'05+ Ford Super Duty Dana 60 Axle Tech & Info

[Sterlingfire]

I ahve used the artec, it went together great. I added an extra plate to the outside that goes from the top, all the way to the bottom of the stock steering arm, all welded in. To break of the knuckle, it would have to shear the whole side, then pull the plate over the arm. The artec kit also clears the busted knuckle brake kit as well. I used the crossover kit as it allows you to play with the mountinig locations of the tie rod and ram.

One this to note, you have to push the tie rod forward if you want to clear the diff. all of the arms shown do that. Also to get 50* you need a 10" ram. I would reccomend doing 1550 shafts at the same time.

Even with my weaver arms I'm looking to double shear. I have in mind almost exactly what you're describing to help hold the knuckle/tie rod together in case of a bad hit.

You want 50° from an 8" ram? Mount the tie rod at about 4.75" from the centerline of the ball joint...
just an FYI, you can get less than 40°of steering out of a 10" ram if it's mounted to the wrong spot on the knuckle. The length of ram matters if it's connected to the knuckle in the right spot.
Also, if you're using the ram as the tie rod, it's irrelevant where your tie rod holes are on the knuckle considering you can offset the ram to clear the cover.

 

[Bebop]

50 degrees with a 8" ram will require very short arms which loses a lot of force.

10" ram is a much better option.

3"x9 or 11 is the definite answer for crawling.

 

[Sterlingfire]

50 degrees with a 8" ram will require very short arms which loses a lot of force.

10" ram is a much better option.

3"x9 or 11 is the definite answer for crawling.

My point was not to use the 8" ram, my point was that the amount of stroke you have can be used instead of destroking the ram by mounting the tie rod in the right spot on the knuckle... with a 10" ram to get 50° steering, you need to mount the tie rod at about 6.75" from the CL of the ball joint.

 

[FLUX]

Sorry if this has been discussed and I missed it.

Has anyone cut the axle tubes and measured the wall thickness? Everyone seems to agree that they are .5" thick. Some local guys that "re-tube" them for mud trucks are saying they are .35" thick.

 

[CDA 455 II]

Sorry if this has been discussed and I missed it.

Has anyone cut the axle tubes and measured the wall thickness? Everyone seems to agree that they are .5" thick. Some local guys that "re-tube" them for mud trucks are saying they are .35" thick.

According to the first post:

Axle Specs:
-Tubes: 3.75" OD, 1/2" wall
-WMS to WMS: 72"
-Ring Gear: 9.75" (a 10" can be sourced from Ford in limited ratios)
-Lug Pattern: 8x170
-Wheel Studs: M14x1.50
-Weight: Approximately 750lbs loaded with brakes
-Axle shafts: 35 spline
-U-joints: 1480
-Tone ring: 60 tooth

 

[CDA 455 II]

Sorry if this has been discussed and I missed it.

Has anyone cut the axle tubes and measured the wall thickness? Everyone seems to agree that they are .5" thick. Some local guys that "re-tube" them for mud trucks are saying they are .35" thick.

Does this pic answer your question?

 

[Bebop]

They are not 1/2 wall that's for sure.

 

[HYDRODYNAMIC]

To clarify as years and models are creating a mix up.
1999-2004 F250 3.5" tube .35" wall metric
1999-2004 F550 3.5" tube .44" wall metric
2005-2010 F250 3.75" tube .35" wall metric
2005-2010 F550 3.75" tube .44" wall metric

 

[Mattygwheeler]

I am building a 05+ d60 for my gmt800 SAS build, what Truss kit is everyone using? I plan a 3 or 4 link(with panhard bar) with 10" coilovers or progressive coil springs. I pretty much have settled on the barnes low pro truss unless there is a better option. With the price of steel in canadistan right now it is cheaper to buy a kit than to build something.

