Quick question about torque

[Guest DaneL24]

When people refer to ft/lbs of torque on a dyno, such as torque at the wheels or at the crank...is distance (feet) measured in radial or circumferential distance? It makes a pretty big difference after all.

[Guest Anonymous]

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[Pop N Wood]

i.e. radial.

[TimZ]

BTW, it's "lb-ft" (lbs x feet), not "ft/lb" (which would be feet per lb).

 

As spiirit pointed out (I hope - I didn't read the whole thing ), torque is a measure of force applied at a distance, not distance travelled per force applied.

[Guest DaneL24]

Thank you SPIIRIT. I actually am familiar with the physics of how torque works, I just didn't know if when they were refering to feet they meant the radius of the flywheel, wheel, etc. (such as with a torque wrench)...or if they meant the actual distance traveled along the circumference of the flywheel, wheel, etc...which I now see that combined with torque this determines horsepower. You did answer my question though...torque is measured as the product of both force and the distance from the fulcrum of a lever that the force is applied. Horsepower is the rate at which a given mass can be moved...or the rate that work can be performed. Perhaps I could have answered my own question...LOL!

 

By the way...my name is just Dane. L24 refers to the 240Z engine specification (you know that...) . No harm done...just clarifying.

[Guest Anonymous]

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[Guest DaneL24]

Actually, no...I didn't start thinking of this after reading an article. The reason I was confused was that it seemed that work/torque had two separate definitions.

 

For example, I have read (from physics books...not magazines) that 5 ft/lbs is the work required to move a 5 pound mass 1 foot...such as lifting a 5 pound object with a pulley 1 foot of distance. It is also equal to the work required to lift a 1 pound object 5 feet. That is why I was thinking of things like rim speed.

 

However, I also know that 5 ft/bs of torque is equivalent to applying 5 pounds of force to a lever arm 1 foot long, or applying 1 pound of force to a lever 5 feet long.

 

Do you see what I mean? Work is defined as the energy required to move an object of a certain mass a certain distance...while torque is force applied to a lever with a fulcrum. These seem to be two completely different things. As far as I know, both work and torque are equal though...because they have the same mathematical definitions (work = force x distance...torque = force x distance...therefore work = torque). Am I mistaken?

 

Also another question, when 1 pound of force is applied...that is equal to the force of one pound of mass x 9.8 meters per second/second (the acceleration due to gravity)...correct?

[Guest Anonymous]

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[David K]

Why not make it easy?

 

torque

 

1. The moment of a force; the measure of a force's tendency to produce torsion and rotation about an axis, equal to the vector product of the radius vector from the axis of rotation to the point of application of the force and the force vector.

2. A turning or twisting force.

 

 

horsepower

 

1. A unit of power in the U.S. Customary System, equal to 745.7 watts or 33,000 foot-pounds per minute.

 

 

Im sure millions of people will have their own little ways of explaining the above 8)

[Guest Anonymous]

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[Pop N Wood]

I think they answered you. Work is force times distance. Power is work per unit time.

 

Torque is the rotational form of force. So for rotating systems, torque times the number of rotations equals work. Torque times RPM is power.

 

If you push (or twist) on something and it doesn't move, you have done no work.

 

Look at the units on each quantity. Force can be measured in terms of pounds (or Newtons outside the US). If you push on something with 100 pounds of force and move it 10 feet, you have done 1000 ft-pounds of work (this is the same as 1355 Newton Meters which is 1355 Joules). If you do the above in 1 second, then your average power was 1000 ft-lbs/sec, which is 1355 Joules/sec, which is 1355 watts, which is the same as 1.8 horsepower.

[74_5.0L_Z]

Let's see if I can muddy the water.

 

 

Work is defined as the integral of Force time distance (actually the dot product). W=integral(F dot ds)

 

Radial distance is defined as the radius X angle (theta measured in radians)

 

ds = R d theta

 

Torque is an applied force about a point at a distance (assuming the force is tangential to the radius).

 

T = F R => F=T/R => W = integral (T dot ds)/R

 

or

 

W = integral (T d theta)

 

Power is the derivative of work with respect to time

 

therefore P = d(W)/dt

 

P = d(integral(T d theta))/dt

 

P = T d theta/dt (assuming constant torque)

 

also d theta/dt can be rewritten (F R)(ds/R)/dt

 

P = F ds/dt which if ds describes rectilinear motion can br rewritten as

 

P = F V

[Mudge]

For example, I have read (from physics books...not magazines) that 5 ft/lbs is the work required to move a 5 pound mass 1 foot...

