2.5L with lower revs/more torque, or 2.0L with higher revs?

[Dimman]

Originally Posted by CSG Mike

How about a better flowing head?

The flow on the FA20 is abysmal.

Do you have a CFM figure?

[CSG Mike]

@Anaxilus: you downplay tractor motors, but do you know how expensive tractors actually are? You can almost buy TWO new vipers for the price of an entry level tractor...

[nonicname]

screw high useless revs! gimme some torque!
I don't care who you are, you all know this car is useless on the st.
you couldn't pass a minivan to save your life without down shifting 3 gears.

[Anaxilus]

Originally Posted by CSG Mike

@Anaxilus: you downplay tractor motors, but do you know how expensive tractors actually are? You can almost buy TWO new vipers for the price of an entry level tractor...

LoL, my apologies. Though California is expensive too, I wouldn't wish it on my worst enemy. The Viper already comes w/ a tractor motor anyway.

[EN2_Squirrel]

I autox the car and i would absolutely love more revs. I will be having them tuned in. For most who dd the car and never really track/autox it the displacement and torque would be so much more fun on the road.

[WolfpackS2k]

I didn't vote in the poll because neither choice seems appealing to me. Additionally there are plenty of other viable options available that aren't listed in the poll.

My vote would be for a 2.5L engine with the same (or even slightly higher) redline

[Suberman]

Originally Posted by Dimman

Adding displacement has nothing to do with bmep.

In layman's terms bmep=torque/L.

Correct, sorry, bmep times bore area is what I should have said. Actually, in the real world bmep tends to decline with bore area because of gas flow characteristics but that's a different issue.

[Suberman]

Originally Posted by ZDan

You are wrong. Power absolutely is the INSTANTANEOUS rate of doing work. In the same sense that SPEED is the INSTANTANEOUS rate of change in position.

To say that power *isn't* instantaneous is to make the same error as Zeno's Achilles and the Tortoise paradox (in which it is reasoned that given a head start, the tortoise can never be caught up with by Achilles).

Power DOES accelerate, and in fact engine TORQUE by itself doesn't tell you enough, whereas POWER DOES. Power tells you exactly how much force you have available to accelerate the car from a given speed.



Wrong again. You can run the exercise I gave above for ANY kind of gearing and you will get the same result, same F.

Ferinstance, with 4.88 gearing, 2nd gear at 7000rpm is 46.6mph or 68.3 ft/s
Power is 173rwhp, so force acting at the contact patches:

F = 173rwhp * 550 / 68.3ft/s = 1393 lb

Same power at lower speed => greater force acting to accelerate the car.

Taking rw torque through the gearing and tire radius:
engine torque = 130 rw lb-ft (173rwhp*5252/7000)
Torque at the wheels (2.188 2nd gear, 4.88 diff):
130 lb-ft * 2.188 * 4.88 = 1388 lb-ft
Force = torque/tire radius = 1388 lb-ft/.995ft = 1395 lb.

SAME (allowing for roundoff error accumulation)

And it's always worth pointing out that with shorter gearing you have to upshift sooner, and LOSE your mechanical advantage over some ranges of speed. Overall power/weight is not changed, so overall acceleration is not changed. Lower gearing will not generally improve 1/4-mile trap speeds, because that is a function of POWER/weight. If you don't change either power or weight, you aren't going to see major changes in trap speeds.

Right, just as POWER is the rate of doing work.

And the torque at the wheels (y'know, where it MATTERS) at a given speed is A DIRECT FUNCTION OF POWER.

Power has EVERYTHING to do with it.

Let's take a different engine, making the SAME 130 lb-ft of torque, but only making half the power as the examples above: 86.5 rwhp.

If the engine is making 86.5 hp out of 130 lb-ft, it is by definition spinning at (86.5/130)*5252 = 3500rpm. Keeping 4.88 diff gearing, this equates to 23.3mph or 34.15 ft/s

Keeping the same 4.88 gearing, actual torque at the rear wheels acting to accelerate the car is:
130 lb-ft * 2.188 * 4.88 = 1388 lb-ft at the wheels, same as above.
1388lb-ft/.995ft = 1395 lb. thrust, same as above

Using F = P/V:
F = 86.5hp * 550 / 34.15ft/s = 1393 lb.

Again, SAME RESULT. Accelerating at the same rate at half the speed takes half the power, but the same torque!

Acceleration is the DIRECT result of force applied at the rear tires' contact patches.

