Landracing Forum
Introductions => Formulas => Topic started by: iskipnw on April 13, 2010, 12:20:42 AM
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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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Depends if you have a high or low wheel rate. Which one are you are you attempting to use? :cheers:
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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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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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What kind if vehicle is it?
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The equation is if you see floating mountain . . . pits . . . highway . . . floating mountain . . . pits . . . highway . . .
You need more ballast.
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Iam trying for 4800-5000 pounds in the roadster, 120 wheel base :?
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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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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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It varies day to day, hour by hour due to weather and tidal (yes, I believe!) tidal factors.
Stan
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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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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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Is Skipping Now --
pm sent.
Stan
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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 :-o 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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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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Here is a basic program I did a few years ago, that has most of that in it.
You can change some of the variables if need be by going into the code.
you will need to have qbasic installed on your system to run these as they run under qbasic.exe in a command window.
I have not gotten around to re-coding this in perl or some other modern language but it works.
===========================
CLS
PRINT
PRINT " Compute Aerodrag and Rolling drag "
PRINT " Compute traction available, and max usable power"
PRINT
PRINT
G# = 32.1725
CL# = .2
airden# = .0619
basetiredragcoef# = .0155
REM air density at Bonneville 4214 ft elevetion
REM 90 deg, 30% humidity, barometer 25.7 in Hg abs
REM for denver 5800 ft elevation
REM temp 85, 30% humidity, barometer 23.4 in Hg abs
REM use a value of .056616
REM basetiredragcoef# = .0155
REM suggested value is .011, range from .007 - .013
REM this value (.0155) used based on coast down tests with subaru
INPUT " Enter frontal area in square ft"; Fa#
PRINT
INPUT " Enter Cd with decimal "; cd#
PRINT
INPUT " Enter speed in mph "; mph#
PRINT
INPUT " Enter weight of car in lbs"; wtlbs#
PRINT
PRINT " Enter percent weight on drive wheels"
INPUT " use whole numbers ie 40% = 40 "; drivewt#
REM INPUT "input Coefficient of lift, default is 0.2 if unsure hit enter"; CL#
PRINT
PRINT " Using coefficient of lift, = "; (INT(CL# * 1000) / 1000)
INPUT " enter portion of lift applied to drive wheels in percent "; liftdrivewt#
PRINT
INPUT " Enter Coefficient of friction for the salt conditions ( .4 - .6)"; salttractioncoef#
REM *********************************
CLS
PRINT
PRINT " Compute Aerodrag and Rolling drag "
PRINT " Compute traction available, and max usable power"
PRINT
PRINT " Frontal area = "; (INT(Fa# * 100) / 100); "ft^2 Cd = "; (INT(cd# * 10000) / 10000)
PRINT " Target speed = "; (INT(mph# * 10) / 10); " MPH"
PRINT " Car weight = "; (INT(wtlbs# * 10) / 10); " lbs"
PRINT " Weight Distribution = "; (INT(drivewt#)); " % on drive wheels"
PRINT " Coefficient of Lift = "; (INT(CL# * 100) / 100); " ratio of lift to drag"
PRINT " Lift applied to drive wheels = "; (INT(liftdrivewt# * 1000) / 1000); " Percent"
