TT_Rex_7

Yea i've looked at those pics, and it really doesnt look like theres alot of material to work with. Although that is from an 86-88. The only way I'll know for sure how much room and so on i'll have is to cut open an rx8 rotor, wich they have for $320.00 wich is even better. How did you come to the conclusion that .9 pounds would be removed from the rotor by enlarging the pockets by that much?

-Alex

KevinK2

my bad ... should be .47 lbs per rotor.

Vc = combustion chamb min volume, cu-in
D=displ per face = 40 cu-in

Vc = D/(CR-1)

Vc (10:1) = 4.44444
Vc (9:1) = 5.0

delta = .555 cu-in per face

rotors are cast iron with steel internal gear inserts ... i used high end of density range for cast iron, .283 lbs/cu-in.

weight loss per rotor = 3 x .555 x .283 = .47 lbs

the_glass_man

iTrader: (2)

I wish Mazda would have went with Aluminum rotors and housings.

jimlab

If each combustion chamber is 654 cc at maximum expansion (2 x 654 cc = 1308 cc or 1.3 liter), then with a 10:1 CR, it would have to be 65.4 cc at maximum compression. 9:1 CR would result in a 72.7 cc combustion chamber at maximum compression.

V1 = volume at maximum expansion
V2 = volume at maximum compression
CR = V1 / V2
654 cc / 65.4 cc = 10:1 CR
654 cc / 72.7 cc = 8.99:1 CR

72.7 cc - 65.4 cc = 7.3 cc.
1 cubic inch = 16.387064 cc
7.3 cc = 0.45 cubic inches by my math.

3 x .45 x .283 = .38 lbs per rotor

KevinK2

if you start with CR = (D + Vc)/Vc, with my definitions, you'll avoid a bad right down the road, where your V1 is assumed constant for 2 different pocket sizes.

DamonB

Some more food for thought. I'm reading a book on datalogging and came across this which I thought interesting:

"...aerodynamic drag force rises with the square of speed but the horsepower absorbed rises with the cube of speed."

I knew about the rate aero forces increase but hadn't seen it related to horsepower. This goes along with the equations KevinK2 posted earlier in that even with aero drag removed the higher the speed the less power is available for acceleration.

jimlab

Ah, I see what you're saying now.

40 + 4.4 = 44.4
44.4 / 4.4 = 10:1

40 + 5.0 = 45.0
45.0 / 5.0 = 9:1

jimlab

Yep.

The formula for fluid friction ("wind resistance") is...

F = 0.5 * Cd * A * rho * v^2

Where Cd is the coefficient of drag, A is frontal area, rho is the air density (1.29gm/l under standard conditions), and v is velocity.

To calculate the power required to maintain a certain velocity, we have...

P = F * v

and with a little substitution...

P = 0.5 * Cd * A * rho * v^3

and finally...

Q = P * 5,252 / RPM

tookwik

I think most of us get it, just have a hard time explaining it.

KevinK2

Well I wasn't clear .... not unusual

My point was that at 120 mph, with no friction or drag, 300 hp at the engine (and at wheels) will give you 1/2 the acceleration (g's) that 300 hp at 60 mph will provide.

Going by known power, don't need to know gear ratios, rpm, etc.

Take a 3000lb FD, with steady speed of 60 mph on flat road (zero hp required w/o friction and drag). Now slam in 300 hp.

Since hp = F x V, the instant push force will be hp/V, and then instant acc'n will be F/M, so combining these:

a = hp/(VxM)

get units right ...

hp = 300 x 550, ft-lbs/sec
V = 60x(88/60), ft/sec^2
M = 3000/32.2, slugs

a = 20.1 ft/sec^2, or .63 g's at 60 mph.

Using same eq'n for 120 mph start, a = .31 g's.

--------------------

Aero drag sucks up hp based on V^3 and a drag factor, as you and jim noted. This does leave less hp to accelerate the car, in my equaton above.

The hp loss due to drag can be expressed as k x V^3, where k is a constant that includes stuff jimlab noted. Now including drag, formula for instant acceleration in my example where hp is at engine (and wheel, since still no friction).

a = (hp - kV^3)/(VxM)

So now a double wammy regarding reduced acc'n at higher speeds.


-----------------

Add friction. Friction in the tires and drive train can include constant hp losses and those ~proportional to velocity. Since the numerator in the equation above is already reduced at high speed due to aero drag, even a constant hp loss will have more effect on acceleration at higher speeds.

I'ts tough to go real fast.

turbojeff

iTrader: (30)

Check my numbers...

If you shift at your shift points you are well past "acceleration peak or peak torque" in every gear except the 1-2 shift because of a large jump in the 1-2 ratios. If I did the math right it shows keeping the engine at or near peak HP not peak torque gives the best overall acceleration.

after 1-2 shift = 4600 rpm, 400 rpm shy of peak torque
after 2-3 shift = 5300 rpm, 300 rpm after peak torque
after 3-4 shift = 5450 rpm, 450 " "
after 4-5 shift = 5400 rpm, 400 " "

Keeping the engine closest to peak HP not peak torque gives the greatest acceleration. When torque curves cross, wheel speed is equal, power to the wheels will be equal in both gears. Since we know that the torque and HP is falling at high rpms it is advantageous to shift at that the intersection of the curve.

