RPM question


donnymac

Defered Ajudication
Mark IV Lifetime
Sep 26, 2008
739
West Texas
Was in a little speed run and picked up a noise that didn't belong. Not sure if it was speed or RPM related. I could back in and out of the noise but never tried to run thru it.

At 180 in 6th gear, RPM should be ballpark what?

Thanks.
 
Was it low buzzing or resonance ?

I know over 100mph shifter cables vibrate and if rubber bumpers aren't aligned in tunnel there's noise (different from rattlesnake noise)

Saw some posts on it early in first few years
 
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Do a search for rattlesnake noise.
 
speedingear.jpg
 
How much NM is needed to reach topspeed in 6th gear ?

Not that I am ever going to reach that though :)
 
I can tell you that when we ran 278.2mph in the mile we were still in 5th gear.... :)
 
Thanks all for the input. Actually sounded like a tin margarita mixer with BBs. Rattler buz is close if made of tin.
 
I can tell you that when we ran 278.2mph in the mile we were still in 5th gear.... :)

I was cruising in 6th just looking to radar at 200. Had about a 6 mile run - plenty of time.
 
How much NM is needed to reach topspeed in 6th gear ?

Not that I am ever going to reach that though :)

NM?

And you will not reach top speed in 6th gear...

and thanks Xcentric. It is amazing how many questions are answered by just reading the book (SAE PT-113, see Figure 24 of report 2004-01-1252 found in PT-113 for the graph Xcentric posted).
 
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Newton Meters, metric for horsepower.
 
Newton Meters, metric for horsepower.

Newton Meters is Torque (foot-lbs in English) . HP (English) is Watts=joules/sec (Metric)
 
Tony is correct.

This is why I asked.

Nautical Miles?

Always good practice to define an acronym the first time it is used as the reader may have no idea of your jargon.
 
Newton Meters is Torque (foot-lbs in English) . HP (English) is Watts=joules/sec (Metric)
Roger that...brain dump. I really know this from engineering school as I had to take the class twice!
 
Is SAE torque in lb-ft, or ft-lb? or does it matter? I was taught that in SAE that Work was measured in ft-lbs, and Torque was measured in lb-ft.
 
Tony is correct.

This is why I asked.

Nautical Miles?

Always good practice to define an acronym the first time it is used as the reader may have no idea of your jargon.

Next time you are in Dearborn around Ford people, mention PDC (generically) in a sentence. It can be a person, place or thing. :rofl
 
Since none of you guys seem to really work, I guess it is hard for you to define.
 
Doesn't matter. They are equivalent. One pound acting on a one foot lever arm around a pivot produces the same amount of torque and work.

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Well Gary, kinda.....:wink

You are correct that the labeling of Torque can be ft-lbf or lbf-ft. It is like 5x2=10 just as 2x5=10.

And because Torque and Work (force through a distance) do have the same units, you have to be careful.

From Wikipedia-
“"Foot-pound" is sometimes also used as a unit of torque (see Pound-foot (torque)). In the United States this unit is often used to specify, for example, the tightness of a bolt or the output of an engine. Although they are dimensionally equivalent, energy (a scalar) and torque (a vector) are distinct physical quantities. Both energy and torque can be expressed as a product of a force vector with a displacement vector (hence pounds and feet); energy is the scalar product of the two, and torque is the vector product."

If you apply 100 ft-lbf through your torque wrench on a head bolt (and it does not rotate) you have applied a "torque" to the fastener but you have not done any "work". The torque can be modeled as a force “couple”, two equal but opposite forces acting at the bolt head in equilibrium and each force located at a distance of 50% of the total length of the torque wrench.

If you apply the same torque to the bolt head and the head turns a certain degree of arc (given as φ in the attached figure) then you have done work on the bolt head.

This result can be understood more simply by considering the torque as arising from a force of constant magnitude F, being applied perpendicularly to a lever arm at a distance r, as shown in the figure. This force will act through the distance along the circular arc s = rφ, so the work done is

W = Fs = Frφ


Notice that only the component of torque in the direction of the angular velocity vector contributes to the work.
 

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That explains why I get tired without doing any work. ;)
 
I prefer fig Newtons. And not too much work.

Another Prof. Bill lesson, thanks. Well under 300 words!