 

The only thing I can liken to that is horsepower, 1 horse can move xx pounds xx distance in xx seconds.

[Guest Anonymous]

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[Guest DaneL24]

Honored to be part of the club...theres still one thing I don't understand.

 

Torque is the measure of force applied a lever of a certain length...and power is the measure of work (distance traveled under a load...such as pulling a 1 pound mass 5 feet being equal to 5 ft/lbs) performed in a certain amount of time.

 

So if torque is force applied to a lever, but does not indicate distance traveled (instead indicates length of lever arm with equivalent force applied)...and power is work performed in a certain amount of time, how is it possible to derive HP from torque with equation HP = torque x RPM / 5250? While work and power are obviously related...the relationship between torque and power is harder for me to understand...because distance has a totally different meaning when you compare torque and power (lever arm length vs distance traveled).

 

Don't we need to somehow convert torque into rotational distance traveled before determining HP...such as 1 ft lb of torque converted to 3.14 ft/lbs of work when one rotation is accomplished? And if one pound of force is equal to 1 pound of mass x 32 feet per second/second (acceleration due to gravity)...this would convert to 10.18 (32 feet / 3.14 feet traveled per rotation) rotations per second after 1 second...and 20.36 rotations per second after 2 seconds. I don't even know why I calculated how fast the rotations would accelerate under 1 ft/lb of force (against 1 lb of mass at the point of force applied) but I did, maybe it means something...just a sudden mental tangent...but that starts to get into power.

 

But anyways...do you see what I mean? How do you start with force applied to a lever and result in work performed by powering a mass a certain distance in a certain period of time...ie converting torque to power. Or did I just do that?

[74_5.0L_Z]

Here is a visual basic routine that I wrote that runs in Excel. The program takes the torque vs rpm curve, gear ratios, shift points, drag coefficients,weight, traction coefficient, etc.

 

The program interpolates torque between given points using cubic spline interpolation.

 

So at time 0 the program uses your engine torque at the launch rpm, multiplies it by the 1st gear ratio, and the rear gear ratio to determine the torque at the back tires. The program then devides the torque by the tire height to determine force at the back tires. This for is then compared to the weight of the car times the traction coefficient. If the force at the back tires is greater than available traction, the force applied to the tires is set equal to the maximum traction limit (good driver).

 

Now here is the important part. The ballistics equation and its derivatives is used to find the position, velocity, and acceleration of the car at a time t+1. At time t+2, the equations are reapplied using the final state from t+1 as the initial state at t+2. This is continued for 2000 cycles to give 20 seconds of acceleration.

 

By the way, this program predicts the ET of my car within 0.1 sec and MPH within .5 mph.

 

If anyone wants a free copy of the excel worksheet, let me know.

 

Sub quarter_mile()

 

t = 0

Dim GR(1 To 5)

Dim rpm(2000)

Dim a(2000)

Dim V(2000)

Dim x(2000)

Dim T_avail(2000)

Dim T_eng(2000)

Dim HP(2000)

Dim T_req(2000)

Dim SP(1 To 5)

Dim fpp(0 To 6)

Dim es(0 To 6)

Dim fx(0 To 6)

spline

es(0) = Val(DialogSheets("Input Dialog").EditBoxes("rpm1").Text)

es(1) = Val(DialogSheets("Input Dialog").EditBoxes("rpm2").Text)

es(2) = Val(DialogSheets("Input Dialog").EditBoxes("rpm3").Text)

es(3) = Val(DialogSheets("Input Dialog").EditBoxes("rpm4").Text)

es(4) = Val(DialogSheets("Input Dialog").EditBoxes("rpm5").Text)

es(5) = Val(DialogSheets("Input Dialog").EditBoxes("rpm6").Text)

es(6) = Val(DialogSheets("Input Dialog").EditBoxes("rpm7").Text)

''es(8) = Val(DialogSheets("Input Dialog").EditBoxes("rpm8").Text)

fpp(0) = Val(DialogSheets("Input Dialog").EditBoxes("fpp0").Text)

fpp(1) = Val(DialogSheets("Input Dialog").EditBoxes("fpp1").Text)

fpp(2) = Val(DialogSheets("Input Dialog").EditBoxes("fpp2").Text)

fpp(3) = Val(DialogSheets("Input Dialog").EditBoxes("fpp3").Text)

fpp(4) = Val(DialogSheets("Input Dialog").EditBoxes("fpp4").Text)

fpp(5) = Val(DialogSheets("Input Dialog").EditBoxes("fpp5").Text)

fpp(6) = Val(DialogSheets("Input Dialog").EditBoxes("fpp6").Text)