Not including the 1-2 shift, the numbers show that the proper shift points are centered more on the HP curve than the torque curve. ~1000rpm before peak HP to ~1000rpm after peak HP. This gives more "area under the HP curve" and gives the greatest acceleration.

jimlab

Did you read the part about torque at the axles in the current gear falling below torque at the axles in the next higher gear being the definition of the ideal shift point?

Of course you're past your torque peak in the current gear. But you're still making more torque at the axles than you would be in the next higher gear, and that means you're achieving the fastest rate of acceleration.

Give it up. Horsepower does not determine the rate of acceleration. End of discussion.

turbojeff

iTrader: (30)

Jim, your shift points show keeping the engine closer to the HP peak. That coupled with the gear spacing is determining the shift points, and your shift points agree with me.

Keeping the engine closest to the HP peak absolutely gives the greatest acceleration. There is absolutely NOTHING wrong with that statement.

What I think you are thinking is that yes the gearing provides the "thrust" force at the contact patch, lower gears multiply this thrust force and give greater acceleration.

Example:
From trap speed and weight we can reasonably calculate engineHP but not engine torque. Because you can have any number or torque * rpm combos that will produce the given HP to move the vehicle down the track but it is the HP that moved it in the given time.

importrx7

Wouldn't the only thing lightweight rotors would do is let the engine rev up fast since its moving a lighter rotor?

Quadulus

Torque is a function of how much air you put into the motor. The weight of the rotors have no influence on torque in an ideal world.

Heavier rotors means higher friction forces on the bearings which lead to greater power losses at speed.

I think the torque your friend is referring to is just due to the flywheel effect the extra rotating mass would give you.

Go with the lightweight rotors.

EVAN616

iTrader: (2)

TT_rex_7

I got tired of reading all this stuff so i skipped and just replied hope it helps.
I personally have milled my own rotars on a 2nd gen N/A car this does make the compretion ratio drop, but this is the result you want. Well, the result i wanted because after the drop in compretion i slapped on a turbo and there you go. I make lots of my own stuff in my shop as well. For the tourque ? i dont think it will change at all since it does nothing to the so called stroke of the engine i think the only way this could be acheved in a rotary would be to remove material from the eshaft journals on the rotars then make your own eshaft "since you seem capable given the right material that is". The new eshaft would have to then make up for the now larger hole through the rotar. You would do this in your new shaft by not making the journal on the shaft bigger but offset further from the center. Making a larger "stroke in a rotary" which would then increase tourque.

Hopefully that isnt too confusing, and i am an engineer btw i read that far and thought i would chime in lol.

I am in the process of making working 3-d models or rotarys so we can test this stuff before resorting to trial and error. So if any or you have full specs or drawings or a rotary apart that would be nice.

jimlab

You're an Engineer, huh?

TT_Rex_7

What compression ratio did you machine it to?!

What I've decided to do is machine 2 s4 TII rotors from an 8.5 to an 8.0:1 compression ratio. Depending on the weight of it afterwards will decide if I'll take more material off on the front and back of the rotors (see the pictures Jimlab posted in a previous post) As long as it weights in at the same weight of the 93-95 rotors I will have no reason to take more material off. Machining an rx8 renesis rotor down to a 9.0:1 compression ratio really doesnt have any benefits. It will be lighter but that wont yeild near as much power as a lower compression ratio where you can run more boost on pump gas. I think trying to take that much material off the rotors is to much of a risk for a street car. I'll be taking measurements from the s4 rotors Monday and hopefully get the programming started so I can put it on 1 of the machines next week!

-Alex

KevinK2

Bingo. As I posted before, you can look at hp to the rear wheels, or torque to the rear wheels as jim and most others do, to get your optimum shift points. Both methods give the same exact shift points. In both cases, you want to keep the value as high as possible at all speeds.

Jim is max'ing torque at the wheels (and not torquing bolts ...), and to do this one must go past the hp peak in each gear, as Jeff is trying to say. If the hp curve is flat beyond the rpm for peak hp, you shift at redline. I'll try to ship my crude excel sheet to los dos, it has both curves.

banzaitoyota

Stan At SDJ offers lightening and balancing. In Fact he owns Daryl's original balancing machine

the_glass_man

iTrader: (2)

Contact info?

Gorilla RE

iTrader: (1)

jimlab-"Give it up. Horsepower does not determine the rate of acceleration. End of discussion."

i have to agree with you ... torque makes the car accelerate horsepower more or less keeps you going till top speed, givin your geared for it.

