Chassis Dyno vs. Engine Dyno
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
Chassis Dyno vs. Engine Dyno
I'll make this as short as possible. I'm wondering how a chassis
dyno actually calculates torque.
If I dyno a motor on an engine dyno and it produces 400 ft.lbs @
4500 RPM and 450 HP @ 5500 RPM, will the chassis dyno create
similar results (taking into account any loss through the driveline
of course)?
My thoughts are yes, because the chassis dyno is taking engine
RPM into account while turning the weighted, calibrated drum.
A few questions then:
- What is the formula for calculating rear wheel torque through
the transmission selected gear, the differential gear and the tire
size? Please don't refer me to an online calculator!
The best I have is my school text which gets me from crank
power to rear axle - but NOT through the tire. I would like to
know how tire sizes affect torque to the ground.
- If my thoughts are correct about chassis dyno's, then you can
pull in any gear and get the same power curves correct?
- Please clear up this myth: In car mags, when they quote
power/torque is it chassis dyno or engine dyno (flywheel, or
wheel)?
Please respond in technical terms. This post is quite basic in
principal, but I do have a solid background in automotive and
engine tech. as well as chassis dyno experience.
Thank you all in advance.
dyno actually calculates torque.
If I dyno a motor on an engine dyno and it produces 400 ft.lbs @
4500 RPM and 450 HP @ 5500 RPM, will the chassis dyno create
similar results (taking into account any loss through the driveline
of course)?
My thoughts are yes, because the chassis dyno is taking engine
RPM into account while turning the weighted, calibrated drum.
A few questions then:
- What is the formula for calculating rear wheel torque through
the transmission selected gear, the differential gear and the tire
size? Please don't refer me to an online calculator!
The best I have is my school text which gets me from crank
power to rear axle - but NOT through the tire. I would like to
know how tire sizes affect torque to the ground.
- If my thoughts are correct about chassis dyno's, then you can
pull in any gear and get the same power curves correct?
- Please clear up this myth: In car mags, when they quote
power/torque is it chassis dyno or engine dyno (flywheel, or
wheel)?
Please respond in technical terms. This post is quite basic in
principal, but I do have a solid background in automotive and
engine tech. as well as chassis dyno experience.
Thank you all in advance.
Registered User
Joined: Oct 2002
Posts: 2,931
From: Upstate NY
Re: Chassis Dyno vs. Engine Dyno
Originally posted by Zero_to_69
I'll make this as short as possible. I'm wondering how a chassis
dyno actually calculates torque.
Assume the chassis dyno is an inertia only dyno (Dynojet, et al).
The roller or drum has a known Polar Moment of Inertia (PMoI).
Issac Newton (my college physics teacher; I'm ancient.) said: F=Ma
With the dyno we're working in rotational terms, so Mass is PMoI and acceleration is rpm/sec^2. All we have to measure is the drum's acceleration, and with 30 or more samples/second, we can get fairly accurate results.
Torque = PMoI x rpm(drum)/sec^2 x constant(to make the units work)
So now we know torque and rpm at the dyno roll, so hp is just torque x rpm/5252. This is drive wheel hp since the drive tires are in contact with the dyno roll. If we also record engine rpm, we can calculate engine torque and hp (at the rear wheels).
If I dyno a motor on an engine dyno and it produces 400 ft.lbs @4500 RPM and 450 HP @ 5500 RPM, will the chassis dyno create
similar results (taking into account any loss through the driveline
of course)?
My thoughts are yes, because the chassis dyno is taking engine
RPM into account while turning the weighted, calibrated drum.
Yes, and for the reasons you stated.
A few questions then:
- What is the formula for calculating rear wheel torque through
the transmission selected gear, the differential gear and the tire
size? Please don't refer me to an online calculator!
The best I have is my school text which gets me from crank
power to rear axle - but NOT through the tire. I would like to
know how tire sizes affect torque to the ground.
Well the torque the dyno measured WAS rear wheel torque.
Example: Assume 400 hp (at the dyno roll) @ 100 mph on 48 inch diameter dyno roll. Dyno roll rpm = 700. Let's assume 5600 engine rpm ( I like strokers that peak power below 6k!).