PRO SERIES Ford Superduty Dana 60 Low Profile Truss

The 1999-present Ford Superduty Dana 60 Truss is ideal for mounting link bracketry to your axle while also adding unsurpassed durability. This kit does require trimming of casting and removal of all factory brackets and is made with a 5/16" top. Additional grinding or trimming may be required...

barnes4wd.com

 

[Sterlingfire]

I am building a 05+ d60 for my gmt800 SAS build, what Truss kit is everyone using? I plan a 3 or 4 link(with panhard bar) with 10" coilovers or progressive coil springs. I pretty much have settled on the barnes low pro truss unless there is a better option. With the price of steel in canadistan right now it is cheaper to buy a kit than to build something.

PRO SERIES Ford Superduty Dana 60 Low Profile Truss

The 1999-present Ford Superduty Dana 60 Truss is ideal for mounting link bracketry to your axle while also adding unsurpassed durability. This kit does require trimming of casting and removal of all factory brackets and is made with a 5/16" top. Additional grinding or trimming may be required...

barnes4wd.com

That's the one I'm using. Fits well and plenty strong. Best bang for your buck IMO

 

['84 Bronco II]

To clarify as years and models are creating a mix up.
1999-2004 F250 3.5" tube .35" wall metric
1999-2004 F550 3.5" tube .44" wall metric
2005-2010 F250 3.75" tube .35" wall metric
2005-2010 F550 3.75" tube .44" wall metric

Are you sure about the tube diameter on the '05+ F450/550 axles? Everything I read in the past was that they were 4" diameter and 1/2" wall thickness, but I have never personally measured one.

I'd be really curious to learn more about the '05+ F450/F550 outers particularly for rear steer conversions and building custom axles using the Super Duty outers. The wider ball joint spread that accommodates the big bell RCVs without any fuss and bigger ball joints are very appealing. I've read Bebop talk about the brakes being an issue, but I am not really sure what the challenge there is if you don't care about the weight? Do 17" wheels not fit over them? Not the end of the world since 20" wheels are pretty popular with most sticky 42/43" tire options.

RDY2ROK I think mentioned running F450/550 outers for his rear steer conversion on The Grape and I know he is running the Traction bead locks that only come in a 17" wheel. It would be cool if he'd chime in with his brake setup and some pictures.

 

[Bebop]

I've read Bebop talk about the brakes being an issue, but I am not really sure what the challenge there is if you don't care about the weight? Do 17" wheels not fit over them? Not the end of the world since 20" wheels are pretty popular with most sticky 42/43" tire options.

The 10 lug and 8 lug rotors are different.
Caliper mounts are different.

There are no problems if you build a brake setup, but the factory 450/550 brakes just don't work with a 8 lug UB.

 

['84 Bronco II]

The 10 lug and 8 lug rotors are different.
Caliper mounts are different.

There are no problems if you build a brake setup, but the factory 450/550 brakes just don't work with a 8 lug UB.

Are the 10 and 8 lugs on the same bolt circle diameter then?

EDIT: I just looked it up and the 10 lug pattern is on a 225mm circle vs. 170 mm for 8 lug, so I am guessing the 8 lug unit bearing moves the rotor relative to the caliper then? I thought the unit bearings were the same other than the lug pattern

 

[CDA 455 II]

Can I safely assume these are 170mm?

And the outer lugs are the 225mm?

 

['84 Bronco II]

Can I safely assume these are 170mm?

And the outer lugs are the 225mm?

D'oh!

I didn't think about spacer.

I forgot there was CAD for the F450/550 axle, do you know where the file is? That would help with a lot of my questions.

 

[Grendel]

Are the 10 and 8 lugs on the same bolt circle diameter then?

EDIT: I just looked it up and the 10 lug pattern is on a 225mm circle vs. 170 mm for 8 lug, so I am guessing the 8 lug unit bearing moves the rotor relative to the caliper then? I thought the unit bearings were the same other than the lug pattern

F450/550 unit bearings are 10 bolt outer, to take the wheel adapter. It's the factory wheel adapter in front that goes to 225MM

You can swap in F350 8 lug wheel bearings - still need to figure out brakes, since the caliper abutment bolt spread is MUCH larger. Re-drilling the big rotor may be an option.

