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Old 02-18-2007, 02:16 PM   #1
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Pull Coil Timer Module on 5,9?


Regarding my Cummins 5.9, circa 1992; the Fuel Shutoff Solenoid seems to be
controlled thru a "Pull Coil Timer Module". What does it do?

It's a black box, about 2" square and 1/2" thick, with three wires, mounted near
the Fuel Solenoid.
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Old 02-19-2007, 06:11 PM   #2
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Aha. Sounds like there are two coils in the solenoid. A strong coil for pulling it
open to start, and a less powerful coil for holding it open while running. The "start"
coil draws a lot of amps and will eventually overheat if it stays on. Makes sense. Thanks!
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Old 02-19-2007, 10:01 PM   #3
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Does it need to "cam over" or something? I don't understand why it would take more force to open it initially versus holding it open. Does the fuel pressure help maybe? I've just never looked that closely at my solenoid. You must not have a "Pull to Shut Off" lever if you have a solenoid. I just turn my key off to kill my engine, but it appears many people must pull the lever. I haven't decided which system I like better. Mine is more "automotive" and convenient, but the other system can run indefinitely without a battery as long as the engine is running....
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Old 02-19-2007, 10:59 PM   #4
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Well, I don't know the details, but it make sense to me that to move the solenoid
plunger a distance requires more force than to hold it still in one spot. (The solenoid
is operated automatically by the ignition switch.)

I drove many of the good old fashioned Caterpillar mechanical 3406, and there was a
knob on the fuel pump that could be turned a quarter turn by hand and the engine
would run without electricity. Very handy, yes.
On my mechanical Cummins 5.9, the solenoid operates a linkage to the fuel pump, and
that linkage can be moved by hand if necessary -- even tied in the on position, I suppose.
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Old 02-19-2007, 11:17 PM   #5
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Now for my $.02

Spring force is calculated as (1/2) k*x^2 where k=the spring constant and x=the distance it is moved from its neutral free arch position (stretched or compressed, basically).

Let;s say that spring needs to move a total of 3 inches. That means that to hold it open it will require 9 times the force the spring puts on the closed valve. The "opening" solenoid will go through an exponential force increase as it travels from 0-3 inches of extension on the spring. The max force required will be at the point at which it is maxed...3 inches. A solenoid holding it there, but not moving it would still require the same amount of force to hold it open. Ahh....but this is assuming linear travel.

Perhaps (and most likely) the linkage travels around a path of a curve of sorts. In this fashion, when the spring is extended 3 inches, the linkage itself is now feeling both a vertical and horizontal component of force. The force derived from the (1/2) k * x^2 equation might no longer be acting entirely to try and close the solenoid. Instead it will be fighting a static force from the bearing or bushing or whatever is holding the pivoting arm. In this way the maximum force the solenoid must overcome would be somewhere in the process of opening while a lesser force may be required to hold it open. This is what I meant by camming over. It is a similar idea to how our air brakes work. When the pushrod is at a 90 degree, the brakes shoes are applying the maximum clamping force for a given air pressure in the chambers.

Darn you, Elliot! Now I want to go look at my solenoid.

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Old 02-20-2007, 12:14 AM   #6
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You're welcome.

Yes, camming over or some such. I'm no engineer; I just go out under the shade tree
and imagine stuff. Here, I imagine a big heave-ho, a click, and a lesser force required
to hold it there after the heave-ho is over. Like a weight lifter getting the barbell all the
way up to where his arms are straight.

And darn you; now I need to come up with some funny faux scientificated formula.
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Old 02-20-2007, 12:47 AM   #7
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Well...if it's the weightlifter theory, then the spring force would be divided into the horizontal and vertical components and the force required would be less.
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Old 02-20-2007, 02:14 PM   #8
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Quote:
Originally Posted by the_experience03
Now for my $.02

Spring force is calculated as (1/2) k*x^2 where k=the spring constant and x=the distance it is moved from its neutral free arch position (stretched or compressed, basically).

Let;s say that spring needs to move a total of 3 inches. That means that to hold it open it will require 9 times the force the spring puts on the closed valve. The "opening" solenoid will go through an exponential force increase as it travels from 0-3 inches of extension on the spring. The max force required will be at the point at which it is maxed...3 inches. A solenoid holding it there, but not moving it would still require the same amount of force to hold it open. Ahh....but this is assuming linear travel.

Perhaps (and most likely) the linkage travels around a path of a curve of sorts. In this fashion, when the spring is extended 3 inches, the linkage itself is now feeling both a vertical and horizontal component of force. The force derived from the (1/2) k * x^2 equation might no longer be acting entirely to try and close the solenoid. Instead it will be fighting a static force from the bearing or bushing or whatever is holding the pivoting arm. In this way the maximum force the solenoid must overcome would be somewhere in the process of opening while a lesser force may be required to hold it open. This is what I meant by camming over. It is a similar idea to how our air brakes work. When the pushrod is at a 90 degree, the brakes shoes are applying the maximum clamping force for a given air pressure in the chambers.

Darn you, Elliot! Now I want to go look at my solenoid.


Wow, you just gave me a headache
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Old 02-20-2007, 09:08 PM   #9
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Old 02-21-2007, 02:01 AM   #10
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