Wheel torque is 400 x 5252/700 = 3001 (I tried for an easy number like 3000). With 4.10's 700 rpm(dyno) = a tire diameter of 24.6 in.
Now TORQUE doesn't get to the ground, only FORCE gets to the ground at the tire contact patch. That force is just the axle torque (lb. ft.) divided by the moment arm or the tire rolling radius (in feet, of course). In the example 3001/(24.6/2)/12
= 2928 lbs. of force at the road surface to accelerate the vehicle.
If the tire diameter (and radius) changes, the force at the road will change assuming the same 3001 lb. ft. axle torque. Remember tires are also 'gears'. The torque is the same; only the thrust force is different.
- If my thoughts are correct about chassis dyno's, then you can
pull in any gear and get the same power curves correct?
Sorta. Drive line losses are usually higher in the lower gears, and because the system is accelerating faster in the lower gears, the inertia of the driveline plays a bigger part in the losses. Just like a 600 rpm/second engine dyno pull shows less torque/hp than a 300 rpm/sec, which shows less than a steady-state dyno run. This leads to:
- Please clear up this myth: In car mags, when they quote
power/torque is it chassis dyno or engine dyno (flywheel, or
wheel)?
Almost always in road tests, the mags are quoting manufacturer's ENGINE ratings, which are flywheel numbers taken at steady state dyno runs (no inertia involved), but "as installed" or driving all accessories and with production intake and exhaust systems and corrected to temp/density numbers that more nearly reflect vehicle conditions than the conditions often used on engine dynos in the aftermarket world.
Please respond in technical terms. This post is quite basic in
principal, but I do have a solid background in automotive and
engine tech. as well as chassis dyno experience.
Thank you all in advance.
I'll make this as short as possible. I'm wondering how a chassis
dyno actually calculates torque.
Assume the chassis dyno is an inertia only dyno (Dynojet, et al).
The roller or drum has a known Polar Moment of Inertia (PMoI).
Issac Newton (my college physics teacher; I'm ancient.) said: F=Ma
With the dyno we're working in rotational terms, so Mass is PMoI and acceleration is rpm/sec^2. All we have to measure is the drum's acceleration, and with 30 or more samples/second, we can get fairly accurate results.
Torque = PMoI x rpm(drum)/sec^2 x constant(to make the units work)
So now we know torque and rpm at the dyno roll, so hp is just torque x rpm/5252. This is drive wheel hp since the drive tires are in contact with the dyno roll. If we also record engine rpm, we can calculate engine torque and hp (at the rear wheels).
If I dyno a motor on an engine dyno and it produces 400 ft.lbs @4500 RPM and 450 HP @ 5500 RPM, will the chassis dyno create
similar results (taking into account any loss through the driveline
of course)?
My thoughts are yes, because the chassis dyno is taking engine
RPM into account while turning the weighted, calibrated drum.
Yes, and for the reasons you stated.
A few questions then:
- What is the formula for calculating rear wheel torque through
the transmission selected gear, the differential gear and the tire
size? Please don't refer me to an online calculator!
The best I have is my school text which gets me from crank
power to rear axle - but NOT through the tire. I would like to
know how tire sizes affect torque to the ground.
Well the torque the dyno measured WAS rear wheel torque.
Example: Assume 400 hp (at the dyno roll) @ 100 mph on 48 inch diameter dyno roll. Dyno roll rpm = 700. Let's assume 5600 engine rpm ( I like strokers that peak power below 6k!).
Wheel torque is 400 x 5252/700 = 3001 (I tried for an easy number like 3000). With 4.10's 700 rpm(dyno) = a tire diameter of 24.6 in.
Now TORQUE doesn't get to the ground, only FORCE gets to the ground at the tire contact patch. That force is just the axle torque (lb. ft.) divided by the moment arm or the tire rolling radius (in feet, of course). In the example 3001/(24.6/2)/12
= 2928 lbs. of force at the road surface to accelerate the vehicle.
If the tire diameter (and radius) changes, the force at the road will change assuming the same 3001 lb. ft. axle torque. Remember tires are also 'gears'. The torque is the same; only the thrust force is different.