 

 

fx(0) = Val(DialogSheets("Input Dialog").EditBoxes("T1").Text)

fx(1) = Val(DialogSheets("Input Dialog").EditBoxes("T2").Text)

fx(2) = Val(DialogSheets("Input Dialog").EditBoxes("T3").Text)

fx(3) = Val(DialogSheets("Input Dialog").EditBoxes("T4").Text)

fx(4) = Val(DialogSheets("Input Dialog").EditBoxes("T5").Text)

fx(5) = Val(DialogSheets("Input Dialog").EditBoxes("T6").Text)

fx(6) = Val(DialogSheets("Input Dialog").EditBoxes("T7").Text)

 

SP(1) = Val(DialogSheets("Input Dialog").EditBoxes("SP12").Text)

SP(2) = Val(DialogSheets("Input Dialog").EditBoxes("SP23").Text)

SP(3) = Val(DialogSheets("Input Dialog").EditBoxes("SP34").Text)

SP(4) = Val(DialogSheets("Input Dialog").EditBoxes("SP45").Text)

SP(5) = 6000

TireRadius = Val(DialogSheets("Input Dialog").EditBoxes("TireDiameter").Text) / 2 / 12

GR(1) = Val(DialogSheets("Input Dialog").EditBoxes("GR1").Text)

GR(2) = Val(DialogSheets("Input Dialog").EditBoxes("GR2").Text)

GR(3) = Val(DialogSheets("Input Dialog").EditBoxes("GR3").Text)

GR(4) = Val(DialogSheets("Input Dialog").EditBoxes("GR4").Text)

GR(5) = Val(DialogSheets("Input Dialog").EditBoxes("GR5").Text)

RGR = Val(DialogSheets("Input Dialog").EditBoxes("RGR").Text)

CdA = Val(DialogSheets("Input Dialog").EditBoxes("CdA").Text)

Csr = Val(DialogSheets("Input Dialog").EditBoxes("Csr").Text)

mu = Val(DialogSheets("Input Dialog").EditBoxes("mu").Text)

Weight = Val(DialogSheets("Input Dialog").EditBoxes("Weight").Text)

n = 1

rpm(0) = Val(DialogSheets("Input Dialog").EditBoxes("launch").Text)

'T_avail(0) = (-0.00003 * rpm(0) ^ 2 + 0.2349 * rpm(0) - 135.84) * RGR * GR(1) * 0.87

T_avail(0) = fp1 * (rpm(0) - es(0)) + fx(0)

T_req(0) = 100 * TireRadius

 

a(0) = (T_avail(0) - T_req(0)) / TireRadius / Weight * 32.174

x(0) = 0

'V(0) = 0-

t = 1

Do While n <= 5 And rpm(t) < 8000 And t < 2000

V(t) = V(t - 1) + a(t - 1) * (0.01)

x(t) = x(t - 1) + V(t - 1) * (0.01) + 0.5 * a(t - 1) * (0.01) ^ 2

If x(t) > 60 And x(t - 1) < 60 Then

sixtyFootTime = ((60 - x(t - 1)) / (x(t) - x(t - 1)) + (t - 1)) * 0.01

sixtyFootSpeed = ((60 - x(t - 1)) / (x(t) - x(t - 1)) * (V(t) - V(t - 1))) + V(t - 1)

End If

If x(t) > 660 And x(t - 1) < 660 Then

eighthMileTime = ((660 - x(t - 1)) / 660 * t + t - 1) * 0.01

eighthMileSpeed = ((660 - x(t - 1)) / 660 * V(t) + V(t - 1))

End If

 

If x(t) > 1320 And x(t - 1) < 1320 Then

quarterMileTime = ((1320 - x(t - 1)) / 1320 * t + t - 1) * 0.01

quarterMileSpeed = ((1320 - x(t - 1)) / 1320 * V(t) + V(t - 1))

End If

 

Fn = 0.5 * Weight

'F_avail = (-0.00003 * rpm(t - 1) ^ 2 + 0.2349 * rpm(t - 1) - 135.84) * RGR + GR(n) / TireRadius * 12

 

 

rpm(t) = V(t) / 3.1415927 / (TireRadius * 2) * RGR * GR(n) * 60

If n = 1 And rpm(t) < rpm(0) Then

rpm(t) = rpm(0)

