I was going to post a pic of the instructions but it says I cant!! Got a new torque wrench and was going to change the plugs (after hearing an exhaust leak like sound) at 93000 miles! I do not completely understand the adjustment for attactchments (using the universal joint). I think it is saying apply the formula for the amount the wrench head is off center from perpendicular. It says the setting for foot #'s will be different with this offset but the destructions are very limited. Ya think I am reading this right need more info (like how to attach the scan) or what I am kinda stumped!!! Many Thanks Craig...

 

Hello Craig,
The sparkplug torque for your 99 Ford is only 96 INCH POUNDS, not foot pounds like we have been used to in the past when the heads were made out of iron. You might have the wrong torque wrench or you can try using 8 foot pounds. Do not use anti-sieze and be very careful not to cross thread the plug at the start. Should screw in several turns with just fingers on the extension. Good Luck,
Fasterhorses

 

Thanks Fasterhorses but the wrench I have is in in #'s and this conversion is not the problem! What I need to know is the compensation setting when using the universal joint. The instructions with the torque wrench have a formula for the difference when the extension is not perpendicular to the head of the wrench. I think the # to use is the offset distance verses the length of the handle of the wrench... Let me see if I can get it typed in with the right format for the algebraic formula!

E-Effective length of extinsion - measured along the centerline of the torque wrench

L - Lever length of the wrench - center of grip to center of drive

T(w) Torque set on wrench

T(E) Torque applied by the extension to the fastener


T(W) = T(E) x L / L+E T(E) = T(W) x L+E / L

x is times / is divide by

There is a diagram that makes it look as if the E is only the distance from the center line of the wrench head to what would be the center of the plug in this case!

00000========O------o

|------ L-----|---E--|

00000========O--|
|
|- -o
Kinda looks like this (hope the format holds) but apparently whether the distance is in a strait line or offset I should use the strait line offset only this is where I am confused. It is, from the instructions, a different presure from the wrench to the plug under these circumstances...


Anyone make any sense of this Many Thanks
Craig

 

In the little digram I did you are looking at the wrench and the plug one in a strait line the other drops down or offsets but E apears to still be measured the same in both!!!!
Many Thanks
Craig

 

The formula doesn't change when using an extension that runs perpendicular to the torque wrench, which means you just use it as you normally would. If you were to add extensions that extended the overall length of the torque wrench, you would need to apply the formula you posted. I'm not aware of any formula for using a universal adapter because there's no way of knowing what your angle will be, and the fact your angle will likely change as you turn the wrench. I was taught to never use a universal adapter with a torque wrench.

 

I find it hard to believe that there is any significant torque loss through the universal joint. If you've got 100 ft*lbs or torque on one side of a universal joint, I highly doubt you have any significant difference on the other side.

And, as was stated above, there is no torque loss whatsoever through an extension. If the torque is greater on one end than the other, something would have to be moving.

Oh, and the official Ford spec per the 99 service manual is 84-180 in*lbs. (10-20 N*m)

-Joe

-Joe

 

Yes I would agree that in most circumstances a little off center variation would not make much difference! Of course my concern is the way these 5.4's tend to blow plugs out I want everything to be right as can be The instructions definitely give the formula (which I had never even thought about first torque wrench and all) and the 3 and 4 plugs are going to need a good bit of offset looks like to me!! I am going to put a tape to it and use some real #'s and see just what kind of difference it does make. That is of course after I find my 5/16 socket that fell last night but of course did not fall all the way through the engine compartment

 

I can certainly understand the concern about the torque on these plugs.... I thought that torque was transmitted 100% around the bend in a universal-joint though.... need to dig out my mechanical power transmission book....

-Joe

 

FasterHorses... Why did you say do no use any anti-seize on the threads of the plugs...? I thought this was common practice when installing new plugs into your engine?

 

I think the instructions are talking about using a torque multiplier and not a universal joint. Just use the wrench as you normally would.

 

Yeh I am not sure what they were talking about its why I ask!!! Along this same line I'll bite what is a torque multiplyer?
Many Thanks
Craig

 

Dude, I was a machinist mate. Now check this out: Use whatever is necessary to get the plug in and go to 132 in-lbs. That is right in the middle of the torque spec. By using a universal and/or an extension, you do change the output torque value. But that is generally for much higher torquings than you require. That's because there is flex in every extension (radial flex). However, at low torques, there isn't much, if any, flex. The loss across a typical universal is around 3% (they tend to flex much more than an extension). That is 3.96 in-lbs at 132in-lbs of torque. Well within spec. So install your plugs, use whatever you need to get to them, and always use a number in the middle of the curve. That's why it's a range...

 

True, the DISPLACEMENT on one end of an extension will be higher if the extension flexes, but the TORQUE will be the same. It's one of the first things they teach us in the core engineering curriculum, as well as mechanical power transmission class for my apprenticeship: The resisting torque on the other end of a shaft will always be equal to the input torque. If they're not, the shaft will move. If you apply 100 ft*lbs of torque to a rubber shaft, sure it may move 1/2 turn, but on the other end of the shaft, it'll be transmitting 100 ft*lbs as well. For every action, there is an equal and opposite reaction, and if the action is 100 ft*lbs of torque, the reaction must also be 100 ft*lbs of torque. Anything less than that and the shaft will move.

Universals get complex.... that's why I have to check the books... the speed of the second shaft varies with the angular position of the joint. If the speeds vary, so will the torque (something has to provide the change in acceleration). That's not a linear relationship though, and it's been years since I took that class, so it's all a lot fuzzy....

 

Run the plug down by hand until it bottoms out, then go another 1/8 to 1/4 turn. Forget all that torque crap, I'll pay good money to see someone get a torque wrench into cylinders 3, 4, 7, and 8 without breaking something (including theirselves).

 

Yeh Quintin I got to agree a bunch of Newtons laws got to apply here just like the 5/16 socket I dropped and have yet to find or how when ya need a flat blade there are only phillips screwdrivers in the box... put it down and reach for a phillips and there are the flat blades but some how the phillips are all gone anyone know what I mean