f150 dana hp 44

[sykanr0ng]

ive got 3.5 running a custom 2" front driveshaft to clear pan.

Thanks, and I apologize in advance for more questions but it's valuable info to me.
Do you think it would work with a 2.5"?
Is your custom shaft a skinny shaft?
Is your axle narrowed?

My quick mental math figures the pinion would move about 1.5" closer to the center on a narrowed axle versus a full width axle.

With a 2 1/2 inch lift it would be another inch closer than with his 3 1/2 inch lift.
Which means the radius of the shaft would need to be an inch smaller to clear.
The radius is half the diameter.
So the shaft diameter would need to be two inches smaller.
So a zero inch shaft.

 

[garberz]

....My quick mental math figures the pinion would move about 1.5" closer to the center on a narrowed axle versus a full width axle.

I'm not sure how you arrived at that figure, but a properly narrowed HP44, using EB length axles, will put the pinion in the same location as a stock LP 44. Even if you just shortened the long side .597", the HP short side increases the total axle width .600". When centered up, it will only move the pinion just over a 1/4" closer to the oil pan. Shortening both sides allows you to optimize the pinion angle and run both Bronco length axles.

Mark

 

[Wyflyer]

Gotcha! When I was doing that mental may it was versus installing a full width with no changing of the wedges or radius arms.
It would correct to say " a full width axle would put the pinion shaft 1 1/2" wider than a narrowed pinion shaft.
In any event I'm getting my head around this finally. Mine has the cast wedges, and I could live with a 3 1/2" lift if it'll clear the explorer 5.0 pan and 4R70W pan

 

[garberz]

Mine has cast wedges also, I have 3 1/2" coils, seems to clear my Explorer pan from my measurements. But my HP is still on the bench.

Mark

 

[mattt]

With a 2 1/2 inch lift it would be another inch closer than with his 3 1/2 inch lift.
Which means the radius of the shaft would need to be an inch smaller to clear.
The radius is half the diameter.
So the shaft diameter would need to be two inches smaller.
So a zero inch shaft.

If the driveshaft is 1" closer with the shorter lift, then why are you doubling the distance required for clearance. If the driveshaft is 1" closer, you need 1" additonal clearance, not double that....right?

There really is no set in stone answer to this question without mocking it up and seeing what your EB would have clearance wise with the drivetrain you're using. A good, smaller diameter front driveshaft to use for situations like this is from an 80's-90's XJ. It will be too long out of the box, but driveshafts are easy to have shortened...not so easy to lengthen. That shaft is 2" diameter, & has the correct 1310 CV on the case end and 1310 joint on the axle end. I used one with excellent results in my AOD'd Bronco.

 

[sykanr0ng]

If the driveshaft is 1" closer with the shorter lift, then why are you doubling the distance required for clearance. If the driveshaft is 1" closer, you need 1" additonal clearance, not double that....right?

There really is no set in stone answer to this question without mocking it up and seeing what your EB would have clearance wise with the drivetrain you're using. A good, smaller diameter front driveshaft to use for situations like this is from an 80's-90's XJ. It will be too long out of the box, but driveshafts are easy to have shortened...not so easy to lengthen. That shaft is 2" diameter, & has the correct 1310 CV on the case end and 1310 joint on the axle end. I used one with excellent results in my AOD'd Bronco.

Because that one inch is to the middle of the driveshaft, the other side then rotates around and needs to be removed too.

 

[mattt]

How do you figure that one inch is to the center of the driveshaft? When you say the other side rotates around....where did the other side rotate to...think about that one? It rotates to where your "other side" rotated from. You are stacking dimensions there by using radius dimension instead of diameter.

 

[sykanr0ng]

How do you figure that one inch is to the center of the driveshaft? When you say the other side rotates around....where did the other side rotate to...think about that one? It rotates to where your "other side" rotated from. You are stacking dimensions there by using radius dimension instead of diameter.

The same way a lathe rotates a part around and peels metal off of both sides.

 

[mavereq]

The same way that putting on 2" taller tires only gives you 1" more clearance under your axle

 

[mattt]

With a 2 1/2 inch lift it would be another inch closer than with his 3 1/2 inch lift.
Which means the radius of the shaft would need to be an inch smaller to clear.
The radius is half the diameter.
So the shaft diameter would need to be two inches smaller.
So a zero inch shaft.

Where exactly would the driveshaft be 1" closer? I ask that because as I think about it more....it depends whereyou are measuring driveshaft clearance? Are you measuring at the axle yoke or next to the Trans? Since the shaft is at an angle you gain less as it gets closer to the trans.

 

[mattt]

The same way that putting on 2" taller tires only gives you 1" more clearance under your axle

Correct, that works for tires, where half the tire is above the axle center line, and the other half is below the axle center line. Not the same with lift.