 

End If

If rpm(t) >= SP(n) And n < 5 Then

n = n + 1

rpm(t) = V(t) / 3.1415927 / (TireRadius * 2) * RGR * GR(n) * 60

shift = 50

End If

fp1 = -fx(0) / (es(1) - es(0)) + (fx(1) / (es(1) - es(0)) - fpp(1) * (es(1) - es(0)) / 6)

fp6 = -(fx(5) / (es(6) - es(5)) - fpp(5) * (es(6) - es(5)) / 6) + (fx(6) / (es(6) - es(5)))

 

If rpm(t) < es(0) Then

T_avail(t) = fp1 * (rpm(t) - es(0)) + fx(0)

 

End If

If rpm(t) >= es(0) And rpm(t) < es(1) Then

i = 1

T_avail(t) = fpp(i - 1) / 6 / (es(i) - es(i - 1)) * (es(i) - rpm(t)) ^ 3 + fpp(i) / 6 / (es(i) - es(i - 1)) * (rpm(t) - es(i - 1)) ^ 3 + (fx(i - 1) / (es(i) - es(i - 1)) - fpp(i - 1) * (es(i) - es(i - 1)) / 6) * (es(i) - rpm(t)) + (fx(i) / (es(i) - es(i - 1)) - fpp(i) * (es(i) - es(i - 1)) / 6) * (rpm(t) - es(i - 1))

End If

If rpm(t) >= es(1) And rpm(t) < es(2) Then

i = 2

T_avail(t) = fpp(i - 1) / 6 / (es(i) - es(i - 1)) * (es(i) - rpm(t)) ^ 3 + fpp(i) / 6 / (es(i) - es(i - 1)) * (rpm(t) - es(i - 1)) ^ 3 + (fx(i - 1) / (es(i) - es(i - 1)) - fpp(i - 1) * (es(i) - es(i - 1)) / 6) * (es(i) - rpm(t)) + (fx(i) / (es(i) - es(i - 1)) - fpp(i) * (es(i) - es(i - 1)) / 6) * (rpm(t) - es(i - 1))

End If

If rpm(t) >= es(2) And rpm(t) < es(3) Then

i = 3

T_avail(t) = fpp(i - 1) / 6 / (es(i) - es(i - 1)) * (es(i) - rpm(t)) ^ 3 + fpp(i) / 6 / (es(i) - es(i - 1)) * (rpm(t) - es(i - 1)) ^ 3 + (fx(i - 1) / (es(i) - es(i - 1)) - fpp(i - 1) * (es(i) - es(i - 1)) / 6) * (es(i) - rpm(t)) + (fx(i) / (es(i) - es(i - 1)) - fpp(i) * (es(i) - es(i - 1)) / 6) * (rpm(t) - es(i - 1))

End If

If rpm(t) >= es(3) And rpm(t) < es(4) Then

i = 4

T_avail(t) = fpp(i - 1) / 6 / (es(i) - es(i - 1)) * (es(i) - rpm(t)) ^ 3 + fpp(i) / 6 / (es(i) - es(i - 1)) * (rpm(t) - es(i - 1)) ^ 3 + (fx(i - 1) / (es(i) - es(i - 1)) - fpp(i - 1) * (es(i) - es(i - 1)) / 6) * (es(i) - rpm(t)) + (fx(i) / (es(i) - es(i - 1)) - fpp(i) * (es(i) - es(i - 1)) / 6) * (rpm(t) - es(i - 1))

End If

If rpm(t) >= es(4) And rpm(t) < es(5) Then

i = 5

T_avail(t) = fpp(i - 1) / 6 / (es(i) - es(i - 1)) * (es(i) - rpm(t)) ^ 3 + fpp(i) / 6 / (es(i) - es(i - 1)) * (rpm(t) - es(i - 1)) ^ 3 + (fx(i - 1) / (es(i) - es(i - 1)) - fpp(i - 1) * (es(i) - es(i - 1)) / 6) * (es(i) - rpm(t)) + (fx(i) / (es(i) - es(i - 1)) - fpp(i) * (es(i) - es(i - 1)) / 6) * (rpm(t) - es(i - 1))

