2008 Rear Diff Cover
I wanted the regular type to add my own, but NOBODY has it! I went three places before I found one that had 4 quarts of the RP. Two places had ONE.

But if you add too much modifier, you're clutches won't work at all and you'll have essentially an open diff. If you get chattering, add a little until it just stops and you should be good. I'm about to refill mine now and see how it does on corners.

Also got my oil change done, but as soon as I started filling it back up, it started raining. I never noticed how long it takes to put in 4 gallons of oil before. Nothing like a nice spring shower to make you take notice....
If I paint it I will use the high temp.
I spent about $20 on bolts & washers, but mine are stainless. The ones from DM aren't.
Ford Trucks for Ford Truck Enthusiasts
. Anyways, I added a little Ford modifier to mine cause it was grabbing around corners, not bad but noticable, put it in and its been fine for about 15,000 or so , good luck..
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When an object such as a diff cover is heated to a temperature Td by the hot oil inside each 1 ft^2 of the object radiates away heat ENERGY at a rate that's given by... R=(5.018x10^-10)(Ed)(Td+459.67)^4 Watts/ft^2 ...where Ed=Emissivity of the diff cover which is a ratio from 0 to 1 and Td=Temperature of the diff cover F.
For example if you've got a diff cover temperature Td=250 F and you assume Ed=1 the above equation gives... R=(5.018x10^-10)(Ed)(Td+459.67)^4=(5.018x10^-10)(1)(250+459.67)^4=(5.018x10^-10)(1)(709.67)^4=127.3 Watts/ft^2. If the surface area of the diff cover is 1 ft^2 then the cover radiates away 127.3 Watts. To put this into "RV heater" perspective 1 Watt=3.4 btu/hr so that 127.3 Watts=432.8 btu/hr and the standard electric heater that I use in my RV is 1,500 Watts which is 5,117 btu/hr.
The E=Emissivity is a physical property of the surface of an object and for Polished Aluminum E=0.04, for Black Anodized Aluminum E=0.88, and for Flat Black Paint E=0.91. So if you leave the diff cover shiny to look pretty it only radiates away (0.04)(127.3)=5.1 Watts but if you paint it with flat black paint it radiates away (0.91)(127.3)=115.8 Watts!
Well every little bit helps but it turns out that the additional 110 Watts of radiated heat that you gain by painting the cover black isn't all that much in the over all scheme of things because the 3 mechanisms for transferring heat ENERGY away from any object are radiation, conduction, and convection.
Since air is a good insulator conduction doesn't play much of a role and if it wasn't for the convective heat transfer to the air flowing across the diff cover, the diff cover temperature would increase to very high values of Td before the "(Td+459.67)^4 leveraging factor" in the "Radiation equation" would be large enough to let radiation carry away all of the heat that's generated inside the diff.
If you make an object black it not only radiates away a maximum amount of heat ENERGY but it also absorbs a maximum amount of heat ENERGY that might be radiated towards it from other object's in its vicinity. That's why an air box and turbo inlet tube should be bright and shiny or painted with a glossy white paint so that they reflect the heat ENERGY from the engine and exhaust manifolds.
Well in the case of the diff cover it's surrounded by ambient air with a temperature Ta and this means that a diff cover absorbs heat ENERGY that's radiated towards it from the surrounding air in an amount that's given by... A=(5.018x10^-10)(Ed)(Ta+459.67)^4 ...and if we assume an air temperature Ta=70 F and Ed=1 then A=(5.018x10^-10)(Ed)(Ta+459.67)^4=(5.018x10^-10)(1)(70+459.67)^4=(5.018x10^-10)(1)(529.67)^4=39.5 Watts.
This means that a black diff cover absorbs (0.91)(39.5)=35.9 Watts from its surroundings and it radiates away 115.8 Watts so that the net radiation from a black diff cover to its surroundings is 115.8-35.9=79.9 Watts.
A polished Aluminum diff cover absorbs (0.04)(39.5)=1.6 Watts from its surroundings and it radiates away 5.1 Watts so that the net radiation from a polished Aluminum diff cover to its surroundings is 5.1-1.6=3.5 Watts.
So now we're only talking about a 79.9-3.5=76.4 Watt difference between a polished Aluminum and a flat black treatment for a diff cover!
In general you can just combine the radiation R and the absorption A and write... Rn=R-A={(5.018x10^-10)(Ed)}{(Td+459.67)^4-(Ta+459.67)^4} Watts/ft^2 ...as the net radiation from the diff cover and to check this equation for a black diff cover with E=0.91 it gives... Rn={(5.018x10^-10)(Ed)}{(Td+459.67)^4-(Ta+459.67)^4}={(5.018x10^-10)(0.91)}{(250+459.67)^4-(70+459.67)^4}={(5.018x10^-10)(0.91)}{(709.67)^4-(529.67)^4}=79.9 Watts/ft^2 which for a 1 ft^2 cover is the same net 115.8-35.9=79.9 Watts that we got by calculating the R and A terms separately!
It's interesting to note that the emissivity of human skin is E=0.97 and that the surface area of the human body is about 20 ft^2 so that if you use a skin surface temperature Ts=90 F and an air temperature Ta=70 F and assume a human running around naked you get... Rn={(5.018x10^-10)(Es)}{(Ts+459.67)^4-(Ta+459.67)^4}={(5.018x10^-10)(0.97)}{(90+459.67)^4-(70+459.67)^4}={(5.018x10^-10)(0.97)}{(549.67)^4-(529.67)^4}=6.1 Watts/ft^2 ...and multiplying this by 20 ft^2 gives 122.4 Watts as the total net power radiated by a human under the assumed conditions which is more than the net 79.9 Watts radiated from a black diff cover!
I spent about $20 on bolts & washers, but mine are stainless. The ones from DM aren't.