ftpsec# = mph# * 88 / 60
hpaero# = (((airden# / G#) / 2) * Fa# * cd# * (ftpsec# * ftpsec# * ftpsec#)) / 550
hpaero1# = (((airden#) * Fa# * cd#) / 35389.75) * (ftpsec# * ftpsec# * ftpsec#)
lbaero1# = (((airden#) * Fa# * cd#) / 64.345 * (ftpsec# * ftpsec#))
hprolling# = ((basetiredragcoef# * wtlbs#) + (.00003 * wtlbs# * mph#)) * ftpsec# / 550
lbdragrolling# = ((basetiredragcoef# * wtlbs#) + (.00003 * wtlbs# * mph#))
hptotal# = hpaero# + hprolling#
BHP# = (hptotal# / 84) * 100
REM salttractioncoef# = .4
maxtractionstatic# = ((wtlbs# * drivewt# / 100) * salttractioncoef#)
maxtractionlimitedpower# = (maxtractionstatic# * ftpsec#) / 550
lftlbs# = CL# * lbaero1#
maxtractiondynamic# = maxtractionstatic# - (lftlbs# * liftdrivewt# / 100)
tractionlimitedpower# = maxtractiondynamic# * ftpsec# / 550
PRINT
PRINT " Using 16% drive train loss"
PRINT " BHP required at "; mph#; "mph ..................... = "; (INT(BHP# * 1000) / 1000); " HP"
PRINT " Aero power required is ....................... = "; (INT(hpaero# * 1000) / 1000); " hp "
PRINT " Rolling resistance is ........................ = "; (INT(hprolling# * 1000) / 1000); " hp"
PRINT " Total power at wheels needed at "; mph#; "mph is = "; (INT(hptotal# * 1000) / 1000); " hp "
PRINT
PRINT "======================="
PRINT " Aerodrag hp ........................................ = "; (INT(hpaero1# * 1000) / 1000); " HP"
PRINT " Aerodrag in lbs .................................... = "; (INT(lbaero1# * 1000) / 1000); " lbs"
PRINT " Estimated lift at 0.2 x drag ....................... = "; ((INT(CL# * lbaero1#) * 1000) / 1000); " lbs"
PRINT " Percent of lift applied to drive wheels ............ = "; liftdrivewt#; " %"
PRINT
PRINT "======================="
PRINT " Compute traction available for acceleration"
PRINT " Assumption Traction Coef on salt "; salttractioncoef#
PRINT
PRINT "======================"
PRINT " Max tractive thrust which can be delivered (static) ...... = "; ((INT(maxtractionstatic#) * 1000) / 1000); " lbs"
PRINT " Max usable hp at speed due to traction limitations........ = "; ((INT(maxtractionlimitedpower#) * 1000) / 1000); " HP"
PRINT " Thrust required to overcome rolling resistance ........... = "; ((INT(lbdragrolling#) * 1000) / 1000); " lbs"
PRINT " Portion of lift forces carried by the driving wheels ......= "; ((INT(lftlbs# * liftdrivewt# / 100) * 1000) / 1000); " lbs"
PRINT
PRINT " Max thrust lbs deliverable at speed due to lift ...........= "; ((INT(maxtractiondynamic#) * 1000) / 1000); " lbs"
PRINT " Max useable power at speed due to traction lost to lift ...= "; ((INT(maxtractiondynamic# * ftpsec# / 550) * 1000) / 1000); " HP"
PRINT " Reserve power available (difference between needed power to reach"
PRINT " target speed, and power which can be transmitted due to traction limits)"
PRINT
PRINT "******************************************************************"
PRINT " THESE VALUES MUST BE POSITIVE"
PRINT " POWER RESERVE (static).....................= "; ((INT(tractionlimitedpower# - hptotal#) * 1000) / 1000); " HP"
PRINT " (how much traction limited power exceeds needed power)"
PRINT
PRINT " Surplus transferable power at speed (dynamic)..............= "; ((INT(maxtractionlimitedpower# - hptotal#) * 1000) / 1000); " HP"
PRINT " Surplus traction thrust after overcoming aero drag ....... = "; ((INT(maxtractiondynamic# - lbaero1#) * 1000) / 1000); " lbs"
PRINT " Surplus thrust avail for acceleration .................... = "; ((INT(maxtractiondynamic# - lbaero1# - lbdragrolling#) * 1000) / 1000); " lbs"
PRINT "******************************************************************"
======================
copyright Larry Ledwick free to use with attribution.
Noticed a small typo in the code and corrected it for lift applied to drive wheels, was with units of hp, and should have been percent of total lift.
Larry
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Isn't there a shortcut formula? Like:
F=MA, or maybe E=Mc2? (See my Stan Back answer in another post)
Seriously, REALLY extensive , BUT the truth seekers will appreciate.
