To keep it simple I have 1/2 bolt head which needs 20 ft lbs of torque exactly. It cannot be reached directly with a socket in a conventional manner. There was just enough room between to reach in and get the bolt in. If the box end of a 1/2" combination wrench is placed over the bolt head to be tightened the open end of the wrench is accessable. The combination wrench is 6" long. A half in drive fits nicely in the open end of the wrench and is 6" c/c.
Question is: What is the formula to figure the amount of torque (rotating force) needed to give 20 ft lbs at the bolt to be tightened, with the torque applied at the open end of the combination wrench.
You've been given the "stinkin" formula numerous times - and you still haven't supplied the information wrt the type of torque wrench or the length of the torque wrench handle you're using - but I think you've given me enough information to understand what's confusing you. It's easy to see why - it's even more difficult to explain, I think.
Again, torque wrenches are calibrated using the length from the drive axis to the center point of the handle. Because of the mechanical advantage of a longer handle torque wrench, less effort is required on the handle to deliver the same torque at the fastener. You are correct that the same force can be applied at the end of the two torque wrenches, but different input forces (you) are required to accomplish it. However, the force you want to apply is at the axis of the bolt and the mechanical advantage comes back into consideration in handle length as it relates to calibration of the wrench indicator (gauge). When the extension is added as a torque multiplier, it becomes a factor in the overall equation and although it is as one with the torque wrench, the reading must be changed to derive applied torque at the fastener. IOW, you are using the formula to recalibrate the torque wrench dial to take into account the change in total length and applied force. The change is less of a factor with a longer handle than with a short one because it represents a smaller percentage of total deviation from overall change (length of swing arc determined mechanical advantage) in relationship to the original calibration of the torque wrench.
In your example of the 14" vs. 16" handles, your answers are correct because of the above (admittedly confusing)explanation. If you extend the handle out to infinity (AND CALIBRATE IT THERE BEFORE ADDING THE EXTENSION), then add the extension, the 6" extension would be such a small overall effect as to not require an adjustment - you would simply set the dial at 20 ft/lbs.
For example - look at a handle of 100" (that is calibrated at that length) - the formula tells you to set the torque wrench to 18.9 ft/lbs. That is the ADJUSTED CALIBRATION of the wrench needed to provide 20 ft/lbs on the fastener with a 6" extension.
What you're not grasping - and perhaps I'm not explaining well - is that a longer torque wrench has more mechanical advantage than a short one, but the fixed length of the extension has less effect on the overall calibration of the longer handle wrench than on the short one.
Does that help? If not, it's the best I can explain it - perhaps someone else wouild like to take a try.... or you can use an incorrect formula if it better suits you needs.
Now inquiring minds should know....... <img src="/forums/images/graemlins/kewl.gif" alt="" />
Frank <img src="/forums/images/graemlins/cheers.gif" alt="" />
