should I use my parking brake?

[Frenchy]

And in the US you have emissions system warranties that I think are federally required to 8yr/80k miles.

Not sure what that has to do with the price of eggs in China, but ok

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[Motorpsychology]

Not sure what that has to do with the price of eggs in China, but ok

The price of a dozen large eggs is about the same in Beijing as it is in much of the US.

 

[ControlNode]

The pawl holds the truck just fine on level to moderate grades ~2% estimated but payload increases the stress on the pawl, and more likelihood of torque lock. Everything is usually just fine, Stellantis had a problem a few years back with certain Rams not engaging Park, but IIRC that was more software related.

I appreciate all the dazzling mathematics, but my eye bones just don't trust this thing:

The math is not dazzling and does not support the idea of trusting that parking pawl on any incline. I always use my parking/emergency brake, but if that happens to start failing, I hope the parking pawl does its job to stop too much movement, or at least enough that with any remaining friction the brakes are providing when combined with the pawl are enough.


Looking more into those numbers though....
I found these numbers and hope they are right:
Sprag wheel radius: 87.35mm
rear axle to ground: ~380mm (387mm from tire spec and giving 7mm of deflection to get to round number)

I'd guess the load after the gearing on the parking pawl should be about ~791 lb assuming 697 lb number from @Dgc333 was right.

Bard says that 30-degree incline would require a 2200lb force to prevent the truck from rolling down the hill. (No friction in the drive-line and tires was not considered) Using those numbers, the force on that pawl tooth would be about 2,565 lb.

Bard says the formula is T = W * sin(θ)
T = tension (lb) needed on a rope holding from rolling down incline
W = weight (lb) of ojbect
θ = angle of incline

 

[ControlNode]

Not sure what that has to do with the price of eggs in China, but ok

You stated the only coverage after bumper to bumper is powertrain, which for most members here is incorrect. If you are saying my mention of emissions warranty is irrelevant because OPs post was for brake parts the OP is having an issue with, so was any mention of the powertrain warranty by others.

 

[CB750F]

Can't help it....

PARKING BRAKE SHOULD ONLY BE USED WHEN DRIFTING.

 

[Frenchy]

You stated the only coverage after bumper to bumper is powertrain, which for most members here is incorrect. If you are saying my mention of emissions warranty is irrelevant because OPs post was for brake parts the OP is having an issue with, so was any mention of the powertrain warranty by others.

And how does that apply to the parking brake?

 

[ControlNode]

And how does that apply to the parking brake?

It's not, but neither is the power train, but yet it was brought up as the only remaining warranty after the bumper to bumper, which was incorrect information, so I was adding additional (correcting for most users here) post in reply to that post, not the OPs post.

Just like all the debate about the parking pawl and such, really not related to the brake issue the OP is having or the really poor advice they got from that tech.

 

[ControlNode]

Can't help it....

PARKING BRAKE SHOULD ONLY BE USED WHEN DRIFTING.

If only this would fit in the Ranger, LOL.

 

[Motorpsychology]

The math is not dazzling and does not support the idea of trusting that parking pawl on any incline. I always use my parking/emergency brake, but if that happens to start failing, I hope the parking pawl does its job to stop too much movement, or at least enough that with any remaining friction the brakes are providing when combined with the pawl are enough.


Looking more into those numbers though....
I found these numbers and hope they are right:
Sprag wheel radius: 87.35mm
rear axle to ground: ~380mm (387mm from tire spec and giving 7mm of deflection to get to round number)

I'd guess the load after the gearing on the parking pawl should be about ~791 lb assuming 697 lb number from @Dgc333 was right.

Bard says that 30-degree incline would require a 2200lb force to prevent the truck from rolling down the hill. (No friction in the drive-line and tires was not considered) Using those numbers, the force on that pawl tooth would be about 2,565 lb.

Bard says the formula is T = W * sin(θ)
T = tension (lb) needed on a rope holding from rolling down incline
W = weight (lb) of ojbect
θ = angle of incline

Somehow wouldn't the axle ratio, 3.73:1, figure into the equation? I visualize the truck on an incline and in Park, the force on the sprag wheel has to be multiplied by a factor of 3.73 somewhere, doesn't it?

 

[Frenchy]

Somehow wouldn't the axle ratio, 3.73:1, figure into the equation? I visualize the truck on an incline and in Park, the force on the sprag wheel has to be multiplied by a factor of 3.73 somewhere, doesn't it?

To an extent yes. That said a certain someone on here is going to think it won't be a problem and yet I have seen them break due to high load from the inclines.

 

[ControlNode]

Somehow wouldn't the axle ratio, 3.73:1, figure into the equation? I visualize the truck on an incline and in Park, the force on the sprag wheel has to be multiplied by a factor of 3.73 somewhere, doesn't it?

Yes, and it was in my math. Load put on the wheel would be divided by the 3.73 to get that load on the prop shaft.

The full formula I used was more like this...
W = vehicle weight (lb)
θ = angle of incline (degrees)
G = rear axle gear (ratio)
T = estimated tire radius (mm)
S = estimated parking sprag radius (mm)
L = Load on sprag tooth (lb)

L=(W*sin(θ))/G*(T/S)

2566≈(4400*sin(30))/3.73*(380/87.35)

 

[Motorpsychology]

Yes, and it was in my math. Load put on the wheel would be divided by the 3.73 to get that load on the prop shaft.

The full formula I used was more like this...
W = vehicle weight (lb)
θ = angle of incline (degrees)
G = rear axle gear (ratio)
T = estimated tire radius (mm)
S = estimated parking sprag radius (mm)
L = Load on sprag tooth (lb)

L=(W*sin(θ))/G*(T/S)

2566≈(4400*sin(30))/3.73*(380/87.35)

Thanks for the explanation, John. It's good to know that my own anecdata are confirmed:

• A: When the load (L) exceeds the friction between the surfaces of the pawl, the force upon the shifter mechanism (which is reduced by the number of moving parts and angles), is insufficient to disengage the pawl from the sprag wheel.
• (see post #22) A call to a friend's dad to drive 4mi/6.80 km to push me 0.55 inches/1.39cm in order to decrease (L) sufficiently to move the pawl out of the sprag wheel resulted in a freed car, but increased exponentially the odds against asking my buddys sister out again.
• While the math is dazzling, the whole park setup in any automatic transmission looks wimpy.

 

[got3fords]

Can't help it....

PARKING BRAKE SHOULD ONLY BE USED WHEN DRIFTING.

Or spinning around a snow-covered parking lot.

 

[got3fords]

Thanks for the explanation, John. It's good to know that my own anecdata are confirmed:

• A: When the load (L) exceeds the friction between the surfaces of the pawl, the force upon the shifter mechanism (which is reduced by the number of moving parts and angles), is insufficient to disengage the pawl from the sprag wheel.
• (see post #22) A call to a friend's dad to drive 4mi/6.80 km to push me 0.55 inches/1.39cm in order to decrease (L) sufficiently to move the pawl out of the sprag wheel resulted in a freed car, but increased exponentially the odds against asking my buddys sister out again.
• While the math is dazzling, the whole park setup in any automatic transmission looks wimpy.

Y'all need to stop with all this math.
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[Motorpsychology]

Y'all need to stop with all this math.
<iframe src="https://giphy.com/embed/9JjnmOwXxOmLC" width="480" height="289" frameBorder="0" class="giphy-embed" allowFullScreen></iframe><p><a href="">via GIPHY</a></p>

Your'e right, James, just pull the lever up or don't!

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