TheSamba.com :: View topic - Breaker bar Length + my weight to get correct Torque?

Wed Nov 07, 2012 9:14 pm

Back when I was building my 63, someone mentioned a site that had the math to figure out the correct length of a breaker bar. So you can torue the axle nut and fly wheel gland nut. Does anyone know where I can find this?

Wed Nov 07, 2012 9:28 pm

1lb weight at the end of a 1 foot bar equals 1 foot pound. 100 pounds at the end of a 3 foot bar equals 300 foot pounds at the nut.

Wed Nov 07, 2012 9:30 pm

Thanks gonebuggy!

Wed Nov 07, 2012 10:03 pm

Here's a post I made a while back in another thread about the gland nut

EverettB wrote: Here's what I do:
Get my breaker bar and big socket.
You need 250 ft. lbs. of force so that is 250 lbs. at 1 foot out from the socket.
1 foot = 12 inches.

Say you weigh 175.
So take 250/175 = 1.43
12 inches x 1.43 = 17.1 inches.
Mark a line on the bar at 17.1 inches.
Get someone to hold the engine, I usually have my wife put her foot on the opposite cylinder head with the engine on the ground.
Stand on the bar with all your body weight at the line you just made. I usually have to brace myself on my car or put a hand on a nearby tool chest.
When the bar stops moving downward from your weight, you just applied 250 lbs.

Don't jump on the bar or you will over tighten it.
It takes less force than you think.

I use this method to do the rear axle nuts too.

Wed Nov 07, 2012 10:10 pm

Thanks Ev! That is the info I was searching for. Now what if you weigh more than the required torque? LOL I am joking by the way. :wink:

Wed Nov 07, 2012 10:14 pm

Substitute the numbers.

Say you weight 300, nice round #.

So 250 ft. lbs. of torque/300 lbs. of all man baby = 0.83
12 inches x 0.83 = 10 inches
So stand at the 10 inch point.

Wed Nov 07, 2012 10:33 pm

Harbor freight has torque wrenches for under 30 bucks ya know.

Wed Nov 07, 2012 11:23 pm

And they are about as accurate as standing on a cheater bar. :roll:

Wed Nov 07, 2012 11:39 pm

RPGreg2600 wrote: Harbor freight has torque wrenches for under 30 bucks ya know.
would you use a harbour freight O ring on the space shuttle, :lol:

Thu Nov 08, 2012 1:17 am

When a shop manual gives you a range of 10lbs, I'm sure if you shoot for the low end you'll be in the ballpark. :P

Thu Nov 08, 2012 8:35 am

I used the breaker bar method for years, and didnt even have the high-tech chart EV posted. But I always had some fear in the back of my mind when it came to the flywheel gland nut, it is just better to use a torque wrench.(If possible). I have a $450+ Blue Point torque wrench, and my chessey $59 Harbour Freight torque wrench is only 2lbs. off at the higher settings. :?

Thu Nov 08, 2012 12:06 pm

Where I work we use calibrated torque wrenches (S/N'd and traceable). We bought some at HF to see how they stacked up w/ the very expensive ones we were buying. Our tolerances are pretty tight (+/- 1%) and the HF ones were shockingly bad. As high as 10% on higher values. They gave them to me because they didn't trust them. On the expensive ones, when they get out of spec. they are also sh!tcanned. Some of them can be corrected, but it is tough explaining that to some people, so we just don't. Our big 3/4" drive, 600lb/ft wrench has been in service for 20+ years! For critical torque, DO NOT trust cheap import tools.

Fri Nov 09, 2012 1:46 pm

We have some of those critical cal torque wrenches on base. Or technical data for aircraft repairs are written with + or - on the torque. Based on the fact that a true torque, torque wrench is hard to come by. Plus people don't respect tools like the should. :?

Fri Nov 09, 2012 3:23 pm

I'm not a math wizard,but it should be mentioned that 250lbs @ 12" out is not equal to 125lbs @ 24" out from center.

That is to say...the further out on a cheater bar,the divided force is exponentially increased.

I'm sure a math major could explain this better than myself...

Fri Nov 09, 2012 4:21 pm

Thomas Pedigo wrote: I'm not a math wizard,but it should be mentioned that 250lbs @ 12" out is not equal to 125lbs @ 24" out from center.

That is to say...the further out on a cheater bar,the divided force is exponentially increased.