F350 dually wheel bearings are course lug, F350 SRW is fine thread.

 

[CDA 455 II]

D'oh!

I didn't think about spacer.

I forgot there was CAD for the F450/550 axle, do you know where the file is? That would help with a lot of my questions.

That pic I found on the internet.

 

['84 Bronco II]

Re-drilling the big rotor may be an option.

This is what I am getting at, but if they are both on a 170mm pattern you are going to have interference with multiple holes no matter how you clock the 8 lug pattern. The best fit I found was lining up two of the 8 lug holes with one of the 10 lug holes. You mught be able to weld plugs in the 4 interfering holes, or maybe for a relatively light crawler, 4 holes at 170mm is enough. I wonder if blank rotors are an option? Probably not likely, but I think I have seen blank rotors for Ford 8.8 and 9" applications.

 

[arse_sidewards]

Drill both patterns and tack weld the "problem" studs to the hub?

I'd be more concerned about the studs spinning or causing problems than the wheel mounting surface not being strong enough.

 

[YeeP]

You want 50° from an 8" ram? Mount the tie rod at about 4.75" from the centerline of the ball joint...

I know this was posted a little while back, but would you care to share how you calculated this? It seems like it may be simple trig, but my mind gets stuck on the arc that the end of the tie rod travels in, vs just a straight line calculation. When I built my steering knuckles, I was trying to come up with a way I could have these numbers.

Thank you

 

[pennsylvaniaboy]

 

[CDA 455 II]

 

['84 Bronco II]

Drill both patterns and tack weld the "problem" studs to the hub?

I'd be more concerned about the studs spinning or causing problems than the wheel mounting surface not being strong enough.

I was talking about drilling F450/550 rotors so you can use all the factory F450/550 brake parts with F250/350 unit bearings (which off road-friendly wheels are readily available for) and avoid using aftermarket brake parts which are mucho dinero.

Alternatively, perhaps if the caliper mounting ears are far enough apart on the F450/550 knuckles, you could fab a caliper adapter bracket similar to the one pictured below and just run all F250/350 brake parts to match the unit bearings.

I know this was posted a little while back, but would you care to share how you calculated this? It seems like it may be simple trig, but my mind gets stuck on the arc that the end of the tie rod travels in, vs just a straight line calculation. When I built my steering knuckles, I was trying to come up with a way I could have these numbers.

Thank you

The formula Sterling fire was using is: (Arm Length from pivot axis) = (1/2 Ram Stroke)/Tan(Desired Steering Angle)

Although he multiplied when he should have divided, and the arm length would need to be 3.36" to get 50° with an 8" ram. This formula does not take into account the king pin inclination or the fact that the angle of your tie rods is changing throughout the steering stroke, but it is close enough for most purposes.

 

[Sterlingfire]

I guessed

I know this was posted a little while back, but would you care to share how you calculated this? It seems like it may be simple trig, but my mind gets stuck on the arc that the end of the tie rod travels in, vs just a straight line calculation. When I built my steering knuckles, I was trying to come up with a way I could have these numbers.

Thank you

But, it's pretty simple. I took what i know about finding the sine/ tangent of a 45°
8÷1.414=5.657
Since it's an isosceles right triangle, and we know the hypotenuse is 8", we know both the sine and tangent are equal...5.657 for both legs on the right angle, being 90°.
You can check the math by doing the opposite.
5.657x1.414=~8
If you wanted to use a 10" ram, you would need a hole at very close to 7-1/16"
10÷1.414=7.06
Most people go with the stock tre position, and need a 9" stroke ram, because the stock tre position is about 6-3/8" from center of pivot... so
6.375x1.414=~9
Obviously this is for only 45° of steering lock to lock. That's why i guessed at the 50° steering number i gave.

