Greenlight

Force = mass x acceleration

Torque = Inertial mass x angular acceleration

If you'll notice both of these equations relate to acceleration, but none of them relate to HP.

HP = Torque (lb ft) x RPM / 5252

Therefore,

(HP x 5252 / RPM) = Torque = Inertial mass x angular acceleration

So, everyone is right.


In order for one vehicle to accelerate faster than another the best combination of low mass and low inertia will be the winner.

Torque is what the crankshaft applies to the transmission, etc. (all rotating components) therefore for a given engine torque low inertial mass is critical in high acceleration.

Force is what the tire applies to the ground (via the friction between the ground, which is not rotating and the tire which is not rotating relative to the ground) therefore for a given amount of force applied to the ground low overall vehicle mass (weight) is critical in high acceleration.

The major resisting forces that are fighting the acceleration is the aerodynamic drag, caused by the air, and the rolling resistance, caused by friction. Because there is low aerodynamic drag at the starting line acceleration is high (unless you spin the tires) and continues to lower as speed (aero drag) increases.

The major resisting inertia is caused by the "mass moment of inertia" of the various components that rotate. The gear ratios also play a role.

Also note that the engine torque continuously varies through the rpm, so the area under the torque curve will ultimately determine the vehicle acceleration.


Hope this helps settle the argument.


Greenlight

AlanP

Your derivation fails to make a distintiction between the angular acceleration of the crank and the angular acceleration of the wheels. The two are not the same and are related by the gear ratio. What you call "low inertia" is really rotational inertia and has absolutely nothing to do with the LINEAR acceleration of a car.

We've had many other (better) derivations in this thread that clearly show that acceleration depends only on HP-over-weight.

neko_cat

That is what you perceive the purpose of this thread to be.
Regardless, your consistently saying something that is simply incorrect.

The gear ratio has nothing to do with the proportionality of torque to acceleration!!!

Look at the plot again. I could EASILY superimpose every gear and show you the plot again. They would all line up perfectly. That any line up is only because I've scaled them. The shape once scaled is a perfect fit. That is the proportionality we speak of. The magnitudes are completely different as we're using different units. Look at the SHAPE Alan. In every gear shape always matches for TQ and G. To show anything in a graph that has different magnitudes requires scaling of the graphed components. There is no reason why I couldn't scale each gear. You state it like it's not possible, of coarse it is. You're just not getting what the plots clearly show. SHAPE not magnitude!

These plots obviously show much more about acceleration of a vehicle than weight and peak HP could ever do. THEY SHOW ACCELERATION with time graphed and have everything else superimposed for your edification. It's easy to see the relationships if you'll look.

These plots are GTECHED. They are derived from acceleration samples over time with a known weight. That being the case they are all RW numbers compensating for everything, wind, rolling resistance, transmission loss, engine friction. All of that falls out because it's measuring where the rubber hits the road acceleration over time.

You can even see the flywheels energy dump at the beginning of each shift as the clutch slips. It's the highest acceleration the car undergoes.

Quit being stubborn, it's right in front of you. I give up.





-mikey

foolio

Engine A makes 240 HP @ 9000 RPM. It makes a completely flat 140 lb-ft torque from 0 to 9000 RPM.

Engine B makes 240 HP @ 5000 RPM. It makes a completely flat 252 lb-ft torque from 0 to 5000 RPM.

Here's the answer:
We take the engine, put it in a box (black, of course) with only the output shaft pokin' out. Assuming you are deaf, can you tell which engine is inside the box?

No. I can make the blackbox engine's output characterics perform like B, by using engine A and putting a reduction gear of 1.8 at the output shaft. Similarly I can make the blackbox perform like A by using an overdrive gear of 5/9 on engine B. qed.

Engine torque is irrelevant. What is useful is how the peak torque number hints at the location of the power band.

AlanP

Take a deep breath and see if you can comprehend six lines of algebra without going off on yet another incoherent rant.

T = torque at the crank (units N*m)
wc = angular speed of crank (units radians/sec)
w= angular speed of wheels (units radians/sec)
F = force on car (units Newtons)
R = radius of wheels (units meters)
P = Power developed by engine (units Joules)
v = speed of car (units m/sec)
m = car's mass (units Kg)
a = car's acceleration (units m/s^2)

P = T*wc (see any high-school physics book)
F*v = T*wc (since P = F*v)
m*a*v = T*wc (since F= m*a)
a = (T*wc)/(m*v) (algebra)
a = (T*wc)/(m*R*w) (since v = R*w)

a = (T/(m*R)) * (wc/w)

Note that wc/w is the gear ratio. The acceleration is therefore proportional to the product of torque and gear ratio.

Listen very carefully: If you plot both "a" and "T" as a function of w on the same graph, they will not have the same shape if you shift gears because (wc/w) changes each time you enter another gear.

Don't you understand?

neko_cat

Actually I do understand and so should you.
If you plot what you've just described a and T will have the same shape after gear change. As you shift gears (wc/w) changes but so too does "a" as a result and it's change is by the same ratio preserving the shape but changing the magnitude of "a". "a" has the same shape but a different amount. This is very visible and proven by the plots. I could show each gear for you if you like to prove it. No matter the gear "a"'s curve will always be the same shape as "T" .

Note that we are discussing T at the RW not at the crank. I've mentioned several times that the Gtech measure acceleration to derive HP. It works in reverse of most peoples way of doing it but should be a perfect method for you.

Regardless of where you measure if I have a crank dyno and know the gears and the cars mass and the tire size I can calculate RW values for everything and show then that "a"'s shape does indeed match "T".

-mikey

neko_cat

Engine torque is completely relevant, you've simplified by making the curve flat.
That's not realistic. The acceleration is identical because it's flat in both cases and all you're doing is regearing to match cases.

Do the same test without flat curves. Which is quicker? Now you need more information. See it's not that simple and basically comparing two cases needs all the information. Once you have all the information it's obvious which will be quicker. It's better to look at each case alone then you'll see the real use of the torque curve. I want to improve acceleration in each gear at this RPM, how do I do that? I increase torque at that RPM area. Yes HP will increase too. Gearing is a multiplier to the entire band it doesn't change the bands shape. HP and TQ curves remain the same after gear changes. They just stretch and scale, the shape remains the same. This parts for Alan.

At the same RPM TQ and HP will be the same regardless of gear. Knowing that will the shape of "a" match TQ in each gear?

-mikey

brockLT1

how about this....

a linear horsepower curve + a flat torque curve + aggresive gearing = the best way to accelerate.