Landa's "Automotive Aerodynamics Handbook" has a slew of helpful math similiar to the above (if you can find the book) but not nearly as all-encompassing and Bonnevile specific.
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Hotrod...could you compute this for us cd=.3 frontal area 21.5 sq ft. lift .2 weight 4900 lbs weight on rear wheels 3000lbs hp 2400
I didn't see any input for tire width how about the difference between 10'' and 6'' :cheers:
JL222 :-D
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I ran it a couple times and worked the speed up, since you did not mention a target speed.
Looks like you have plenty of power, but need traction (weight on the drive wheels) --- assuming my calculations are correct??
Tire width is largely irrelevant at bonneville as you have very little of the grip you would get with rubber on asphalt where the rubber "gears" itself to the road due to irregularities and glues itself to the road due to adhesion (ie sticky rubber) it is almost all classic sliding friction from what I can tell.
===========================
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .2999
Target speed = 300 MPH
Car weight = 4200 lbs
Weight Distribution = 0 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 lbs
Using 16% drive train loss
BHP required at 300 mph ..................... = 1242.067 HP
Aero power required is ....................... = 961.016 hp
Rolling resistance is ........................ = 82.319 hp
Total power at wheels needed at 300 mph is = 1043.336 hp
=======================
Aerodrag hp ........................................ = 961.016 HP
Aerodrag in lbs .................................... = 1201.27 lbs
Estimated lift at 0.2 x drag ....................... = 240 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 14 lbs
Max usable hp at speed due to traction limitations........ = 11 HP
Thrust required to overcome rolling resistance ........... = 102 lbs
Portion of lift forces carried by the driving wheels ......= 120 lbs
Max thrust lbs deliverable at speed due to lift ...........= -106 lbs
Max useable power at speed due to traction lost to lift ...= -85 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= -1128 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= -1032 HP
Surplus traction thrust after overcoming aero drag ....... = -1307 lbs
Surplus thrust avail for acceleration .................... = -1410 lbs
******************************************************************
Press any key to continue
==========================
Looks like your top speed should be near 285 mph based on the following.
==========================
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .2999
Target speed = 285 MPH
Car weight = 4200 lbs
Weight Distribution = 71 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 lbs
Using 16% drive train loss
BHP required at 285 mph ..................... = 1072.284 HP
Aero power required is ....................... = 823.951 hp
Rolling resistance is ........................ = 76.767 hp
Total power at wheels needed at 285 mph is = 900.7190000000001 hp
=======================
Aerodrag hp ........................................ = 823.951 HP
Aerodrag in lbs .................................... = 1084.146 lbs
Estimated lift at 0.2 x drag ....................... = 216 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 1491 lbs
Max usable hp at speed due to traction limitations........ = 1133 HP
Thrust required to overcome rolling resistance ........... = 101 lbs
Portion of lift forces carried by the driving wheels ......= 108 lbs
Max thrust lbs deliverable at speed due to lift ...........= 1382 lbs
Max useable power at speed due to traction lost to lift ...= 1050 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= 150 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= 232 HP
Surplus traction thrust after overcoming aero drag ....... = 298 lbs
Surplus thrust avail for acceleration .................... = 197 lbs
******************************************************************
Press any key to continue
===========================
This is only theory, it would be good to get some real test cases to compare what it predicts to what people actually experience. To the best of my knowledge it is reasonably correct but limitations to how precise you can know some variables such as percent of lift applied to drive wheels is unknown, I assume 50% on a car but that could be quite wrong on a car which is using a wing for down force.
Larry
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what?? more power= more ballast :roll:
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Thanks Hotrod ... The 222 car has been timed at 285 mph in the Ist mile at Bville but a blower tubr connector blew off 2 sec or approximately 1100 ft before timing light, computer indicated speed was 312 mph with 1'' of tire growth input.
Car weighs 4900 not 4200 and there is no tire slip.
Target speed is as fast as it will go with 2400 hp.
