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Torque Adapter Calculator

4.4K views 16 replies 9 participants last post by  newsh  
#1 ·
A torque adapter adds length to the torque wrench and therefore lees torque should be dialed-in to match the torque specifications. Should you want let's say for example 30 ft-lbs using a 1.5 inch torque adapter, then set the torque wrench at 27.1 ft-lbs because the extra length provided by the torque adapter provides more torque. Here's the calculator to properly dial-in your setup:

http://www.cncexpo.com/TorqueAdapter.aspx
 
#2 ·
Hi Terry,

If this is in relation to using a torque adapter on barrel base nuts, why do you assume Meriden didn't allow for this in the workshop manual specification? After all, there is no other way of torquing barrel base nuts, available torque adapters wouldn't have varied greatly in length, the torque figure applied doesn't need to be absolutely precise?

Regards, ;)
 
#3 ·
EXCELLENT POINT that I did not consider. And what do you know about this. DID Meriden allow for the torque adapter??? Seems like it's required to know a real Meriden mechanic for certain.

I am here to learn from all members and there are some real ambiguities that exist. 8)
 
#7 ·
In a similar thread here or on another forum this was discussed extensively. Apparently if you can mount the torque adapter at a 90 degree angle to the shaft of the torque wrench there is no leverage advantage and you can take a direct reading from the torque wrench. My T140 has standard size fasteners for the base nuts . I put a 3/8" allen bit the torque wrench and I have a double ended box wrench (ring spanner ) that is 3/8" on one end and 7/16" on the other . I mount the 3/8" end at 90 degrees on the torque wrench and use the 7/16" on the smaller base nuts. For the larger twelve point nuts I just use a 1/2" combination wrench using the open end at 90 degrees on the 1/2 drive on the torque wrench and use the box end on the nuts.
I don't suggest that every one try this just to torque base nuts but it was fun finding a solution to something that really is not a problem. I just have a collection of old wrenches.
. As Stuart suggests , if you have any wrenching experience ,tightening the base nuts evenly should not really present a challenge.
 
#8 · (Edited)
Hi,

if you can mount the torque adapter at a 90 degree angle to the shaft of the torque wrench there is no leverage advantage and you can take a direct reading from the torque wrench.
Between you, me and the gatepost, ... there is a difference ... :rofl

The lever length is still the measurement between the moving (not rotating) end of the torque wrench and the centre of the fastener. The torque wrench and any adapter at 90 degrees form the two shorter sides of a right-angled triangle, so the difference (increase) in torque is the setting on the wrench divided by the cosine of the included angle between the line of the lever and the line of the wrench.

In practice, the shorter the adapter and the longer the torque wrench, the smaller the included angle, so the closer the cosine is to 1 and the smaller the increase in torque.

A potential problem only arises when people read half-explanations on internet forums, find only a foot-long ring wrench they think they can use as an "adapter" and a short torque wrench, then cuss 'cos they muller something (the crankcase threads? :eek:) because they're actually applying a much higher torque than the internet half-explanation says ...

Hth.

Regards,
 
#12 ·
Hi Terry,

Errr ... no ...

"25 lb. ft.", the "25" is a product of multiplying pounds force ("lb.") by distance ("ft.") - easiest to understand(?) is 25 lb.ft. is 25 lb. applied one ft. from the centre of a cylinder base stud (in this case) and one lb. applied 25 ft. from the centre of the stud.

Because your lever is only 4.45 inches between centres, it's only 0.37 feet; as the "25" is a product, you must increase the force applied to the end of the lever with your Baggage Scale/Spring Balance; i.e. 25 is divided by 0.37 to get the pounds force that must be applied to your lever to apply the equivalent of 25 lb. one ft. from the centre of a stud.

Hth.

Regards,
 
#13 ·
Stuart, If you look at my diagram you will see that the effort required (taking into account a 25% reduction for a “wet” thread), is 50 lbs, which exerts a torque of 18.7 lb/ft on the nut.

25 lb/ft divided by 0.37 x 0.75 = 50 lb.

The required 50lbs at the end of the short spanner, is about as much as can be achieved, by pulling on such a short spanner.

I use the spring balance on the spanner, only for consistency.
 
#14 ·
Hi Terry,
If you look at my diagram you will see that the effort required
is 50 lbs, which exerts a torque of 18.7 lb/ft on the nut.
(y) That wasn't clear given, "18.45 LB/FT", "18.54 LB/FT" and "18.75 LB/FT" are also on your drawing.

Btw, "25 lb/ft" = "25 pounds force PER foot", the "/" indicates "divided by". Not the same at all as "25 lb.ft".

Regards,