 

[Sterlingfire]

Derp, i see '84 Bronco II answered above. My guess was a little off after all
My number will give you more than 45 but less than 50...i don't think going to 3.xx" is very possible, but I'm also not sure how long u joints or even cv's will last going a full 50° anyway. Way better off going with a 10" ram IMO

 

['84 Bronco II]

Derp, i see '84 Bronco II answered above. My guess was a little off after all
My number will give you more than 45 but less than 50...i don't think going to 3.xx" is very possible, but I'm also not sure how long u joints or even cv's will last going a full 50° anyway. Way better off going with a 10" ram IMO

Now that I think about it, the Ackerman (or Anti-Ackerman) geometry in the knuckle throws a wrench into the calculations, and it is not as simple as what I posted which would represent the tie rod attachment point being at a 90° to the axis of the axle when pointed straight (i.e., no Ackerman effect).

 

[Sterlingfire]

Now that I think about it, the Ackerman (or Anti-Ackerman) geometry in the knuckle throws a wrench into the calculations, and it is not as simple as what I posted which would represent the tie rod attachment point being at a 90° to the axis of the axle when pointed straight (i.e., no Ackerman effect).

Yes, but i think you're right in that it is not enough to try to throw into the calcs. Your equation will get you close enough as long as you're not trying to go over 50° IMO. I'm not sure how you could even account for Ackerman though. It does add a bit of a loop in, especially if you want it properly set up...
My equation for figuring 45° steering is close enough to get you in the ballpark, and not likely disappointed by what you end up with just the same. Still not equating Ackerman though...

 

[Bebop]

Now that I think about it, the Ackerman (or Anti-Ackerman) geometry in the knuckle throws a wrench into the calculations, and it is not as simple as what I posted which would represent the tie rod attachment point being at a 90° to the axis of the axle when pointed straight (i.e., no Ackerman effect).

No shit

 

[YeeP]

The formula Sterling fire was using is: (Arm Length from pivot axis) = (1/2 Ram Stroke)/Tan(Desired Steering Angle)

Although he multiplied when he should have divided, and the arm length would need to be 3.36" to get 50° with an 8" ram. This formula does not take into account the king pin inclination or the fact that the angle of your tie rods is changing throughout the steering stroke, but it is close enough for most purposes.

Understood. The "king pin inclination", Im assuming is the caster angle. That being said this became increasingly obvious to me as being a necessary factor to consider when I was building the steering ram mount. The farther I steering the knuckle away from 0 degrees of turn, the higher it got when measuring height from the ground. I will take this info and see what I can come up with on my setup.

I guessed


But, it's pretty simple. I took what i know about finding the sine/ tangent of a 45°
8÷1.414=5.657
Since it's an isosceles right triangle, and we know the hypotenuse is 8", we know both the sine and tangent are equal...5.657 for both legs on the right angle, being 90°.
You can check the math by doing the opposite.
5.657x1.414=~8
If you wanted to use a 10" ram, you would need a hole at very close to 7-1/16"
10÷1.414=7.06
Most people go with the stock tre position, and need a 9" stroke ram, because the stock tre position is about 6-3/8" from center of pivot... so
6.375x1.414=~9
Obviously this is for only 45° of steering lock to lock. That's why i guessed at the 50° steering number i gave.

Just confirming my understanding here. So your assumption is that you know the guy had a ram capable of 8" stroke. So your calc begins with that ram fully extended, thereby this line becomes the hypotenuse of the triangle. The 45º angle in this scenario is the desired steering angle to achieve. The calculations would be a little more complicated if it had not been, only based on the fact that we know it's an isosceles triangle.

Your math looks good to me, thanks for the explanation, I wish I had thought of this before I pulled the trigger on the weld on brackets I got, but so far my rough measurements indicate I will be able to achieve 43º of steering with a 9" ram.

---------------------

Side note: JHF seems to be the only creator of weld on brackets that makes a statement about achievable steering angles. I asked multiple manufacturers when I was looking to by some why they didn't at least give the measurement from ball joint center, to rod end center, so it could be calculated.

This thread should include that data. I have an unused set of JHF arms in my garage as well as the ones I bought from Motobilt. I'll take some measurements tonight and add them here.

 

[Bebop]

Understood. The "king pin inclination", Im assuming is the caster angle.

Nope