Need to vent...stupid people

[ryan34]

u shoulda told him u have a 7 second corvette that u use as a weekend car and the formula is just ur daily driver

i kno how u feel, my town is crawling with ricers and mustang, they always ride my (_!_), i just say whatever and move one with my day, they'll end up in a telephone poll some place

Haha, i should've. Prolly shoulda said (insert car name) twin turbo, forged internals, and rearend, rotarized pistons and gamified valves, then see what his reponse would've been.

[MarkFL]

I know this isn't really the point of the thread, but that's not an accurate statement. Column average horsepower wins races.

Anywho... I'm not sure how I'd respond to a complete stranger initiating a conversation with me by saying, "my truck would annhialate your car." What an asinine red-neck thing to do. I don't think I'd take very kindly to it...

I'm certainly no expert at automotive kinematics, but...

if we assume no slipping or loss of traction and ignore drag, then acceleration is proportional, if not equal, to torque per mass. Seems to me acceleration, or how quickly velocity changes with time would win drag races.

btw, I am unfamiliar with column average horsepower, would love to know what that is.

[Craig 94 TA GT]

Wouldn't a similar calculation to column average horsepower be calculated by:
finding the integral of the closest formula of the horsepower curve, i.e. the area under the curve?

[MarkFL]

are you saying it is the discrete version of the average value of the power function P(t) over an interval [a, b], given by

A = (1/(b - a)) times the definite integral of P(t) evaluated from a to b?

hmmm, I would think the area under the torque curve would be more meaningful to the velocity function, given the defined relationship between acceleration and velocity.

[Jay 02 TA ws6]

I'm certainly no expert at automotive kinematics, but...

if we assume no slipping or loss of traction and ignore drag, then acceleration is proportional, if not equal, to torque per mass. Seems to me acceleration, or how quickly velocity changes with time would win drag races.

btw, I am unfamiliar with column average horsepower, would love to know what that is.

We should probably start another thread, because this is definitely off-topic

But... briefly stated...

what you said is correct, assuming the two cars racing have only one gear. In a given gear, at a given ratio, then yes, the car with the most torque is going to win (assuming he doesn't run out of gear and redline before the finish line).

HP and TQ aren't very different... at any given RPM, a car producing more TQ is also producing more HP by definition. The trick is, HP is RPM dependant. So, thereby using RPMs to create mechanical advantage - let's say a car with a 7K RPM shiftpoint is racing one with a 6K RPM shiftpoint - that car can affectively have 6/7th as much torque, but by gearing the car with a 6/7ths shorter gear, can put the same amount of torque to the pavement and accelerate equally as quickly (or they can use the same gears, and just remain in the lower gears longer - basically creating the same affect). You see something like this when you see a Ferrari racing a Corvette... 400HP/400TQ for the Vette, 400HP, 300TQ for the Ferrari, yet they perform very similarly.

That wasn't as short as I hoped.

[Jay 02 TA ws6]

btw, I am unfamiliar with column average horsepower, would love to know what that is.

Essentially, the average area under the HP curve per gear the car is in at its shift points while it accelerates. 1000-3000 RPMs is rather meaningless, because when you bang the upshift off redline, you land on 4000 RPMs in the next gear. Other than 1st gear, 4000-6000 RPMs is the most important part of the powerband for you, and your average area under the curve here is what's going to win you the race. You could then say column average TQ is most important, but remember that RPMs play the role as stated above (and HP is RPM dependent)... if you can spin the higher RPMs, you can use shorter gears (or stay in lower gears longer), put more power to the pavement, and accelerate more quickly.

[MarkFL]

Thank you, Jay for that very understandable explanation.

...The trick is, HP is RPM dependant...

yes...absolutely correct...as per the work-energy theorem.

Imagine an object of mass m free to move in space outside of any gravity fields and whose velocity is initially zero. Now also imagine a net unbalanced constant force F being exerted on the object. The object begins to move in accordance with Newton’s laws of motion. For a constant force, we will have constant acceleration, giving a linear rise in velocity and a quadratic change in position, according to the calculus.

While the work being done by the force per unit of distance remains constant, the work being done per unit of time is increasing with the velocity, i.e. Fv. This means the power output is proportional to the velocity, with the constant of proportionality being the magnitude of F.

So, for constant torque, power must rise linearly.

[fastTa]

Torque is nothing without HP. TQ and HP, although related, are two different animals. TQ is a measure of force and HP is a measure of work, in this case mechanical power.

Neither TQ or HP solely win a race.....it's an ideal balance of both.

[Jay 02 TA ws6]

Torque is nothing without HP. TQ and HP, although related, are two different animals. TQ is a measure of force and HP is a measure of work, in this case mechanical power.

Neither TQ or HP solely win a race.....it's an ideal balance of both.

We'll agree to disagree. And I'm sure we'll talk about it on AIM too. haha

[fastTa]

We'll agree to disagree. And I'm sure we'll talk about it on AIM too. haha

I'll be on later.

[MarkFL]

my $0.02...

power is the rate of work per time, or more generally the time rate of energy transfer. ofr us that's how quickly the potential chemical energy of the fuel can be converted to kinetic energy of our cars.

torque is the time rate of change of angular momentum, the vectorial cross-product of the force and displacement vectors. It is the angular version of force...i.e. force is the time rate of change of linear momentum.

for our purposes

torque = (16,500/pi)*(power/angular shaft speed)

you can't have one without the other...

angular shaft speed is rpm/ tranny gear ratio. Of course the gearing in the differential will also affect the torque at the pavement.

[WhiteSmoke]

I feel your pain ryan... but i'de have to say living in NYC surrounded by riceburners with 23 HP thinking they can take a 95 LT1 reving those lawnmowers at ya and asking to race on a daily bases gets right up there with the worst of them =P So next time do as i do, lay a nice patch and hope they get lost in your smoke =D