Physics Q; remotely aviation

Let'sgoflying!

Let'sgoflying!

Touchdown! Greaser!
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Dave Taylor
When pulling slowly & steadily on a torque wrench, the mind has free time to explore related topics. This is one that occurred to me last week.

Let’s say you are pulling on a 2 ft bar attached to a socket which is on a nut you are tightening.
If the tightening torque req’d is 20 ftlbs, applying 20lbs of force at 12” from the nut center will get 20ftlbs.
What lesser force is required at 24” from the nut center, half the force, or 10lbs?

How much force is needed at 6”?

I will explain the conundrum in my mind, in a while.
 
Torque = force * distance.

Using in-line extensions will not change the torque applied to the nut/bolt contrary to what some will claim.
spin_prod_206296701

Using a universal joint will reduce the effective torque applied to the nut/bolt as a function of the cosine of the angle between the axis of the wrench and the axis of the nut/bolt.
 
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Same principals as weight and balance calculations.
 
Yeah, neglecting any complicatedness of the engineering of the wrench itself...

Torque = r (times) F
If you double the length of the lever arm but want to keep the torque the same, the force needed is half the pounds.
And if the lever arm is half as long, you'll need double the force.

This assumes a spherical cow in a vacuum. :)

[Edit: sorry, that's a physics joke...]
 
Let'sgoflying! said:
When pulling slowly & steadily on a torque wrench, the mind has free time to explore related topics. This is one that occurred to me last week.

Let’s say you are pulling on a 2 ft bar attached to a socket which is on a nut you are tightening.
If the tightening torque req’d is 20 ftlbs, applying 20lbs of force at 12” from the nut center will get 20ftlbs.
What lesser force is required at 24” from the nut center, half the force, or 10lbs?

How much force is needed at 6”?

I will explain the conundrum in my mind, in a while.
Click to expand...
make sure the force is constant....and no acceleration is involved.o_O
 
Capt. Geoffrey Thorpe said:
Torque = force * distance.

Using extensions will not change the torque applied to the nut/bolt contrary to what some will claim.

Using a universal joint will reduce the effective torque applied to the nut/bolt as a function of the cosine of the angle between the axis of the wrench and the axis of the nut/bolt.
Click to expand...

Depends on what you mean by extension.
 
At 24”, 10 lbs of force on the end of the torque wrench. Assuming you had the space, at 20’, 1 lb of force. At 6”, 40 lbs of force.
 
So....explain the “conundrum” here?
 
The old prop on the Navion required 400 foot pounds of torque. I specified that as my wife hanging 3' out on the metal bar that I had threaded through the prop nut.
 
Capt. Geoffrey Thorpe said:
spin_prod_206296701
Click to expand...

Then you are correct as those do not extend the torque arm and do not act as a force multiplier. However the items illustrated in Post 4 could also be considered extensions and definitely do affect torque. While I’m positive that you know that, I was concerned that your post might confuse the less mechanical knowledgeable.
 
I use a Craftsman model 1019 Laboratory Edition Signature Series torque wrench; the kind used by Caltech high energy physicists and NASA engineers. It had been calibrated by top members of the state and federal Departments of Weights and Measures, to be dead-on balls accurate.
 
flyingron said:
I use a Craftsman model 1019 Laboratory Edition Signature Series torque wrench; the kind used by Caltech high energy physicists and NASA engineers. It had been calibrated by top members of the state and federal Departments of Weights and Measures, to be dead-on balls accurate.
Click to expand...

Do you have a certificate of validation?
 
wrbix said:
So....explain the “conundrum” here?
Click to expand...

I drew a graph of lbs of force needed vs arm (distance from applied force to center of the nut) to achieve 20ftlbs.
It appears to be a curve (a non-linear relationship).
I had thought with the simple 'half the distance, double the force' sounded like a linear relationship. I guess not.
Thanks.
 
Let'sgoflying! said:
I drew a graph of lbs of force needed vs arm (distance from applied force to center of the nut) to achieve 20ftlbs.
It appears to be a curve (a non-linear relationship).
I had thought with the simple 'half the distance, double the force' sounded like a linear relationship. I guess not.
Thanks.
Click to expand...
If for every double distance you require half the force, you won't have a linear curve for sure. Simple math.
 
and....I was thinking if we get to .001mm from the nut center, the required force would be exponential... which fits with a non-linear situation.
It’s been a few decades since I used this part of my brain.
 
Let'sgoflying! said:
and....I was thinking if we get to .001mm from the nut center, the required force would be exponential... which fits with a non-linear situation.
It’s been a few decades since I used this part of my brain.
Click to expand...

Yep. Think bicycle spokes. They are laced for a reason. If you put radial spokes on the rear wheel (which you might do for the front) you would have the case of exponential force applied to the spokes transferred from the pedals.
 
