Technical Actual meaning of inch pounds

SORRY,, PROBABLY TITLED THIS INCORRECT.. Am I correct in in saying that if you were going to torque something such as preload on a pinion and the spec calls for 14 inch pounds, the truest way to explain that would be it would take a one pound force at exactly 14 inches from the exact center of the pinion. IOW, If you made a devise that attached to the pinion shaft and measured out from the absolute pinion center exactly 14 inches and applied a 1 pound weight you would be spot on. Now if that is correct , would using a 1-7/8 wrench that is perhaps 8 inches long with a torque wrench attached to the end ( 8 inches from center of pinion) where there is a 1/2 square to attach a ratchet ( or in this case a torque wrench) give you an incorrect reading ? Saw a video where a guy did just this, had what looked like a fan clutch wrench that fit his pinion nut which is 1-7/8 ,had a torque wrench attached to the end of that wrench .

 

An inch pound is a force of one pound applied one inch from the center. A foot pound is a force of one pound applied twelve inches from the center. Force needs to be in a straight line. Depending on the configuration, wrench adapters can change the value. In the wrench situation you described you would need to adjust the torque wrench value applied to compensate for the wrench extension beyond the torque wrench. Search torque wrench adapters for diagrams and the math involved.

 

a buddy use to ask how many eeeks equaled an inch pound

 

If the spec for tightening something was say 100 inch pounds, and you only had a foot pound torque wrench, could you divide 100 by 12?? or 8 1/2 ??? This is how I've done it with a old beam type torque wrench and I guess it worked?? What you guys say??

 

Yes, 1 ft. lb. is equal to 12 in. lb.

 

Yup. If you used a crows foot socket that is only about an inch or so off center, it will give you less of an error. Keep in mind that the preload torque is usually specified as a range, so you don't need to be dead nuts on the right number, for it to work fine, if you are just checking bearing preload.

 

A good torque wrench is not that expensive or hard to come by why overthink it.

here is a good description and if you need a conversion I will find you a equation.

Similarly, an inch-pound (or pound-inch) is the torque of one pound of force applied to one inch of distance from the pivot, and is equal to 1/12 of a pound-foot. It is commonly used on torque wrenches and torque screwdrivers for setting specific fastener tension.

Here is a simple article that does not dig too deep into quantum theory. LOL

https://sciencing.com/convert-inchpounds-footpounds-8368608.html

 

The "correct" way of stating that is "Pound-Inch", in actual Engineering terms.
Same as, Pound-Feet, the correct way of stating that value...Engineering term wise.

Look it up...

Mike

 

Mike, you are correct. But it seems mechanics and engineers don't always speak the same language.

 

I'm an engineer, and I always say inch pounds or foot pounds when I'm talking about tightening bolts, amongst folks who work on cars.

Because that is what they understand.

 

helpful on-line calculator for offset torque wrench adaptors
just click on the blue link of each example
Note; it doesn't show the formula used. IIRC I've seen those before.

http://www.fordservicecontent.com/renderers/torquewrench/wrench_formula_main_en.asp

 

Like I said...

 

You know what they say about getting info off the internet........

No one has mentioned proper lubrication when torquing.

 

I don't mean to hyjack the thread, but have a related question. When setting a bearing preload to an inch pound spec, is the spec based on the torque required to start the nut turning from a standstill or to keep it moving once in motion?

 

Years ago I got a Snap-on adaptor to torque a bolt that was hard to get to on my ladder truck. It also stated that I had to measure the length of my torque wrench in order to figure the adjustment rate! I never could understand why? A torque wrench measures torque, it make no difference in the length of the wrench! In my opinion!
For example, it you need to torque your rod nuts 45 ft/lbs, it makes no difference if your wrench is 18 inches long or 24 inches long, you still get 45 ft/lbs of torque. You have to pull harder on the shorter wrench, but the twisting torque is the same, if the wrench is set the same!
I though and still think that any adaptor should have a fixed correction value, based solely on the adaptors length!
I must be wrong as I have seen the correction factor, on all adapters, refer to wrench length.
Maybe some one here could enlighten me?








Bones

 

Needing the length of the torque wrench makes sense to me.

Explaining why will take some thought, though.

Do you understand the concept of a "free body diagram", from physics class?