- If my thoughts are correct about chassis dyno's, then you can
pull in any gear and get the same power curves correct?
Sorta. Drive line losses are usually higher in the lower gears, and because the system is accelerating faster in the lower gears, the inertia of the driveline plays a bigger part in the losses. Just like a 600 rpm/second engine dyno pull shows less torque/hp than a 300 rpm/sec, which shows less than a steady-state dyno run. This leads to:
- Please clear up this myth: In car mags, when they quote
power/torque is it chassis dyno or engine dyno (flywheel, or
wheel)?
Almost always in road tests, the mags are quoting manufacturer's ENGINE ratings, which are flywheel numbers taken at steady state dyno runs (no inertia involved), but "as installed" or driving all accessories and with production intake and exhaust systems and corrected to temp/density numbers that more nearly reflect vehicle conditions than the conditions often used on engine dynos in the aftermarket world.
Please respond in technical terms. This post is quite basic in
principal, but I do have a solid background in automotive and
engine tech. as well as chassis dyno experience.
Thank you all in advance.
My $.02
Last edited by OldSStroker; Jan 5, 2003 at 11:23 AM.
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
Thanks SS,
That's a pretty valuable $0.02! I have a couple more questions
for you if you don't mind.
If you could elaborate on the formula with respect to why certain
valvues are used:
Wheel torque = (flywheel HP * 5252)/700
Where are the 5252 and 700 values coming from? 5252 is the RPM cross over point for Torque and horsepower on the graph
so I can understand why that value is used. What about 700...
where and why is that there? How is 700 derived?
Next:
"Wheel torque is 400 x5252/700 = 3001 (I tried for an easy number like 3000). With 4.10's 700 rpm(wheel) = a tire diameter of 24.6 in. "
How did you arrive at 4.10:1 gear and 24.6" tire?
How do I plug those values into the formula?
My goal by using the formula is to calculate power/tq. at the
wheel when given flywheel values. The formula I require must
account for differential ratio, and tire diameter. I think that's
what you have given me, I just can't break it down.
I would also like to calculate distance traveled and MPH
when given engine RPM, transmission gear, differential ratio
and tire OD".
Using circumference of the tire, and wheel RPM, I should be
able to nail down the amount of rev's my tires will rotate per
1/4 mile.
This is what I have from my school book:
Axle Shaft Torque = Engine Torque * (Driven Gear/Drive Gear) *
Differential Ratio
IE:
Alxe Shaft Torque = 400 ft.lbs * 2 * 3.73
Axle Shaft Torque = 2984 ft./ lbs.
I can also substitute Engine RPM in the same formula to find
alxe output RPM by inverting the gear ratios (IE: 2:1 = 0.5)
**assuming 100% efficient driveline components
in either case**
The killer is, the fomula doesn't account for tire diameter!
If you know the missing link, or if it's already part of your first
response, please shed some light!
Thanks again!
BTW - This forum rocks and I haven't found a better source for
credible info anywhere else. Good job everyone!
That's a pretty valuable $0.02! I have a couple more questions
for you if you don't mind.
If you could elaborate on the formula with respect to why certain
valvues are used:
Wheel torque = (flywheel HP * 5252)/700
Where are the 5252 and 700 values coming from? 5252 is the RPM cross over point for Torque and horsepower on the graph
so I can understand why that value is used. What about 700...
where and why is that there? How is 700 derived?
Next:
"Wheel torque is 400 x5252/700 = 3001 (I tried for an easy number like 3000). With 4.10's 700 rpm(wheel) = a tire diameter of 24.6 in. "
How did you arrive at 4.10:1 gear and 24.6" tire?
How do I plug those values into the formula?
My goal by using the formula is to calculate power/tq. at the
wheel when given flywheel values. The formula I require must
account for differential ratio, and tire diameter. I think that's
what you have given me, I just can't break it down.
I would also like to calculate distance traveled and MPH
when given engine RPM, transmission gear, differential ratio
and tire OD".
Using circumference of the tire, and wheel RPM, I should be
able to nail down the amount of rev's my tires will rotate per
1/4 mile.
