Physics: Load transfer during braking...
Thread Starter
Team Owner
Joined: May 2002
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From: San Diego , CA Double Yellow DirtBags 1985..Z51..6-speed
Physics: Load transfer during braking...
3400 lb Corvette. Physics say:
1.0g braking: F 2195 lb, R 1205 lb, 60-0mph in 120 feet.
1.2g braking: F 2294 lb, R 1106 lb, 60-0mph in 100 feet.
0g braking: F 1700 lb, R 1700 lb, (duh.)
Do these numbers sound reasonable to you guys?
The fronts are doing about 65% of the work.
1.0g braking: F 2195 lb, R 1205 lb, 60-0mph in 120 feet.
1.2g braking: F 2294 lb, R 1106 lb, 60-0mph in 100 feet.
0g braking: F 1700 lb, R 1700 lb, (duh.)
Do these numbers sound reasonable to you guys?
The fronts are doing about 65% of the work.
Team Owner
Joined: Dec 2000
Posts: 90,675
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From: SE NY
Cruise-In II Veteran
I don't know the official numbers but those shound ball park.
Actually those numbers are kind of interesting as a typical launch pulls about 1.1g in the first 60', more on a very tight track with good air.
I notice the front end lift when I launch but didn't realize the weight transfer was that great. But it does explain how I can launch that hard on GY Eagle street tires.
Thanks for doing the math.
Actually those numbers are kind of interesting as a typical launch pulls about 1.1g in the first 60', more on a very tight track with good air.
I notice the front end lift when I launch but didn't realize the weight transfer was that great. But it does explain how I can launch that hard on GY Eagle street tires.
Thanks for doing the math.
Last edited by 65Z01; May 18, 2005 at 02:42 AM.
Melting Slicks
Joined: Apr 2003
Posts: 2,150
Likes: 6
From: NS
Originally Posted by Mr6spd
What number did you use for the height of the center of gravity when you made those calculations?
It all depends on the CG height in relation to the moment center height. It's not a simple percentage.
eg. If the CG was at same height as the moment center (not possible) there would be 0 weight transfer no matter how many Gs you pull under braking. The reason for this is there would be no moment arm (torque) to cause the weight transfer.
Generally, the lower the CG, the less weight transfer under braking/acceleration (or lateral around turns). Think of why an F1 car has such mind blowing performance.
Thread Starter
Team Owner
Joined: May 2002
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From: San Diego , CA Double Yellow DirtBags 1985..Z51..6-speed
CG = 14" I used. Raise that and the rear brakes will lock up.
Anybody know a general coef of friction value for brake rotors and hot racing pads?
I'm not pulling these numbers out of my but by the way. I'm figuring optimal line pressures, pad force, etc. Just checking GM's work, that's all.
Anybody know a general coef of friction value for brake rotors and hot racing pads?
I'm not pulling these numbers out of my but by the way. I'm figuring optimal line pressures, pad force, etc. Just checking GM's work, that's all.
Team Owner
Joined: Sep 2001
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From: Athens AL
C7 of the Year - Unmodified Finalist 2021
C4 of Year Finalist (performance mods) 2019
I'd like to see the formulas and logic trail you used to figure those numbers, I dont believe I've ever looked at this.
Racer
Joined: Jan 2005
Posts: 312
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From: Las Vegas / Scotland
Originally Posted by Atok
It all depends on the CG height in relation to the moment center height. It's not a simple percentage.
eg. If the CG was at same height as the moment center (not possible) there would be 0 weight transfer no matter how many Gs you pull under braking. The reason for this is there would be no moment arm (torque) to cause the weight transfer.
Regardless of the point of the CofG the turning moment exterted by the tires rotation would still cause weight transfer - for every action there is an equal and opposite reaction.
Interesting subject
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Team Owner
Joined: Mar 2001
Posts: 54,124
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From: Cincinnati, Oh USA
Rotor pad friction isn't important because it is possible to lock the wheel. What is important and what limits maximum deceleration is the coefficient of friction the tires have with the road surface.
Drifting
Joined: May 2003
Posts: 1,730
Likes: 1
From: Charlotte NC
St. Jude Donor '05
Sum of the moments = I(bar)Alpha - Ma(bar)D
You can throw Ibar alpha out because you'll just find it at a instant of time where the car isn't rotating about its axis with any angular acceleration
Find the sum of the forces in the X direction by simple F=ma, F in Y is 0 because its not acceleration into the ground or up into the air.
I can figure it out for you if you give me the hieght from the tire to the CG in hieght.
You can throw Ibar alpha out because you'll just find it at a instant of time where the car isn't rotating about its axis with any angular acceleration
Find the sum of the forces in the X direction by simple F=ma, F in Y is 0 because its not acceleration into the ground or up into the air.
I can figure it out for you if you give me the hieght from the tire to the CG in hieght.
Last edited by Alvin; May 18, 2005 at 12:55 PM.
Drifting
Joined: May 2003
Posts: 1,730
Likes: 1
From: Charlotte NC
St. Jude Donor '05
oh, also would need weight (MG) of the subject in question at the front and rear tires OR a good CG in the x direction, distance to each tire in X
Heck I need the distance in X direction of each tire in relation to Cg anyway.