I'm sure a math major could explain this better than myself...

No, this is incorrect, as long as your breaker bar (lever) is perpendicular to the nut being tightened.
It would be strange for it not to be when doing an axle or gland nut... and in most other cases as well.

Sat Nov 10, 2012 2:15 am

EverettB wrote: Thomas Pedigo wrote: I'm not a math wizard,but it should be mentioned that 250lbs @ 12" out is not equal to 125lbs @ 24" out from center.

That is to say...the further out on a cheater bar,the divided force is exponentially increased.

I'm sure a math major could explain this better than myself...

No, this is incorrect, as long as your breaker bar (lever) is perpendicular to the nut being tightened.
It would be strange for it not to be when doing an axle or gland nut... and in most other cases as well. So, as an example, 50lbs of force on a 5' breaker bar is equal to 250 fp of torque? (Of coarse, perpendicular to that being turned)

Does a bus with with 50fp of torque and 14" tires, pull the same as a bus with 100fp of torque and 28" tires?

Sat Nov 10, 2012 9:05 am

Thomas Pedigo wrote: EverettB wrote: Thomas Pedigo wrote: I'm not a math wizard,but it should be mentioned that 250lbs @ 12" out is not equal to 125lbs @ 24" out from center.

That is to say...the further out on a cheater bar,the divided force is exponentially increased.

I'm sure a math major could explain this better than myself...

No, this is incorrect, as long as your breaker bar (lever) is perpendicular to the nut being tightened.
It would be strange for it not to be when doing an axle or gland nut... and in most other cases as well. So, as an example, 50lbs of force on a 5' breaker bar is equal to 250 fp of torque? (Of coarse, perpendicular to that being turned)

Yes

Thomas Pedigo wrote: Does a bus with with 50fp of torque and 14" tires, pull the same as a bus with 100fp of torque and 28" tires?
This question involves other factors, including tire contact patch, tire weight, etc.

Sun Nov 11, 2012 7:52 am

Everett is correct^^

Regarding the engine torque question, what you say is effectively correct.
Both buses will accelerate at the same rate (assuming of course that the buses have identical mass, rolling friction, etc).
As the bus with 100 ft lbs of torque is more powerful, it will be able to overcome losses (hill grades, wind resistance, rolling friction) better, so it will ultimately be able to go faster than the 50 foot pound bus, but their initial accelerations will be the same.

Regarding the initial question about torques, it should also be understood that published torque values are correct for clean, dry fasteners.
If a fastener is burred or rusty, it will require more torque to achieve the corrrect tightness. If the fastener is lubricated in any way, one would require less torque to reach the required tightness.
These values can differ significantly, so for any critical torques (like head bolts or gland nuts) ensure they are clean and dry.
For less critical torques (axle nuts) not such a big deal.

Sun Nov 11, 2012 2:33 pm

Riff Raff wrote: Everett is correct^^

Regarding the engine torque question, what you say is effectively correct.
Both buses will accelerate at the same rate (assuming of course that the buses have identical mass, rolling friction, etc).
As the bus with 100 ft lbs of torque is more powerful, it will be able to overcome losses (hill grades, wind resistance, rolling friction) better, so it will ultimately be able to go faster than the 50 foot pound bus, but their initial accelerations will be the same.

The one BIG difference is obviously the transmission and consequent gearing. Your example is only applicable for a direct drive scenario. Initial acceleration cannot be the same due to the fact that the engine develops maximum torgue only at a given value. The torque values are a curve, not a flat line unless it is a constant torque motor such as a electric motor. Typically, the RPM where the two engines develop maximum torque will not be X for the 50 lb-ft and 2X for the 100 lb-ft. And even if it were, the rolling diameter of the wheels is not Y & 2Y even though R is r &2r. There are way too many variables in this case to make blanket statements. The maximum torque cannot overcome the mechanical limitations of a short tire vs. tall tire. Even if the maximum torque values were identical the 28" tire would clearly have the advantage in speed.

Sun Nov 11, 2012 6:01 pm

I agree in principle with your statements.
My goal was to illustrate in a elementary fashion (constant torque, constant gearing and 14" diameter tires and 28" diameter tires (so y and 2y)) how torque functions for moving a vehicle rather than snugging a fastener.
In a "real life" situation, there are many more variables that need to be considered.