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Old 07-09-2005, 09:44 AM   #1 (permalink)
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Tech Week: Brakes and maths

Since I have the 4.6 sat on the bench and a drum-braked 109 sat there waiting for it, naturally my thoughts turn to braking.

Now I've been and looked at all the usual options - buy a di$c conver$ion kit, graft on RR discs, do nothing and die horribly, and now I'm onto looking for more original solutions.

The problem I've hit on is that I've no idea how to calculate the braking force required to stop two tonnes of vehicle on a 45 degree slope, or indeed from 70mph.

I can find physics stuff that will tell me how much energy is wasted in stopping a vehicle, how to calculate the mechnical advantage of the pedal / caliper etc. but nothing that calculates the torque required at the friction surface itself - annoying since this seems to be a commonly published spec from brake manufacturers
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Old 07-09-2005, 09:55 AM   #2 (permalink)
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Would it be torque, or pressure exerted against the pads?

Would one use the outer diameter of the swept area to calculate this torque, or the inner diameter of the swept area? Swept area (or surface area of the pad) * pressure exerted against the rotor=sq. in. * psi=pounds of force exerted?

Thinking on Saturday hurts!
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Old 07-09-2005, 10:10 AM   #3 (permalink)
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http://www.magtrol.com/tensioncontrol/bcselect.htm

Here is some help. Perhaps from this you can derive the correct formula.

(Google is my friend)
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Old 07-09-2005, 10:13 AM   #4 (permalink)
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You're right - thinking on Saturday is painful

This is typical of the sort of data (if any at all) that I'm turning up: http://www.knottuk.com/brakes4.htm#top
Quote:
Torques ranging from 300 Nm to 25000 Nm
(2600lb.in. to 220000 lb.in)
I guess it's the actual torque the unit can transmit (or resist) to the wheel.

Note: I'm not planning on using drums but this was the 1st example I could find.
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Old 07-09-2005, 10:13 AM   #5 (permalink)
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http://www.magtrol.com/tensioncontrol/bcselect.htm

Another one with some narrative. Even Pendy could understand this one.
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Old 07-09-2005, 10:14 AM   #6 (permalink)
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http://www.mech.uq.edu.au/courses/me.../brakes/s1.htm

My last contribution.
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Old 07-09-2005, 10:34 AM   #7 (permalink)
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How handy!

Throwing some numbers in, assuming a LR weighs 3000kg and is travelling at about 70mph:

Force = Mass * velocity = 3000kg * 30m/s = 90,000 N/m

Assuming a 14" (35cm) disc:

Torque = (Force * Distance) / 2 = (90,000 * 0.35) / 2 = 15,750 N/m

Assuming I've not made some terrible error there, it seems like quite a low number compared to some of the numbers kicking round on the web, but I guess some of them are for stopping 40-ton trucks.

I know this doesn't factor in the fact that there's more than one brake disc on the vehicle, but then 4x is a good safety margin IMHO, especially for something as critical as brakes.

Damn my brain hurts!
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Old 07-09-2005, 10:59 AM   #8 (permalink)
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Quote:
Originally Posted by fridgefreezer
How handy!

Throwing some numbers in, assuming a LR weighs 3000kg and is travelling at about 70mph:

Force = Mass * velocity = 3000kg * 30m/s = 90,000 N/m

Assuming a 14" (35cm) disc:

Torque = (Force * Distance) / 2 = (90,000 * 0.35) / 2 = 15,750 N/m

Assuming I've not made some terrible error there, it seems like quite a low number compared to some of the numbers kicking round on the web, but I guess some of them are for stopping 40-ton trucks.

I know this doesn't factor in the fact that there's more than one brake disc on the vehicle, but then 4x is a good safety margin IMHO, especially for something as critical as brakes.

Damn my brain hurts!
Only thing I want to say is 3000kg is a really loaded down Rover. I would think that a 109 would be more like 2000-2250kg (4500-5000lbs).
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Old 07-09-2005, 11:15 AM   #9 (permalink)
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so now you know the math what are you going to bolt on your truck to make it stop?
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Old 07-09-2005, 01:53 PM   #10 (permalink)
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Other than the mud/oil soak issue SIII 6 banger brakes are pretty effective.

