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Author Topic: Ballast Equations  (Read 17013 times)
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iskipnw
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« on: April 12, 2010, 11:20:42 PM »

Before I consume mass amounts of brain cells has anyone derived an equation to determine optimum rear wheel ballast, knowing Cd, Frontal Area, drag force, rear wheel HP.

At some point the effects of drag force will overcome the wheels ability to grab the salt and will start spinning without proper ballast.  Can it be as simple as F=UN, where U = coefficient of friction between tire and salt, N = Normal force which is equal and opposite to weight on tire, and F = tangential force of wheel on ground as it is being driven; which must be greater than the drag forces pushing on the car.
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Ken
bvillercr
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« Reply #1 on: April 12, 2010, 11:55:09 PM »

Depends if you have a high or low wheel rate.  Which one are you are you attempting to use? cheers
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jl222
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« Reply #2 on: April 13, 2010, 01:01:03 AM »


  I assume your asking about a bville salt car and some cars naturally have more weight on the rear tires [ like lakesters]
 and the fastest of them all has weight added plus a rear wing. Also thrust will be limited by tire contact area.
  The formula 1 cars have at least 65% static weight on the rear wheels + wings.
  Tune To Win by Carroll Smith has a lot on weight-traction - and load transfer, but a long wheelbase [And or] a solid suspended Bville car with narrow tires will not get much load transfer.


                                      JL222
   

                             
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iskipnw
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« Reply #3 on: April 13, 2010, 08:26:01 AM »

Yes, this car is a Bonneville Salt flat car and the wheel rate is low since the rear axle is of a solid mount design.
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Ken
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« Reply #4 on: April 13, 2010, 09:04:03 AM »

What kind if vehicle is it?
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Dean Los Angeles
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« Reply #5 on: April 13, 2010, 09:27:26 AM »

The equation is if you see floating mountain . . . pits . . . highway . . . floating mountain . . . pits . . . highway . . .

You need more ballast.
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mkilger
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« Reply #6 on: April 13, 2010, 10:16:11 AM »

Iam trying for 4800-5000 pounds in the roadster, 120 wheel base  huh
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maguromic
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« Reply #7 on: April 13, 2010, 12:08:33 PM »

In LSR we need net traction and acceleration can be slow, so a heavy car is fine.  In my opinion what you want is to have the center of mass forward of the center of aero pressure.  If you got really heavy, the drag from the front tires and the HP it takes to roll it all would take a toll.  Each car is different and it’s still a drag race to the quarter.  Tony
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john robson
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« Reply #8 on: April 13, 2010, 01:55:55 PM »

I suggest that maximum forward thrust is equal to the coefficient of friction multiplied by weigh on the rear axle. When forward thrust equals resistance from all operative factors power can only spin the drive tires. You will need more weight or less resistance to gain speed. I have heard it said that the coefficient of friction on the salt is .45 to .5. Does anybody actually know?
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Stan Back
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« Reply #9 on: April 13, 2010, 02:21:04 PM »

It varies day to day, hour by hour due to weather and tidal (yes, I believe!) tidal factors.

Stan
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Member of the San Berdoo Roadsters – California's most-exclusive roadster club.
Celebrating 65th anniversary of racing on the salt.
iskipnw
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« Reply #10 on: April 13, 2010, 02:56:57 PM »

The car is a street roadster.  I believe the weight should be around 3600-4000 lbs when all is said and done, but I would like to work some math to make my mind at ease.  The closest research I have found is with tractor ballast which I can use as a starting point.  The variables that I have heard are:

Rolling Resistance on compact salt, which will vary is about .06
Coefficient of friction is about 0.43, again it will vary.
I can extract HP and Torque and factor driveline inefficiencies
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Ken
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« Reply #11 on: April 13, 2010, 03:52:50 PM »

Ballast can help in 3 ways , resists aero lift , adjusts the CG to reduce spins and makes the car easier to drive while it's traction limited . It also reduces acceleration past the point of wheel slip , fewer hp/pound . For fast cars that are fully traction limited ballast makes aero drag a smaller % of traction leaving more for acceleration . Ballast may also reduce traction , more pounds per aquare inch at the contact patch .
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Stan Back
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« Reply #12 on: April 13, 2010, 04:57:33 PM »

Is Skipping Now --

pm sent.

Stan
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Member of the San Berdoo Roadsters – California's most-exclusive roadster club.
Celebrating 65th anniversary of racing on the salt.
jl222
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« Reply #13 on: April 13, 2010, 07:33:36 PM »

Yes, this car is a Bonneville Salt flat car and the wheel rate is low since the rear axle is of a solid mount design.

 wheel rate is how many lbs it takes for a suspension to move 1'', being solid mounted your wheel rate is how many lbs it takes
to compress your tires 1'' divided by 2 shocked Thats extreamely high or infinity as Carroll Smith would say. Resulting in excessive wheelspin and a need for more ballast, and don't mistake wheelspin for hp.

                                      JL222

 

   
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Rex Schimmer
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« Reply #14 on: April 25, 2010, 09:12:13 AM »

It is a balance of forces. You have aero drag and rolling resistance that are pointing one way, and forward thrust of the wheels against the salt pointing the other when the two are equal you won't go any faster. Aero drag is pretty much a constant related to the frontal area of your car and its' coefficient of drag and the velocity that you are going and can be calculated in pounds by using this equation: Cd x 1/2(density of air in lb/cu. ft)(velocity, in feet/sec)squared x (frontal area in square feet)= pounds drag. To get velocity in ft/sec from miles/hour divide  the mph by .681 and the density of air (as sea level) is .00238 lbs/cu.ft.

So lets say your roadster has a frontal area of 12 square feet and a Cd of .75 and you want to go 200 mph. The drag force would be:
.75(.5)(.00238)(200/.681)sq x(12)=923.75 pounds force. From this number you can look at the coefficient of friction for the tire/salt and come up with an estimate of how much weitht you need on the drive wheels. If the coefficient of friction was .5 you would need to have about 1850 lbs on the drive wheel. I have not included rolling resistance which is related to tires, salt conditions and car weight and can be a much more difficult thing to calculate because of the various data required that is difficult to obtain accurately.
All of this assumes that the tire is on the ground 100% of the time which as John (JL222) says it dependent on wheel rate and suspension. Nothing is easy or exact but these calculations can get you to a starting point.
Rex

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Rex

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