This is what I have from my school book:
Axle Shaft Torque = Engine Torque * (Driven Gear/Drive Gear) *
Differential Ratio
IE:
Alxe Shaft Torque = 400 ft.lbs * 2 * 3.73
Axle Shaft Torque = 2984 ft./ lbs.
I can also substitute Engine RPM in the same formula to find
alxe output RPM by inverting the gear ratios (IE: 2:1 = 0.5)
**assuming 100% efficient driveline components
in either case**
The killer is, the fomula doesn't account for tire diameter!
If you know the missing link, or if it's already part of your first
response, please shed some light!
Thanks again!
BTW - This forum rocks and I haven't found a better source for
credible info anywhere else. Good job everyone!
Last edited by Zero_to_69; Jan 5, 2003 at 09:41 AM.
Registered User
Joined: Oct 2002
Posts: 2,931
From: Upstate NY
[QUOTE]Originally posted by Zero_to_69
If you could elaborate on the formula with respect to why certain
valvues are used:
Wheel torque = (flywheel HP * 5252)/700
Where are the 5252 and 700 values coming from? 5252 is the RPM cross over point for Torque and horsepower on the graph
so I can understand why that value is used. What about 700...
where and why is that there? How is 700 derived?
The basic horsepower formula is: HP = Torque x rpm/5252
1 hp = 33,000 lb-ft/min (as defined by James Watt)
Next: revs/min. We need to convert to rotational distance or radians. There are 2 x PI radians in one revolution, so 2 x PI = 6.283
HP = lb-ft x (rev/min) x (6.283 radians/rev) x (hp/33000 lb-ft/min)
Cancelling the units, we have left; HP = lb-ft x (engine speed) x (6.283/33000) or HP=lb-ft x rpm/5252
To get the '700 rpm' we convert the surface speed of the roller from 100 mph.
48 in / 12 ft/in = 4 ft. (diameter dyno roll)
Circumference = PI x diameter or 3.1415 x 4 = 12.57 ft.
Rev/min= 100 miles/hr x hr/60 min x 5280 ft/mile x rev/12.57 ft.
or RPM = 100 /60 x 5280 / 12.57 or 700 rev/min of the drum.
If HP = torque x rpm/5252, then torque = hp x 5252/rpm and rpm = hp x 5252/torque
"Wheel torque is 400 x5252/700 = 3001 (I tried for an easy number like 3000). With 4.10's 700 rpm(wheel) = a tire diameter of 24.6 in. "
How did you arrive at 4.10:1 gear and 24.6" tire?
How do I plug those values into the formula?
I chose 5600 engine rpm at 100 mph, so the formula: Tire Diameter = (Trans Ratio x MPH x 336 x Rear Gear Ratio) / (RPM)
Tire Dia = 1.00 x 100 x 336 x 4.10/5600 = 24.6. I chose a standard rear gear of 4.10
My goal by using the formula is to calculate power/tq. at the
wheel when given flywheel values. The formula I require must
account for differential ratio, and tire diameter. I think that's
what you have given me, I just can't break it down.
I'll repeat, torque at the axle is torque at the wheel, but what drives the vehicle is the force at the road/tire patch
Think of the distance from the axle center to the ground at the drive wheel. This is the "ft" in "lb-ft" axle torque.
Remember that gears multiply torque (and divide rpm) but don't change power (assuming no frictional/inertia losses). The power at the 48 in dyno roll was 400hp @ 700 rpm, and it was 400 hp @ 5600 at the flywheel (no losses). The torques however were 3001 lb-ft at the 700 rpm and 375 lb-ft @ 5600 rpm.
If you are looking for wheel horsepower, as most are, the big problem is determining losses through the drivetrain, both frictional and inertial. There are lots of threads on this in many forums.
I'm not sure why you want to know torque at the wheels.
I would also like to calculate distance traveled and MPH
when given engine RPM, transmission gear, differential ratio
and tire OD".