This force is a DIRECT function of POWER and SPEED.

This force is the SAME as what you get if you take engine torque through the trans/diff and rear tire radius.

Feel free to not understand, but do NOT discourage others from understanding!

TNX...

http://en.wikipedia.org/wiki/Power_(physics)

[ame="http://en.wikipedia.org/wiki/Acceleration"]Acceleration - Wikipedia, the free encyclopedia[/ame]

[StormTrooper]

I don't know what celica you were driving but it wasn't the last gen....or maybe your BRZ/FRS has some sort of problem.

In almost every way the fa20 pulls harder. I live at 4500 ft and it still feels better than my 2000 celica ever did at sea level.



That being said I'd like more power, I think a well designed header will cure most of the power issues with this car.

[ZDan]

Originally Posted by Suberman

http://en.wikipedia.org/wiki/Power_(physics))

I fixed the linkie for you (it was missing a closing parenthesis)
Glad to see you are reading up on these mechanical concepts!
I hope you paid particular attention to this part:
Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity [P=F*V] , or the product of a torque on a shaft and the shaft's angular velocity.

P=FV, F=P/V

The engine POWER (torque*rpm) is the same as the POWER at the rear tires' contact patches (force*velocity), if you go by rwhp to account for losses.

The same is NOT true of engine TORQUE. Knowing how much torque an engine makes doesn't tell you the force it can apply at the tires' contact patches. You must also know the rpm, or know the gearing and tire diameter and road speed, both of which are roundabout ways to arrive at POWER. Engine torque by itself tells you nothing.

Acceleration - Wikipedia, the free encyclopedia

Yes, F=ma, and a=F/m.

The force "F" acting to accelerate the car's mass "m" at acceleration "a" can be calculated from power and speed thusly: F=P/V. If you want this force in pounds, multiply rwhp by 550 and divide by speed in ft/s. It's that simple.
You can also get the force "F" by taking engine torque, multiplying by trans gear ratio, multiply by diff gearing, and dividing by tire radius.
It's the *same* "F".
There are several examples of this in my previous posts you may want to review.

I do performance modeling as part of my job. I've also done it for my cars. I did a spreadsheet model for my RX-7 before I went to the Texas Mile last year, and predicted its standing mile speed within 1mph.
(see posts #3, #5, and #7 here: http://www.rx7club.com/v-8-powered-r...-mile-1012319/ )
The thrust FORCE I used in the model was F=P/V. Force = power/velocity.
If anybody wants the spreadsheet I used PM me your email and I'll send it.

Engine POWER absolutely determines how quickly you can accelerate from a given speed. Engine torque, on the other hand, tells you nothing. Engine torque with rpm *does* tell you enough, but this *is* engine power! Engine torque, plus knowing the gear ratios, and tire diameter, and road speed, is a very roundabout way of knowing the rpm, and hence POWER.

Knowing only engine torque, gear ratios, and tire diameter *does* give you the force at the contact patches, but does not tell you the SPEED at which you have that force at the contact patches! This is kind of a critical omission. Knowing either engine rpm or road speed fills in the missing blank, giving you POWER.

Again, torque by itself doesn't tell you enough. POWER does.
You cannot calculate/predict performance without knowing POWER (directly or indirectly).

[EAGLE5]

Originally Posted by WolfpackS2k

I didn't vote in the poll because neither choice seems appealing to me. Additionally there are plenty of other viable options available that aren't listed in the poll.

My vote would be for a 2.5L engine with the same (or even slightly higher) redline

Yes, I want the 6 liter, quad turbo V12 that spins to 14k rpm, has gobs of torque off idle that plateaus immediately, hits the power peak 5rpm before redline, and weighs 100lbs less than the FA20 to improve the car's balance.

[EAGLE5]

Originally Posted by nonicname

screw high useless revs! gimme some torque!
I don't care who you are, you all know this car is useless on the st.
you couldn't pass a minivan to save your life without down shifting 3 gears.

You should buy a Mustang or Camaro.

Or shift.

One or the other.

[ZDan]

If you had a 150 lb-ft 2.0 liter making 250hp at 8750rpm with a 9250rpm redline, and a 187.5 lb-ft 2.5 liter making 250hp at 7000rpm with a 7400rpm redline, both having identical power curve characteristics spread over their different rpm ranges, with the 2.5 taller-geared to give the same shift points, they would have identical acceleration characteristics.

Either way is fine with me!