal to the rate of
change of momentum of the body. This can be written in equation form
as:

F = d/dt(M*V)

where:

F = sum of all the external forces acting on a body
M = mass of the body
V = velocity of the body
d/dt = time derivative

For a constant mass system, this reduces to the more familiar equation:

F = M*A

where:

F = sum of all the external forces acting on a body
M = the mass of the body
A = the resultant acceleration of the body due to the sum of the forces

For top speed, the same basic equation applies but we only care about
one point, that being th point at which the car stops accelerating.
For drag strip acceleration, you have to worry about the summation of all
the points on the RPM curve between each shift. In mathematics, this
summation is known as the integral of the curve. If you read the article
I noted earlier (https://mustangsandmore.com/ubb/DanJonesTorqueVsHP.html),
it derives the above then goes on to relate crankshaft horsepower to rear
wheel torque. Plagarizing from my own article: "Knowing the basic physics
outlined above (and realizing that acceleration can be integrated over time
to yield velocity, which can then be integrated to yield position), it would
be relatively easy to write a simulation program which would output time,
speed, and acceleration over a given distance. The inputs required would
include a curve of engine torque (or horsepower) versus rpm, vehicle weight,
transmission gear ratios, final drive ratio, tire diameter and estimates
of rolling resistance and aerodynamic drag. The last two inputs could be
estimated from coast down measurements or taken from published tests.
Optimization loops could be added to minimize elapsed time, providing optimal
shift points, final drive ratio, and/or gear spacing. Optimal gearing for
top speed could be determined. Appropriate delays for shifts and loss of
traction could be added. Parametrics of the effects of changes in power,
drag, weight, gearing ratios, tire diameter, etc. could then be calculated.
If you wanted to get fancy, you could even take into account the effects of
the rotating and reciprocating inertia (pistons, flywheels, driveshafts,
tires, etc.). Relativistic effects (mass and length variation as you
approach the speed of light) would be easy to account for, as well, though
I don't drive quite that fast."

>I did some research and found out that you were right and I was wrong. I
>stand corrected.. However... it is possible to have two comparative engines
>with the same amount of Torque, and one can be more powerful than the other..

Correct. In the HP versus Torque article, I given an example of how a
lower torque engine can have more power and therefore better acceleration
than a higher torque but lower power engine.

>Because the maximum torque for a given displacement is constant,

I think you mean "if" the torque is constant. Displacement does
not necessarily determine torque.

>the power per cubic inch is directly proportional to engine speed.

The cubic inch part is not important. The important part is that
the same torque, developed at two different speeds, will result in
two different horsepower numbers. The one developed at higher RPM
will make more power.

>Since engine maximum speed is inversely proportional to engine stroke,
>the shorter the stroke, the more power per cubic inch the engine can
>develop.

Two main issues here. The first is a structural issue. A shorter stroke
implies a slower piston speed and therefore less load on a crankshaft for
a given RPM, all other things equal. Secondly, for a given displacement,
a shorter stroke has a larger bore which means more room for larger valves
and the potential for better breathing. There are lots of other issues
involved but those are the two basic ones.