That's easy: MPH = (RPM x Tire Diameter) / (336 x Rear Gear Ratio x Trans Ratio)
Distance = miles/hr x 5280 ft/mile x time (in hours)
Using circumference of the tire, and wheel RPM, I should be
able to nail down the amount of rev's my tires will rotate per
1/4 mile.
That's even easier: look up the revs/mile value for the tire you want to use. Try this calculator:
http://www.miata.net/garage/tirecalc.html
This is what I have from my school book:
Axle Shaft Torque = Engine RPM * (Driven Gear/Drive Gear) *
Differential Ratio
IE:
Alxe Shaft Torque = 400 ft.lbs * 2 * 3.73
Axle Shaft Torque = 2984 ft./ lbs.
I can also substitute Engine RPM in the same formula to find
alxe output RPM by inverting the gear ratios (IE: 2:1 = 0.5)
**assuming 100% efficient driveline components
in either case**
The killer is, the fomula doesn't account for tire diameter!
If you know the missing link, or if it's already part of your first
response, please shed some light!
You meant "Engine torque" not "Engine RPM" in the formula.
Once again, look at the part about wheel torque and force above. Don't confuse force (lb) with torque (force acting over a distance).
Sorry to drone on so long. Hope I haven't confused you.
If you could elaborate on the formula with respect to why certain
valvues are used:
Wheel torque = (flywheel HP * 5252)/700
Where are the 5252 and 700 values coming from? 5252 is the RPM cross over point for Torque and horsepower on the graph
so I can understand why that value is used. What about 700...
where and why is that there? How is 700 derived?
The basic horsepower formula is: HP = Torque x rpm/5252
1 hp = 33,000 lb-ft/min (as defined by James Watt)
Next: revs/min. We need to convert to rotational distance or radians. There are 2 x PI radians in one revolution, so 2 x PI = 6.283
HP = lb-ft x (rev/min) x (6.283 radians/rev) x (hp/33000 lb-ft/min)
Cancelling the units, we have left; HP = lb-ft x (engine speed) x (6.283/33000) or HP=lb-ft x rpm/5252
To get the '700 rpm' we convert the surface speed of the roller from 100 mph.
48 in / 12 ft/in = 4 ft. (diameter dyno roll)
Circumference = PI x diameter or 3.1415 x 4 = 12.57 ft.
Rev/min= 100 miles/hr x hr/60 min x 5280 ft/mile x rev/12.57 ft.
or RPM = 100 /60 x 5280 / 12.57 or 700 rev/min of the drum.
If HP = torque x rpm/5252, then torque = hp x 5252/rpm and rpm = hp x 5252/torque
"Wheel torque is 400 x5252/700 = 3001 (I tried for an easy number like 3000). With 4.10's 700 rpm(wheel) = a tire diameter of 24.6 in. "
How did you arrive at 4.10:1 gear and 24.6" tire?
How do I plug those values into the formula?
I chose 5600 engine rpm at 100 mph, so the formula: Tire Diameter = (Trans Ratio x MPH x 336 x Rear Gear Ratio) / (RPM)
Tire Dia = 1.00 x 100 x 336 x 4.10/5600 = 24.6. I chose a standard rear gear of 4.10
My goal by using the formula is to calculate power/tq. at the
wheel when given flywheel values. The formula I require must
account for differential ratio, and tire diameter. I think that's
what you have given me, I just can't break it down.
I'll repeat, torque at the axle is torque at the wheel, but what drives the vehicle is the force at the road/tire patch
Think of the distance from the axle center to the ground at the drive wheel. This is the "ft" in "lb-ft" axle torque.
Remember that gears multiply torque (and divide rpm) but don't change power (assuming no frictional/inertia losses). The power at the 48 in dyno roll was 400hp @ 700 rpm, and it was 400 hp @ 5600 at the flywheel (no losses). The torques however were 3001 lb-ft at the 700 rpm and 375 lb-ft @ 5600 rpm.
If you are looking for wheel horsepower, as most are, the big problem is determining losses through the drivetrain, both frictional and inertial. There are lots of threads on this in many forums.
I'm not sure why you want to know torque at the wheels.
I would also like to calculate distance traveled and MPH
when given engine RPM, transmission gear, differential ratio
and tire OD".