End If

If rpm(t) >= es(5) And rpm(t) < es(6) Then

i = 6

T_avail(t) = fpp(i - 1) / 6 / (es(i) - es(i - 1)) * (es(i) - rpm(t)) ^ 3 + fpp(i) / 6 / (es(i) - es(i - 1)) * (rpm(t) - es(i - 1)) ^ 3 + (fx(i - 1) / (es(i) - es(i - 1)) - fpp(i - 1) * (es(i) - es(i - 1)) / 6) * (es(i) - rpm(t)) + (fx(i) / (es(i) - es(i - 1)) - fpp(i) * (es(i) - es(i - 1)) / 6) * (rpm(t) - es(i - 1))

End If

If rpm(t) >= es(6) Then

 

T_avail(t) = fp6 * (rpm(t) - es(6)) + fx(6)

 

End If

T_eng(t) = T_avail(t)

HP(t) = T_eng(t) * rpm(t) / 5252

'T_req(t) = (0.5 * 0.31 * 1.8 * 0.002377 * 100 ^ 2 / 2.54 ^ 2 / 144 * V(t) ^ 2 + 100) * TireRadius

T_req(t) = (CdA * 0.002377 * V(t) ^ 2 / 2 + Csr * V(t)) * TireRadius

 

If shift > 0 Then

T_avail(t) = 0

shift = shift - 1

Else

T_avail(t) = T_avail(t) * RGR * GR(n) * 1

End If

 

 

 

'T_avail(t) = (-0.00003 * rpm(t - 1) ^ 2 + 0.2349 * rpm(t - 1) - 135.84) * RGR * GR(n) * 0.87

a(t) = (T_avail(t) - T_req(t)) / TireRadius / Weight * 32.174

'If a(t) > 0.5 * 32.174 + 0.221 * a(t - 1) Then

'a(t) = 0.5 * 32.174 + 0.221 * a(t - 1)

'End If

If a(t) > mu * 32.174 Then

a(t) = mu * 32.174

End If

 

Sheets("Sheet4").Range("b10").Cells(t, 2) = T_req(t)

Sheets("Sheet4").Range("b10").Cells(t, 3) = T_avail(t)

Sheets("Sheet4").Range("b10").Cells(t, 4) = a(t)

Sheets("Sheet4").Range("b10").Cells(t, 5) = V(t) * 3600 / 5280

Sheets("Sheet4").Range("b10").Cells(t, 6) = x(t)

Sheets("Sheet4").Range("b10").Cells(t, 7) = rpm(t)

Sheets("Sheet4").Range("b10").Cells(t, 8) = n

Sheets("Torque&HP").Range("b10").Cells(t, 1) = t * 0.01

Sheets("Torque&HP").Range("b10").Cells(t, 2) = rpm(t)

Sheets("Torque&HP").Range("b10").Cells(t, 3) = T_eng(t)

Sheets("Torque&HP").Range("b10").Cells(t, 4) = HP(t)

t = t + 1

 

Loop

DialogSheets("Input Dialog").EditBoxes("QM_time").Text = quarterMileTime

DialogSheets("Input Dialog").EditBoxes("EM_time").Text = eighthMileTime

DialogSheets("Input Dialog").EditBoxes("SF_time").Text = sixtyFootTime

DialogSheets("Input Dialog").EditBoxes("QM_Speed").Text = quarterMileSpeed * 3600 / 5280

DialogSheets("Input Dialog").EditBoxes("EM_Speed").Text = eighthMileSpeed * 3600 / 5280

DialogSheets("Input Dialog").EditBoxes("SF_Speed").Text = sixtyFootSpeed * 3600 / 5280

'Sheets("Sheet9").Activate

End Sub

[Guest Anonymous]

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[Guest DaneL24]

Yes...that is pretty much what I was thinking. Just forgot to multiply pi by 2 when I calculated the distance traveled around the circle (was thinking diameter, not radius). I understand now.

[Guest Anonymous]

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[Pop N Wood]

The 5252 is not a constant. You use that number because you chose to express torque in foot pounds and power in horsepower. If you rated your engine in newton-meters and kilowatts (like the rest of the world) the number would be 9549. The number is simple a conversion factor between units, just like 1 foot = 12 inches.

 

There is nothing nebulous about power. There are very precise physical relationships between force, energy and power. Although judging from the number of erroneous web tutorials I have read, it is not hard to imagine why the average motorhead's understanding of power is nebulous.

 

I wouldn't read too much into the idea that torque is what is commonly measured on the dyno. That is just the easiest way to do it. Under certain circumstances it is easier to measure power first and to derive torque from that. It doesn't change the realtionship between the two, nor should it alter your interpretation of their meaning.