[CS]

WOW - Torque that's freakin' amazing.. Cool write up!

Eric, yours was also very interesting, although I have a hard time not seeing it on paper.

 

[CS]

By the way - what does the dyno measure then? I know Vipers dyno in fourth cause the ratio is 1:1 - does that cancel the torque multiplication factor? In what gear would the RWTQ be 2456?

In my mind, if the ratio is 1:1 and the rear is a 3.07 then the dyno is measuring power at the 3.07 ratio - why then is the RWTQ 1000 ft/lbs less when measured on a dyno?

 

[CS]

By the way - what does the dyno measure then? I know Vipers dyno in fourth cause the ratio is 1:1 - does that cancel the torque multiplication factor? In what gear would the RWTQ be 2456?

In my mind, if the ratio is 1:1 and the rear is a 3.07 then the dyno is measuring power at the 3.07 ratio - why then is the RWTQ 1000 ft/lbs less when measured on a dyno?

Good question. The # that the dyno reports should actually be called flywheel torque as measured at the rear wheels. You are correct in that a stock viper making 500 lb.ft torque (at the flywheel) is actually putting down 1500 lb.ft. at the rear wheels (minus losses).

So how does the dyno *know* how to factor out such things as gearing (trans & diff) and wheel/tire size? It does that by a ratio of drum speed to known engine speed. They're measuring your rpm, right? And the dyno knows at any instant how fast the drum is turning and that allows them to factor out the mechanical advantage (torque multiplier) due to gearing etc.

Wow, ok I think I got it :thumb:! So, Eric, does that mean when we ask how many RWHP/TQ a car is making, we are really referring to the amount of power being transferred to the wheels from the crank, rather then the actual amount of power at the wheels?

And if torque is a quantifiable, tangible 'thing', then is horsepower simply a function of torque relative to engine speed?

Sorry I missed your call last night, I didn't hear my damn phone! I will definitely get it tonight.

 

[CS]

By the way - what does the dyno measure then? I know Vipers dyno in fourth cause the ratio is 1:1 - does that cancel the torque multiplication factor? In what gear would the RWTQ be 2456?

In my mind, if the ratio is 1:1 and the rear is a 3.07 then the dyno is measuring power at the 3.07 ratio - why then is the RWTQ 1000 ft/lbs less when measured on a dyno?

Good question. The # that the dyno reports should actually be called flywheel torque as measured at the rear wheels. You are correct in that a stock viper making 500 lb.ft torque (at the flywheel) is actually putting down 1500 lb.ft. at the rear wheels (minus losses).

So how does the dyno *know* how to factor out such things as gearing (trans & diff) and wheel/tire size? It does that by a ratio of drum speed to known engine speed. They're measuring your rpm, right? And the dyno knows at any instant how fast the drum is turning and that allows them to factor out the mechanical advantage (torque multiplier) due to gearing etc.

Wow, ok I think I got it :thumb:! So, Eric, does that mean when we ask how many RWHP/TQ a car is making, we are really referring to the amount of power being transferred to the wheels from the crank, rather then the actual amount of power at the wheels?

Correct. The actual torque at the rear wheels is (trans gear ratio) X (rear gear ratio) X flywheel torque. Talking about 4,000 to 5,000 lb.ft in first gear.

Isn't the transmissions gear ratios directly affected by the rear gear ratio (3.07)?

What I mean by that is in first gear, is the motor turning 3.07 times for every one time the wheels turn? What about second gear? Is fourth the only gear (at least in a T-56) that retains a true 3.07:1 ratio between revolutions of the motor and tire?

 

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Jerome - if the rolling circumfrence remains constant, then the dyno doesn't know wether or not it's a lo-pro 19" or a 16" wheel/tire combo. If the total circumference is 3.14*19"= roughly 60" for both, then it should register the same. In fact, I would say that if anything, the 19" lo-pro, real stiff sidewalled tire would give a bigger reading, because of the low amount of sidewall flex and reduced loss of energy resulting from it.

But that's just something that popped into my head, so it could be /images/graemlins/bs

 

[CS]

Ok - here's a wild stab in the dark

maybe the bigger sidewalls twist more and this twisting when it reaches its limits provides a counter force that actually helps the power - kinda like a pendulum effect... kind of...

maybe not

:help:

:screwy:

:leaving:

Sidewall flex = stored energy :nod: Think wrinkle walls

Keep goin' Barry, you are on a roll. :thumb:

The dyno spins counterclockwise and the tire is spinning the opposite direction. When the operator hits it, the sidewall flex is going to pull power out of the reading because it doesn't trasmit the power as accurately to the drum.

 

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BTW - why is 5252 the magic number in the relationship between horsepower and torque?

 

[CS]

Jerome,

The severe wrinkle-wall phemomenon on a drag slick helps increase net torque at the tire by shortening the rolling radius (effective lever arm) between the axle and the ground. After a few revolutions, the tire begins to grow circumferentially. This raises the axle center from the ground and lowers torque, but the tire is covering more distance in one revolution than at rest, or low rpms which accounts for the huge mph gain. A shorter tire will accelerate the dyno rolls more quickly than a tall tire. I believe that should raise the indicated power levels.

Steel belted street tires exhibit very little "growth" and maintain circumference. HP street tires (Pilots, not E/Ts) have very stiff sidewalls and there's not much energy wind-up in them at recommended inflation pressures. That's why we have to run such low pressures at the strip to make them work well. Flexing stiff sidewalls creates heat and wastes energy. That's why you should run plenty of air in them on the chassis dyno. Some of the "cheater" boys' cars were moving around laterally a good bit - wasting power.

Dean - that makes perfect sense, I wasn't thinking of it in a decreasing-diameter sense, more of a loss of power when the tires twisted it created almost a slack in the tires, not transmitting all of the power.

 

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By the way Barry, that stored up energy in the tire isn't ever released as power measured on the dyno. That's because as that tire flexes, it remains flexed althrought the run; the power transferred the the wheels increases and increases throughout the run and the tension on the sidewall isn't released until after the power is disengaged at the end of the run.

That stored energy in the sidewall doesn't show up on the dyno.

 

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Dean - that makes perfect sense, I wasn't thinking of it in a decreasing-diameter sense,

No Clint, remember it's a decreasing RADIUS issue. The only (easily changeable) distance that affects output torque is the effective loaded radius of the tire.

Ahh, good point Dean. /images/graemlins/freak

 

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Same OD (and weight) = same reading by laws of physics.

However, I think your point was that due to the greater sidewall flex of the 16" combo, the reading will be different. Assuming you pump up the 16" combo adequately, the flex will be neglibible and the difference will be lost in the noise of any series of dyno pulls. And if you did not pump up the 16" combo, the reading would be less, because energy is wasted in the tire flex instead of being transferred to the drum (there is no traction issue here like with the dragster from rest).

That's what I'm saying Jerome.