I can do the problem no sweat in about 5 minutes if someone wants to cough up all the givens.
Heck I need the distance in X direction of each tire in relation to Cg anyway.
I can do the problem no sweat in about 5 minutes if someone wants to cough up all the givens.
Pro
Joined: Feb 2002
Posts: 533
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From: Augusta Ga
Does the C4 have 50% front and 50% rear weight distribution to begin with? If it doesn't than 1700lbs front and 1700lbs rear at zero braking doesn't work. I don't know I just thought that the C5 was closer to a 50/50 distribution than the C4, but still not complete 50/50.
Thread Starter
Team Owner
Joined: May 2002
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From: San Diego , CA Double Yellow DirtBags 1985..Z51..6-speed
Originally Posted by jfb
Rotor pad friction isn't important because it is possible to lock the wheel. What is important and what limits maximum deceleration is the coefficient of friction the tires have with the road surface.
This doesn't take into account rotational inertia of the car or wheels. Also assumes no movement of the CG during braking, which isn't true either.
Thread Starter
Team Owner
Joined: May 2002
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From: San Diego , CA Double Yellow DirtBags 1985..Z51..6-speed
I don't have it in front of me, I'm at work. I forget what the wheelbase is. CG halfway in between, 14" up. See what you come up with.
Drifting
Joined: May 2003
Posts: 1,730
Likes: 1
From: Charlotte NC
St. Jude Donor '05
Ok, I wrote the formula because I have to leave here for the rest of the day.
X= Distance between two tires (positive feet)
a= either = -32.2ft/sec^2 or -38.6ft/sec^2 Depending on if you want 1.0 g or 1.2 g braking
d= distance to CG from point of interest in j direction.
Fft = force on front tire in j direction
Fft = (-mad - mg(x/2))/-x
Or partially filled out
Fft= (- 3400lbm(a)(14/12') - (3400lbm(-32.2ft/sec^2)(x/2)) / (-x)
take Fft/Mg and you should have the percentage of wieght on the front wheels.
That should work. Someone ought to check to make sure all my units cancel
X= Distance between two tires (positive feet)
a= either = -32.2ft/sec^2 or -38.6ft/sec^2 Depending on if you want 1.0 g or 1.2 g braking
d= distance to CG from point of interest in j direction.
Fft = force on front tire in j direction
Fft = (-mad - mg(x/2))/-x
Or partially filled out
Fft= (- 3400lbm(a)(14/12') - (3400lbm(-32.2ft/sec^2)(x/2)) / (-x)
take Fft/Mg and you should have the percentage of wieght on the front wheels.
That should work. Someone ought to check to make sure all my units cancel
Last edited by Alvin; May 18, 2005 at 02:14 PM.
Racer
Joined: Feb 2005
Posts: 398
Likes: 1
From: Kelowna BC
Hey CentralCoaster,
I have a cool Mpeg on braking and the difference speed makes to stopping distances.
Based on the accents and that they drive on the wrong side, I think it is Australian.
I could send it to you if I knew how, but I dont.
Any easy way of posting something like that here?
Fred
I have a cool Mpeg on braking and the difference speed makes to stopping distances.
Based on the accents and that they drive on the wrong side, I think it is Australian.
I could send it to you if I knew how, but I dont.
Any easy way of posting something like that here?
Fred
Thread Starter
Team Owner
Joined: May 2002
Posts: 24,337
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From: San Diego , CA Double Yellow DirtBags 1985..Z51..6-speed
Well, I've got the stock bias spring rate figured out, and have some different springs to try out. Oh boy this sure beats working!
Race Director
Joined: Sep 2000
Posts: 17,298
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From: The Top of Utah
I have some numbers from my 1984 Corvette Specifications Manual, that may be of assistance.
Wheelbase: 96.2"
Tread, front: 59.6"
Tread, rear: 60.4"
Weight: 3192# (auto)
Front: 1630#
Rear: 1562#
Brake line hydraulic pressure @ 100" pedal pressure:
Front: 1305#
Rear: 667#
I don't see the center of gravity or roll center listed, but I'll look more closely. Lots of fine print. I have the manual trans weights, also, This thing even lists the weight each option adds to the car weight. There is also anti dive and anti squat percentages as designed into the suspension geometry plus spring rates. I don't know what year you are dealing with, CentralCoaster, but these numbers ought to be pretty representative of 1987 and earlier C4s.
RACE ON!!!
Wheelbase: 96.2"
Tread, front: 59.6"
Tread, rear: 60.4"
Weight: 3192# (auto)
Front: 1630#
Rear: 1562#
Brake line hydraulic pressure @ 100" pedal pressure:
Front: 1305#
Rear: 667#
I don't see the center of gravity or roll center listed, but I'll look more closely. Lots of fine print. I have the manual trans weights, also, This thing even lists the weight each option adds to the car weight. There is also anti dive and anti squat percentages as designed into the suspension geometry plus spring rates. I don't know what year you are dealing with, CentralCoaster, but these numbers ought to be pretty representative of 1987 and earlier C4s.
RACE ON!!!