That's easy: MPH = (RPM x Tire Diameter) / (336 x Rear Gear Ratio x Trans Ratio)
Distance = miles/hr x 5280 ft/mile x time (in hours)
Using circumference of the tire, and wheel RPM, I should be
able to nail down the amount of rev's my tires will rotate per
1/4 mile.
That's even easier: look up the revs/mile value for the tire you want to use. Try this calculator:
http://www.miata.net/garage/tirecalc.html
This is what I have from my school book:
Axle Shaft Torque = Engine RPM * (Driven Gear/Drive Gear) *
Differential Ratio
IE:
Alxe Shaft Torque = 400 ft.lbs * 2 * 3.73
Axle Shaft Torque = 2984 ft./ lbs.
I can also substitute Engine RPM in the same formula to find
alxe output RPM by inverting the gear ratios (IE: 2:1 = 0.5)
**assuming 100% efficient driveline components
in either case**
The killer is, the fomula doesn't account for tire diameter!
If you know the missing link, or if it's already part of your first
response, please shed some light!
You meant "Engine torque" not "Engine RPM" in the formula.
Once again, look at the part about wheel torque and force above. Don't confuse force (lb) with torque (force acting over a distance).
Sorry to drone on so long. Hope I haven't confused you.
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
That's a great reply. Yes, I meant to type Engine Torque instead
of RPM in the equation. I will have to edit that for future readers!
I'm cool with everything you posted except the torque and force
concept at the contact patch of the tire. I need to sit back and
wrap my head around that for a while.
I might come back with a summary of my thoughts on this and
have you check me
Thank you very much for the data. I will be putting it to good use.
The reason I want to know tq./hp. at the wheel was to prove my
understanding of chassis and engine dyno's. It's also nice to
know the multiplied torque through all the gears just for curiousity.
The reason for wanting the other information is to optimize my
gear and tire combination for 1/4 racing. Some quick calculations
and trial runs at the track will give me a better idea of what
size of slick to purchase to round out my 5500 RPM power peak,
3.73 gears when finishing in 1:1 transmission gear (3rd gear for
the TH350).
I can't remember exactly how much RPM I have left, but I know
I'm shifting into 3rd a good distance before the finish line and
if memory serves me correct, I have 1500 RPM of motor left.
I don't have the car readily accessible to measure the current
rear tire OD, but I know I have to go with a slightly smaller tire
to get the best bang in the 1/4 with my current motor power
peak.
Thanks again and have a good weekend.
of RPM in the equation. I will have to edit that for future readers!
I'm cool with everything you posted except the torque and force
concept at the contact patch of the tire. I need to sit back and
wrap my head around that for a while.
I might come back with a summary of my thoughts on this and
have you check me
Thank you very much for the data. I will be putting it to good use.
The reason I want to know tq./hp. at the wheel was to prove my
understanding of chassis and engine dyno's. It's also nice to
know the multiplied torque through all the gears just for curiousity.
The reason for wanting the other information is to optimize my
gear and tire combination for 1/4 racing. Some quick calculations
and trial runs at the track will give me a better idea of what
size of slick to purchase to round out my 5500 RPM power peak,
3.73 gears when finishing in 1:1 transmission gear (3rd gear for
the TH350).
I can't remember exactly how much RPM I have left, but I know
I'm shifting into 3rd a good distance before the finish line and
if memory serves me correct, I have 1500 RPM of motor left.
I don't have the car readily accessible to measure the current
rear tire OD, but I know I have to go with a slightly smaller tire
to get the best bang in the 1/4 with my current motor power
peak.
Thanks again and have a good weekend.
Administrator
Joined: Nov 1998
Posts: 71,110
From: Hell was full so they sent me to NJ
Just think of the "torque" as the twisting of the axles, and the force acting on the contact patch of the tires as the translation of that torque through the "lever arm" which is the tire rolling radius.
