About primary locking tools

[thirdjld]

damn good info

[JohnK]

impact wrench - don't you risk knocking the crank out of alignment?

[FBobPilot]

I think Baker needs to make a primary locking bar that works better with their Attitude Adjuster, might grind down mine in the middle to help with the teeth engagement.

[Karl H.]

impact wrench - don't you risk knocking the crank out of alignment?

The service manual (TC88) says:
When LOCTITE cannot be broken under normal conditions with conventional methods, apply moderate heat or use an air impact wrench ONLY TO LOOSEN compensating sprocket nut.

[runamuck]

my no frills locking bar from a leftover piece of straight metal has done the trick several times. no need to shop for an "official tool"

[rbabos]

I think Baker needs to make a primary locking bar that works better with their Attitude Adjuster, might grind down mine in the middle to help with the teeth engagement.

In a case like that, using the comp's hex and a wrench is the best option. Torques values end up the same whether using the hex or a locking bar.
Ron

[rbabos]

my no frills locking bar from a leftover piece of straight metal has done the trick several times. no need to shop for an "official tool"

Let's face it, they all start out as a 20' length from a metal supply oufit. A couple of cuts and 1000% markup, instant tool.
Ron

[clawdog60]

 

Just got to be careful not to loosen up them magnets on the rotor.. I've taken off about a dozen comp sprockets with an air impact and haven't had a problem YET. 

I've always heard that there is a possibility that you can loosen the magnets too. I think the key is a lot of air pressure, and heavy duty air impact. You know, just rip the nut off, rather than clanking away at it. I haven't loosened any magnets yet either.

[clawdog60]

The key is not using a cheap ass impact and beating the hell out of everything. A good impact will pull the comp quickly and you can't even feel the impact on the rotor.

[clawdog60]

Nice force illustration Karl.

[JimB]

Impressive math there.... when I replaced the comp on my 07 SG, I got a piece of oak 1 x 4" cut it to fit inbetween the teeth, wedge it in there & used a long breaker bar.
Everything worked just fine. Oh... I did remove the adjusted first.

[lqqk_out]

Where is the force illustration? I'm not seeing it.

[Karl H.]

These pics were originally included into my first post but got lost:

[attach=0]

[attach=1]

[FSG]

 

[lqqk_out]

Thank you! 

[kd]

There's a sticky in the tool section with the dimensions of the bar reqquired for each model  I have been meaning to add the one I made for my 49 tooth EVO Industries basket.

https://harleytechtalk.com/htt/index.php/topic,1086.0.html

[lqqk_out]

I'm trying to understand the equations. Can you explain where the distance constants 50 & 25, which I'm guessing are in the units of meters comes from?
I'm guessing it might have something to do with a locking bar length being about 25 cm?

[lqqk_out]

Result: Bending forces (FS) at tranny shaft (and crank) are more then ten times (!) higher using the wedge. Simply spoken the wedge method results in a gun tackle effect.

Example: Torquing down the comp nut to spec with 165 ft-lb (TC88) results in a bending force on tranny shaft of about 16,500 lb using the wedge. But breaking the nut (red loctite!) with lets say three times the spec. torque results in a force on tranny shaft of about 50,000 lb

I'm not following this and empirically it doesn't make any sense to me. If we are to assume a gun tackle effect, wouldn't that give us twice the mechanical advantage? So, 165 ft-lbs results in 330 lbs?
I'm just not seeing how 165 ft-lbs of force can result in 16,500 lbs on the tranny shaft. That's over 8 tons of force! That's much more than your average bottle jack.

[Karl H.]

The purpose of this calculation was to show that the wedge method applies significantly more load to the primary shaft. The result of the comparison doesn'n depend on the very exact dimensions.

The ,,25" results from 1/r = 1/40mm = 1/0,04m = 25 [1/m]

The ,,50" results from 1/d = 1/20mm = 1/0,02m= 50 [1/m]

Karl

[Karl H.]

...If we are to assume a gun tackle effect, wouldn't that give us twice the mechanical advantage? So, 165 ft-lbs results in 330 lbs?...

Right! If the clutch sprocket rotates freely the load on the shaft is two times the force in one chain strand: Fs = 2 x F

[a_disalvo]

Anyone have a picture or diagram of the bars used? Something we can make or is it store bought? Thanks, Frank

[FXDBI]

Anyone have a picture or diagram of the bars used? Something we can make or is it store bought? Thanks, Frank

http://harleytechtalk.com/htt/index.php/topic,1086.msg1321324.html#new

[Norton Commando]

Anyone have a picture or diagram of the bars used? Something we can make or is it store bought? Thanks, Frank

Here's my home-made lock bar to suit a 2012 FLHTK.  It's made from mild steel bar 1/4" thick by 1" wide by 5-15/16" long.  It performed perfectly when I changed out my compensator. [attach=0]

[Norton Commando]

Nice force diagram there Karl.

However, to find the force component using trigonometry you need to use a right triangle, and not an acute scalene triangle as shown in the diagram.

Jason

[Admiral Akbar]

Nice force diagram there Karl.

However, to find the force component using trigonometry you need to use a right triangle, and not an acute scalene triangle as shown in the diagram.

Jason

Ignore the vertical line.  It's throwing you off. Draw the triangle.