For an obscene amount of money roversnorth sells a kit to bolt on D90/RRC style discs to a series axle. I would love to see someone replicate that with junkyard parts.
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Old 07-09-2005, 02:14 PM   #11 (permalink)
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Yeah that disc kit is pretty costly. I sourced some junkyard Range Rover axles to go on my Series for lots less than the kit...not hooked up the brakes yet.

Fridgefreezer, If you don't have a brake servo on that 109 you might try one of those too, It makes a difference.
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Old 07-09-2005, 04:32 PM   #12 (permalink)
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Quote:
Originally Posted by fridgefreezer
How handy!

Throwing some numbers in, assuming a LR weighs 3000kg and is travelling at about 70mph:

Force = Mass * velocity = 3000kg * 30m/s = 90,000 N/m

Assuming a 14" (35cm) disc:

Torque = (Force * Distance) / 2 = (90,000 * 0.35) / 2 = 15,750 N/m

Assuming I've not made some terrible error there, it seems like quite a low number compared to some of the numbers kicking round on the web, but I guess some of them are for stopping 40-ton trucks.

I know this doesn't factor in the fact that there's more than one brake disc on the vehicle, but then 4x is a good safety margin IMHO, especially for something as critical as brakes.

Damn my brain hurts!
Sorry, but your physics isn't good enough

Force = mass x accel

Torque = force x radius (for disc brake, use mean radius of brake pad).

Off the top of my head, I need to use SI units.

1. Find the braking force acting at the tyre contact point with the ground.

1.1 Find kenetic energy of vehicle when braking commences

given 3000kg at 30m/s
now kinetic energy = 0.5 mass x velocity squared
ie. KE = 0.5 x 3000kg x 30m/s x 30m/s

then KE = 1350000 Joules or 1350kJ

1.2 Find change in potential energy while braking

The stopping distance was not given - for this example I will use 50m down a 45 degree slope
Change in potential energy decending 50m down 45 deg slope

now change in PE = mass x gravity accel x change in height
ie. delta PE = 3000kg x 9.81m/s^2 x (50m x sin45)
then delta PE = 1040508 Joules (say 1041kJ)

1.3 Combine KE and change in PE to get total braking energy

ie. E = 1350kJ + 1041kJ
then E = 2391kJ

1.4 Find braking force to overcome KE and change in PE

Now Work = Force x distance
re-arrange to get Force = work done by brakes / stopping distance

this is equivalent to braking force = (KE + PE) / braking distance

ie. braking force = 2391kJ / 50m
then braking force = 47.82 kiloNewton (say 47.82kN)

1.5 Find braking force at tyres to resist weight of vehicle on slope

given 3000kg and 45 deg

ie braking force at tyres to hold on slope = 3000kg x 9.81m/s^2 x sin45
then braking force = 20810 Newton or 20.81kN

1.6 Find total braking force at tyres

Now total braking force at tyres is braking force to overcome KE and change in PE plus braking force to resist weight of vehicle on slope

ie total braking force = 47.82kN + 20.81kN

Then total braking force = 20.81kN

Now you have to determine how much braking should be proportioned between the front tyres and the rear tyres. This depends upon the height of the centre of gravity and the wheel base.

Use this proportion of the total braking force calculated above to determine the braking force at each front and rear tyre.

Once you have braking force at tyre:

then braking force at brake disc = braking force at tyre x tyre radius / mean radius of brake pad

Then divide by 2 because there are 2 pads (each side of disc)

Then divide by coefficient of friction force for brake pad material on brake disc to get the force required from the pistons on one side of the brake caliper.

Hope this helps.
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Old 07-09-2005, 04:45 PM   #13 (permalink)
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Speed only matters if you're worried about the heat (power) your brakes can dissipate.

What you really need is brakes more than capable of locking a wheel. Take a rover weight of 2.5 ton in a hard stop with about 70% of that transferred to the front axle. The total weight on the front axle is (0.7*2.5 = 1.75 ton)

A good friciton figure for the grip of a road tyre on asphalt would be about 0.8. So your front tyres have (0.8*1.75 = 1.4 ton) of grip on the road.