"Exponential" has to be one of the most misused words in the English language. Exponential means that how fast something increases is proportional to its current value. That's not the case here. It's not linear, but there are many kinds of non-linearity that aren't exponential. This one is an inverse relationship, as in inversely proportional.

Assuming all you do is change the lever arm, the force required is always going to be the needed torque divided by the lever arm. I.e. the force is inversely proportional to the lever arm.
 
lol! the thought of my possibly inaccurate use of that word crossed my mind just as I was typing it. Not sure if I was too tired or didn't care, figured I'd take a chance with it.
 
Let'sgoflying! said:
lol! the thought of my possibly inaccurate use of that word crossed my mind just as I was typing it. Not sure if I was too tired or didn't care, figured I'd take a chance with it.
Click to expand...
Jeez, Dave. You’ve been around here long enough to know that taking that kind of a chance will bring out the nit pickers. LOL
 
azure said:
"Exponential" has to be one of the most misused words in the English language. Exponential means that how fast something increases is proportional to its current value. That's not the case here. It's not linear, but there are many kinds of non-linearity that aren't exponential. This one is an inverse relationship, as in inversely proportional.

Assuming all you do is change the lever arm, the force required is always going to be the needed torque divided by the lever arm. I.e. the force is inversely proportional to the lever arm.
Click to expand...
Well if we're going to pick nits... an exponential function is one where the function can be described using... an exponent (see conics). In this case the exponent is (-1):
y = x^(-1) <---- Hyperbola
thus I don't feel that "exponential" was used incorrectly at all. Exponential != Exponential Growth.

If it's not linear, and it's not logarithmic... there's a good chance it's exponential.
 
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Mtns2Skies said:
Well if we're going to pick nits... an exponential function is one where the function can be described using... an exponent (see conics). In this case the exponent is (-1):
y = x^(-1) <---- Hyperbola
thus I don't feel that "exponential" was used incorrectly at all. Exponential != Exponential Growth.

If it's not linear, and it's not logarithmic... there's a good chance it's exponential.
Click to expand...


No, an exponential function will have the variable as the exponent. That is, it will be of the form

y=a^x​

where a is a constant.
 
Half Fast said:
No, an exponential function will have the variable as the exponent. That is, it will be of the form

y=a^x​

where a is a constant.
Click to expand...
Yep you're right (sheepish grin). I was trying to cut Dave some slack :).

I'm out of school for 3 months and this is what happens. :eek:
 
No, an "exponential" function has the form:
y = (A) e^(Bx)
... which has the special property, like azure said, where the rate of change of y(x) is proportional to y itself.
This describes things like population growth (going up), or nuclear isotope decay (going down).

Something like:
y = A x^(n)
... is called something else, like a polynomial if n is an integer. Or a "power law"

Something like:
y = A B^x
... okay, I'm not sure what that's called, but I wouldn't call it an exponential.
 
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Geez....how did this go exponential?....without first going logarithmic....o_O
 
kath said:
No, an "exponential" function has the form:
y = (A) e^(Bx)
... which has the special property, like azure said, where the rate of change of y(x) is proportional to y itself.
This describes things like population growth (going up), or nuclear isotope decay (going down).

Something like:
y = A x^(n)
... is called something else, like a polynomial if n is an integer. Or a "power law"

Something like:
y = A B^x
... okay, I'm not sure what that's called, but I wouldn't call it an exponential.
Click to expand...


Common vernacular often has e as the constant in an exponential equation, but by the strict mathematical definition it doesn’t have to be e. Y = 10^x is also an exponential equation.

In nature, e is very often the constant.
 
Yep, @Half Fast nailed it.

My definition is just a putting into plain words of the differential equation that defines an exponential function, which is of the form dy/dx = ay, where "a" is a constant. The solution is y(x) = e^(ax).

And yes, even there the presence of the constant "a" means the base really isn't necessarily e, rather it's e^a, so it can be any number, as Half Fast said.
 
kath said:
Something like:
y = A B^x
... okay, I'm not sure what that's called, but I wouldn't call it an exponential.
Click to expand...
Oh, but it is! You can write it as y = A e^((ln B)*x). Wouldn't you call that an exponential? ;)
 
azure said:
Oh, but it is! You can write it as y = A e^((ln B)*x). Wouldn't you call that an exponential? ;)
Click to expand...


Bear in mind that a true mathematician would never just write 1+1=2 when instead he could write

((sin(x))^2 + (cos(x))^2) - e^((pi)i) =2​
 
azure said:
"Exponential" has to be one of the most misused words in the English language. Exponential means that how fast something increases is proportional to its current value. That's not the case here. It's not linear, but there are many kinds of non-linearity that aren't exponential. This one is an inverse relationship, as in inversely proportional.

Assuming all you do is change the lever arm, the force required is always going to be the needed torque divided by the lever arm. I.e. the force is inversely proportional to the lever arm.
Click to expand...
I literally think there are other words that are misused exponentially more often.
 
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