300ft-lb at the flywheel, through a 2.66 1st gear and a 3.73 rear axle is a 9.92X (2.66 x 3.73) multiplication of torque through the gears. That gives you about 2,977 ft-lb twisting the rear axles, if there were no losses in the drivetrain. But you have about 12.5% losses in an M6, so you will only be seeing about 2,605 ft-lb at the axles. (I know you don't have an M6, but I have the numbers handy. You also have to take the "torque multiplication" of your convertor into account on the automatic tranny)
That 2605 ft-lb of torque at the axles is resisted by the horizontal force acting on each of two tires, so each tire needs to transmit 1,302 ft-lb of torque. Since torque is the product of force x distance, the force at each tire contact point is (1,302 ft-lb) / (rolling radius of the tire in feet) = pounds of force propelling the vehicle at each tire contact patch. If you have a stock tire (25.65/2) / 12 = 1.06875 ft, so the force at each tire patch is 1,302 / 1.06875 = 1,218 pounds.
If you switch to a 28" drag tire, the rolling radius becomes (roughly) (28.0/2) / 12 = 1.1667 ft, and the force at each tire is 1,302 / 1.1667 = 1116 pound. Of course the smaller force will be present for a larger number of revolutions of the tire, so the total work delivered in each gear will be the same.
300ft-lb at the flywheel, through a 2.66 1st gear and a 3.73 rear axle is a 9.92X (2.66 x 3.73) multiplication of torque through the gears. That gives you about 2,977 ft-lb twisting the rear axles, if there were no losses in the drivetrain. But you have about 12.5% losses in an M6, so you will only be seeing about 2,605 ft-lb at the axles. (I know you don't have an M6, but I have the numbers handy. You also have to take the "torque multiplication" of your convertor into account on the automatic tranny)
That 2605 ft-lb of torque at the axles is resisted by the horizontal force acting on each of two tires, so each tire needs to transmit 1,302 ft-lb of torque. Since torque is the product of force x distance, the force at each tire contact point is (1,302 ft-lb) / (rolling radius of the tire in feet) = pounds of force propelling the vehicle at each tire contact patch. If you have a stock tire (25.65/2) / 12 = 1.06875 ft, so the force at each tire patch is 1,302 / 1.06875 = 1,218 pounds.
If you switch to a 28" drag tire, the rolling radius becomes (roughly) (28.0/2) / 12 = 1.1667 ft, and the force at each tire is 1,302 / 1.1667 = 1116 pound. Of course the smaller force will be present for a larger number of revolutions of the tire, so the total work delivered in each gear will be the same.
Last edited by Injuneer; Jan 5, 2003 at 08:43 AM.
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
You boys are good!
Thanks for clearing that up Fred.
BTW: I have an '80 Z28 with TH350 transmission (heavily modified
internals), 2600 Stall, 3.73 gears and using a street tire at the
moment (255/60/15 Daytona's). Not very efficient, but it's
producing some decent 60' times with a best of 1.98 secs.
The plan now is to do some weight transfer mods. IE: battery
relocate, replacing the front bumper with an aluminum bar,
solid motor mounts, adjustable shocks 50/50 rear, 90/10 front
and Weld lites wrapped with a good race slick. I'm shopping
around for Hoosier and Goodyear tires at the moment.
I hope to be in the 1.7's 60' times with the above mods and
mid to low 12's in the 1/4 mile with a 3761 lb. race weight and
355 CID N/A motor.
Thanks for your time guys...only 3 1/2 months before racing
season!!!
Thanks for clearing that up Fred.
BTW: I have an '80 Z28 with TH350 transmission (heavily modified
internals), 2600 Stall, 3.73 gears and using a street tire at the
moment (255/60/15 Daytona's). Not very efficient, but it's
producing some decent 60' times with a best of 1.98 secs.
The plan now is to do some weight transfer mods. IE: battery
relocate, replacing the front bumper with an aluminum bar,
solid motor mounts, adjustable shocks 50/50 rear, 90/10 front
and Weld lites wrapped with a good race slick. I'm shopping
around for Hoosier and Goodyear tires at the moment.
I hope to be in the 1.7's 60' times with the above mods and
mid to low 12's in the 1/4 mile with a 3761 lb. race weight and
355 CID N/A motor.
Thanks for your time guys...only 3 1/2 months before racing
season!!!
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