Gorilla RE

iTrader: (1)

jimlab-"Give it up. Horsepower does not determine the rate of acceleration. End of discussion."

i have to agree with you ... mr. webster...torque(to twist. a force that produces or tends to produce rotation or torsion {an aoutomobile engine delivers to drive the shaft}. a measure of effectiveness of sucha force that consists of the product of the force and the perpendicular distance that from the line of action of the force to the axis of rotation. a turning or twisting force.) makes the car accelerate horsepower more or less keeps you going till top speed, givin your geared for it.

Gorilla RE

iTrader: (1)

ooops dont know how that happend....

MontegoBlue

I'd like to weigh in on the torque/hp debate as a graduating mechanical engineer at the University of Michigan. (Note: this post got very long somehow, skip to the last 2 paragraphs if you get bored)

Let's start at the beginning of the process - combustion. We've agreed that torque is generated when the center of pressure from expanding gases pressing against the piston/rotor face acts with some moment arm about the centerline of the crank/eccentric shaft. Each combustion event occurs in basically the same way, regardless of engine speed - some air is mixed with a mist of fuel and sucked into the combustion chamber, where it's compressed and ignited. Why does torque change with engine speed then, if each combustion event uses (basically) the same volume of air (ignoring turbocharging)? There are a number of physical reasons, chiefly the amount of resistance the piston/rotor encounters at different times during the intake and exhaust cycles, which depends on the airflow characteristics of the intake and exhaust paths. (Think about the Harley guys who weld tuneable restrictors into their exhausts to increase low-end torque. The effect that the restriction has on engine performance changes dramatically as airflow increases) Also interesting to consider is the difference in the rate of change in pressure during combustion (which is a function of fuel/air mixture and ignition) and the rate of change in volume of the combustion chamber, which is a function of engine speed.

The point of all this is that "power" and "torque" are describing the same energy, but with different parameters. Power is just torque multiplied by rotational velocity. Imagine a perfect engine with a perfectly flat torque curve - power output would increase linearly with engine speed. This is desirable because most of the forces that resist the acceleration of your car also increase linearly with velocity - viscous friction in the air and in the lubricated gear contact in the driveline. (as a first approximation, anyway)

Jimlab is correct when he insists that it all boils down to F=ma. F, in this case, is the SUM of the forces on the car - the force at the wheels (a funtion of engine torque and gear reduction) minus the external forces on the car (viscous friction and tire rolling resistance, which are more or less a function of speed). Since the mass of the car is constant, the acceleration of your car is always proportional to the sum of the forces on it. From a dead stop, you can accelerate relatively quickly at any given RPM, because the force at the tires is very high (because you're in first gear) and the external forces are very low (because they depend on velocity, which is zero.) In this case, all that is resisting acceleration is inertia - the inertia of the engine internals and the inertia of the car. At very high vehicle speed, you can't accelerate as quickly at the same RPM - not because the inertia has changed, but because F is now very low. The force at the tires is much lower because you are now in 5th gear, and the frictional forces are much higher because you're moving at high speed. At some point (the top speed), the frictional forces equal the forces at the tire and you can't accelerate any more (because F is zero). The reason that a lightened flywheel (or rotor, or driveshaft, or wheel, or anything else) produces a noticeable change in acceleration is that it reduces the inertia of the system, as we've said. At low speeds, when the external forces are very low, inertia is the dominant parameter in limiting acceleration. At high speeds, when the external forces are very high, they swamp the inertia.

The bottom line is that the conclusion is correct that lightened internals do not change torque or power output. They decrease inertia, thereby increasing the acceleration provided by a given force.

For me, the best way to think about all this is to consider an energy balance. The first law of thermodynamics is conservation of energy - and your engine is producing energy whenever it's running (power is just the energy produced per unit time). The energy that the engine produces has to go somewhere, and the only two places it can go are to accelerate the car or to be dissipated to heat by friction.

As for the question of whether to shift at the hp peak or the torque peak, just think of it mathematically. Because your goal is acceleration, you want to maximize the average torque over the time of your acceleration. But when you look at a dyno graph, you don't see torque as a function of time, you see it as a function of SPEED, in RPMs. This is what can be misleading. Because power = torque x speed, THE MAXIMUM POWER IN THE SPEED DOMAIN DELIVERS THE MAXIMUM TORQUE IN THE TIME DOMAIN, which leads to maximum acceleration. What does this mean? When you shift at the power peak on the dyno graph, you've gotten the most average torque during the time of your acceleration in that gear, and therefore the highest average rate of acceleration. Both turbojeff and jimlab have the right idea, but it's very hard to talk about this topic without getting confused by jargon.

As for lightening the rotors, it's an absolute last resort. You can achieve very useful reductions in inertia by lightening rotational components that are not routinely subjected to huge stresses and temperature cycles, like the flywheel or pulleys - or your wheels. Don't mill down the face of the rotor - remember that unlike in a piston engine, the shape of the combustion chamber is dynamic. If you change the rotor face geometry, you'll move the center of pressure, and most likely DECREASE your torque by reducing the average distance between that center and the E-shaft, regardless of compression ratio.

I apologize for the long post, I hope it makes some sense...