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Something new: Torque vs. Power!

El Feo

New member
Joined
Feb 29, 2016
Location
Holiday in Cambodia
Moto(s)
Odd Duck
Name
Tuco
The premise/hypothesis is torque (not power)=acceleration.

Consider two motorbikes with the same horsepower. Blue (you're mah boy Baloo!) with 100 ft-lbs which redlines at 10k. And Red (ala Chevelle's seeing red again) with 50 ft-lbs yet redlines at 20k.

Both have the same gearing. That is, ten thousand RPMs yields 50mph on both bikes. We'll constrain the discussion to one gear only.

To simplify, two bikes with same HP and gearing; one with twice the torque.

Quiz question:

Which is quickest to 50 mph? And by how much relative to the other?

Extra credit:

Consider we changed the gearing of Red to redline at 50mph. Now, which is quickest?
 

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I'll say "Same-Same", but if I think the HP equation needs RPM of peak torque output, NOT redline RPM.

100lbs-ft x 10,000 = 1,000,000
1,000,000 / 5252 = 190.4hp

50lbs-ft x 20,000 = 1,000,000
1,000,000 / 5252 = 190.4hp
 
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I'll say "Same-Same", but if I think the HP equation needs RPM of peak torque output, NOT redline RPM.

100lbs-ft x 10,000 = 1,000,000
1,000,000 / 5252 = 190.4hp

50lbs-ft x 20,000 = 1,000,000
1,000,000 / 5252 = 190.4hp

Same peak hp, but the 20k'er is geared twice as tall- running out to redline takes you to 100mph, whereas the 10k'er only goes to 50.
 
I have to imagine that these conversations look better in countries that don't cling to retarded units of measurement.
 
I have to imagine that these conversations look better in countries that don't cling to retarded units of measurement.

I'm old school: Foot/pounds.
Newer school is pounds/feet.
Euro is Newton/meters.

I have no clue what a Newton is and I can't buy a meter stick at the hardware store. Yes it's an old system but it has better visuals in a country that sells pints and quarts.
 
for part A) the blue bike wins (assuming both bikes have same mass)

f=ma

for comparison purposes, f is the same as T because of the way the question is presented.

T=ma, a=T/m

T is a constant per the chart, so a is constant for both bikes but aRed is half of aBlue, so Blue will get to 50mph first and in half the time

this kind of scenario is why harley or similar seems fast around town where rpms remain low, and a sportbike is king for high speed on the open road.
 
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I'm old school: Foot/pounds.
Newer school is pounds/feet.
Euro is Newton/meters.

I have no clue what a Newton is and I can't buy a meter stick at the hardware store. Yes it's an old system but it has better visuals in a country that sells pints and quarts.

Actually it's foot-pounds or pound-feet :afm199

Force times distance

A Newton is a unit of force. A meter is something you put money into when you park.
 
I'm old school: Foot/pounds.
Newer school is pounds/feet.
Euro is Newton/meters.

I have no clue what a Newton is and I can't buy a meter stick at the hardware store. Yes it's an old system but it has better visuals in a country that sells pints and quarts.

It's a unit of multiplication, not division. Force x Distance. So it's not foot/pounds (feet per pound). It's pound x feet, which is abbreviated lbs-ft.
 
The premise/hypothesis is torque (not power)=acceleration.

You can "apply" a million ft-lb of torque at a wheel very slowly, let's say 1"/sec. Obviously, you won't get much acceleration out of that.

Rather than prolonged arguing, read the whole article that Russ linked in post #7, and its conclusion.
 
Serious. The high torque bike does with the same gearing. Your high rev bike will bog down forever until it gets to the last 4000 rpm of usable power.
 
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The chef may not be singing "torque, torque, torque" but the subject matter of this brief clip is torque-y...

[youtube]B-OFXUaMIv8[/youtube]
 
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Thrust at the contact patch is horsepower * 550 / (speed * 88/60) [because 1 horsepower = 550lb-ft/sec, and speed must be converted to ft/sec]

Horsepower is torque * RPM / 5252, or with the original gearing: torque * (speed * 10000/50) / 5252 = torque * speed * 200/5252 [note that speed units cancel]

So thrust at the patch is 550 * (torque * speed * 200 / 5252) / (speed * 88/60) = 550 * torque * 200 / (5252 * 88/60)

For BLUE, thrust at the contact patch is a constant 1428lb at any speed.

For RED (original gearing), thrust at the contact patch is a constant 714lb at any speed--half of blue.

For RED (modified gearing), thrust at the contact patch is a constant 1428lb at any speed--same as blue.

Since thrust is constant, acceleration is also constant, so v = a * t. So time to speed is inversely proportion to thrust: Double the thrust, half the elapsed time.

Therefore BLUE will accelerate to a given speed in half the time of RED with original gearing.

With modified gearing RED and BLUE will accelerate to a given speed in equal time.
 
If I was on the blue bike I would win because I would have pulled a spark plug wire on the red bike when you weren't looking. Did I say that out loud?
 
Well hell :mad No use posting because all the good answers are taken :laughing

Love reading this stuff :thumbup
 
I've always enjoyed the adage "People shop HP but we all drive Torque"
 
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