 

The torque wrench is calibrated to read correctly based on it's beam length. The adapter length relative to the beam length determines what's actually applied to the fastener.
https://www.engineersedge.com/manufacturing_spec/torque_wrench_1.htm

 

Yes, I have a basic concept of forces. I just had high school physics. I understand leverage and multiplication of force. But to me, the torque wrench, as a measuring “ gauge” takes the length out of the equation.
Why do you have to take the length of the wrench into consideration with a , say ,three inch adapter but not if you put it directly on the nut?








Bones

 

When you use a torque wrench, you are using a lever and a force, to generate torque at the bolt. But this torque is not pure torque--there is also a side force, equal to the force that you are applying to the handle of the wrench. We usually ignore this force, because we are only interested in the torque that's applied to the bolt. But when you add an extension to the torque wrench, this force now acts on that extension lever, affecting the final torque that's applied to the bolt. And the angle of the extension relative to the handle of the wrench, and the length of the torque wrench handle, come into play in figuring out the actual torque applied to the bolt.

Draw a few free body diagrams of those torque wrenches and extensions in gatz's post. See if you can figure it out...

 

Preload is measured while rotating, the breakaway torque will always be a little higher

Whoops!!


Sent from my iPhone using H.A.M.B.

 

For you non-engineers, the term "gogher" is translated to "higher."



Sorry, man, just busting ya a little, we all know typing on phones SUCKS!

 

I understand that the wrench is calibrated, but the end results, after it has been set at X ft/lbs is, that much “ twist” is on the wrenches drive. 45 ft/lbs on a 16 inch wrench is the same as 45 ft/lbs on a 24 inch wrench.
The adapter only senses how much torque is applied to the female drive .

I would be in total agreement, if we were using a bar x number of inches long and adding weights to it, the length would be paramount! But it’s the job of the torque wrench to measure the torque applied to the female drive of the adapter.










Bones

 

"I would be in total agreement, if we were using a bar x number of inches long and adding weights to it, the length would be paramount! But it’s the job of the torque wrench to measure the torque applied to the female drive of the adapter."
Yes, the torque wrench will read the force it applies to the adapter. Your not adding weights, but you are pushing on the torque wrench. But if you want to apply a specific torque to the fastener at the end of the adapter, the value that you set the torque wrench to get your desired fastener torque is relative to their lengths. It's just the physics of it. I'll leave the rest of the debate to the real engineers. I was just a military engineer.

 

Well, now I am really confused. I use Gatz’s formul used 90 lbs for torque, with a 12 in wrench and a four inch straight adaptor, it said I need to set my 12 in wrench to 68 ft/lbs.
Then I made my wrench twice as long, 24 inches, and now I need 77 ft/lbs! What! I would have though with a longer wrench you would need to set lower than a short wrench.

This makes absolutely no sense to me!






Bones

 

Figure out what force you have to apply to the handle of the torque wrench, for the two different lengths. See if it makes sense when you think of it as applying that force to the torque wrench, and dividing that force by the actual length of the wrench, including the extension.

draw diagrams!

 

In my line of thinking , using 90 ft/lbs of torque, you put 90 lbs of weight out at 12 inches. If you add 12 inches of (24) you would reduce your weight to 45 lbs, to get the same torque.
Using this theory, by my calculations, for a four inch adapter, you would need to reduce your torque by 16.66 percent, no matter the length of the wrench. That comes to 15 lbs reduction at 90, or 75 ft/lbs. This figure falls in between Gatz’s formula of 68 and 77 Ft/lbs, for a 4 inch adapter with a 12 and 24 inch wrench.





Bones

 

The modern day world is now using newton metres.

 

I used my torque wrench last week. it was 35 Ncm. Must be a metric bolt. and not very big. I tell ya, you engineers.........

 

I double checked it and added different figures, but, I did find a correlation to the ratios. The longer the torque wrench, the closer you get to the desired torque! Quite the opposite as I expected.
But since “ they” say wrench length is a factor, as the wrench gets longer, the comparablely, short adaptor becomes less of a factor. If I use my six foot torque wrench, I only have to reduce it by 3 lbs. But here’s the kicker, to get rid of those last three pounds, I have to find a torque wrench that is 725 inches long! So it’s kind of a sliding scale!








Bones

 

Ok, while I’m beating this dead horse....... I have probably ten torque wrenches, some large( six ft” ) some small. How to you figure the adapter correction for my Tee shaped small torque wrenches?







Bones