The torque to be able to lock the front wheels is the force (1.4 ton = 14,000 N) multiplied by the wheel radius (say 0.4m) (14,000*0.4 = 5,600 Nm).
Divide this by two and you get the torque for one front wheel (2,800 Nm)

Divide this by the radius of your brake drum (or distance from hub to piston centre on a disc brake, say 0.1m) and you get the force on the brake shoes or pads. (2800/0.1 = 28,000N = 2.8 ton)

Divide the force on the brake pads by the friction factor (0.3-0.5 range) and you get the actuation force needed. This last bit only works for disc brakes as drum brakes can be self energising (the shoes get pulled in harder by the drum rotation). (28,000/0.3 = 93,300N = 9.3 ton).

Divide the force by the area of your brake pistons (only count one in opposing pairs) and you get the line pressure needed. (front brakes 2*41.3mm pistons = 2,679 mm^2), (93,300/2,679 = 34.8 MPa = 5,200 PSI).

How you acheive that 5,000psi line pressure depends on what sort of booster setup you have.

I have put a lot of assumptions in here to simply the process, but unless I've made some big maths errors, this is basically the process.
The differences in reality will come from how high the centre of mass of your vehicle is (changes how much weight is transferred forwards in a stop), how big your wheels are, what pads you are running, how big your brake pistons are and how the front/back brake bias is.
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Old 07-09-2005, 07:34 PM   #14 (permalink)
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Quote:
Originally Posted by evilfij
For an obscene amount of money roversnorth sells a kit to bolt on D90/RRC style discs to a series axle. I would love to see someone replicate that with junkyard parts.
Or you can get just the pieces you need (hub and swivel housing) direct from BCB an source your own discs and calipers.






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Old 07-09-2005, 08:05 PM   #15 (permalink)
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Now how to replicate the hub and swivel with junkyard parts.
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Old 07-10-2005, 06:08 AM   #16 (permalink)
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Or you can get just the pieces you need (hub and swivel housing) direct from BCB an source your own discs and calipers.



-Jeff
Jeff,
I thought the hubs were standard coiler parts?
So the hubs and swivel housing are custom?
Does it use leaf or coil hub bearings?

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Old 07-10-2005, 10:09 AM   #17 (permalink)
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Jeff,
I thought the hubs were standard coiler parts?
So the hubs and swivel housing are custom?
Does it use leaf or coil hub bearings?

Pete
Yes the hubs and swivel housing are custom. It uses leaf bearings. I suspose you could use coiler hubs, but you would have to use 5 bolt drive flanges or source some 5 bolt locking hubs but there isn't much selection. Im going to have those M.A.P.'s broached to work with the new SuperLongs and will be good to go.

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Old 07-10-2005, 06:18 PM   #18 (permalink)
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Man guys thought fluid mechanics and dynamics were done for the summer..... really bringing up some bad memories here.
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Old 07-11-2005, 01:39 AM   #19 (permalink)
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Man guys thought fluid mechanics and dynamics were done for the summer..... really bringing up some bad memories here.
Summer, yeesh.
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Old 07-11-2005, 11:08 AM   #20 (permalink)
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The BCB idea is not bad, but:

1) I'm in the UK and P&P on big lumps of metal is a PITA
and
2) I'm not using Rover axles
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Old 07-11-2005, 01:55 PM   #21 (permalink)
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A mate of mine here (in the UK also), cut and welded some RR axles onto his series 3 88",

it was "just" a case of cutting off the radius arm mounts, and welding on the leaf spring pads.

can't remember how he did the steering though...
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Old 07-11-2005, 11:08 PM   #22 (permalink)
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Originally Posted by Diesel Jim
A mate of mine here (in the UK also), cut and welded some RR axles onto his series 3 88",

it was "just" a case of cutting off the radius arm mounts, and welding on the leaf spring pads.

can't remember how he did the steering though...
Most people take the coils when they take the axles.
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