JL222
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No matter how you calculate it, what you are doing will help, if and when you run out of horsepower somewhere between the 4 and 5 mile in a roadster you will become a very good driver or be along for the ride. With what it appears you are wanting to do, start with a pretty high gear and work back if needed...
Roadsters just love to dance when they run out of gear or horspower.....Good Luck
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Sorry about the miscue on the weight -- I was working from home and trying to do that at the same time.
Looks like my numbers agree pretty well with your actual times (given the lack of precision for some of the input values especially).
=========================
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .3
Target speed = 305 MPH
Car weight = 4900 lbs
Weight Distribution = 61 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 percent
Using 16% drive train loss
BHP required at 305 mph ..................... = 1319.58 HP
Aero power required is ....................... = 1010.209 hp
Rolling resistance is ........................ = 98.238 hp
Total power at wheels needed at 305 mph is = 1108.447 hp
=======================
Aerodrag hp ........................................ = 1010.209 HP
Aerodrag in lbs .................................... = 1242.06 lbs
Estimated lift at 0.2 x drag ....................... = 248 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 1494 lbs
Max usable hp at speed due to traction limitations........ = 1215 HP
Thrust required to overcome rolling resistance ........... = 120 lbs
Portion of lift forces carried by the driving wheels ......= 124 lbs
Max thrust lbs deliverable at speed due to lift ...........= 1370 lbs
Max useable power at speed due to traction lost to lift ...= 1114 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= 6 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= 107 HP
Surplus traction thrust after overcoming aero drag ....... = 128 lbs
Surplus thrust avail for acceleration .................... = 7 lbs
******************************************************************
Press any key to continue
=========================
Looks like you need to add some weight! ---- see below for additional weight.
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .3
Target speed = 350 MPH
Car weight = 5900 lbs
Weight Distribution = 70 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 percent
Using 16% drive train loss
BHP required at 350 mph ..................... = 1987.786 HP
Aero power required is ....................... = 1526.567 hp
Rolling resistance is ........................ = 143.173 hp
Total power at wheels needed at 350 mph is = 1669.74 hp
=======================
Aerodrag hp ........................................ = 1526.567 HP
Aerodrag in lbs .................................... = 1635.608 lbs
Estimated lift at 0.2 x drag ....................... = 327 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 2065 lbs
Max usable hp at speed due to traction limitations........ = 1927 HP
Thrust required to overcome rolling resistance ........... = 153 lbs
Portion of lift forces carried by the driving wheels ......= 163 lbs
Max thrust lbs deliverable at speed due to lift ...........= 1901 lbs
Max useable power at speed due to traction lost to lift ...= 1774 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= 104 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= 257 HP
Surplus traction thrust after overcoming aero drag ....... = 265 lbs
Surplus thrust avail for acceleration .................... = 112 lbs
******************************************************************
Press any key to continue
================
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .3
Target speed = 370 MPH
Car weight = 6500 lbs
Weight Distribution = 70 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 percent
Using 16% drive train loss
BHP required at 370 mph ..................... = 2350.117 HP
Aero power required is ....................... = 1803.503 hp
Rolling resistance is ........................ = 170.594 hp
Total power at wheels needed at 370 mph is = 1974.098 hp
=======================
Aerodrag hp ........................................ = 1803.503 HP
Aerodrag in lbs .................................... = 1827.875 lbs
Estimated lift at 0.2 x drag ....................... = 365 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 2275 lbs
Max usable hp at speed due to traction limitations........ = 2244 HP
Thrust required to overcome rolling resistance ........... = 172 lbs
Portion of lift forces carried by the driving wheels ......= 182 lbs
Max thrust lbs deliverable at speed due to lift ...........= 2092 lbs
Max useable power at speed due to traction lost to lift ...= 2064 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= 90 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= 270 HP
Surplus traction thrust after overcoming aero drag ....... = 264 lbs
Surplus thrust avail for acceleration .................... = 91 lbs
******************************************************************
Press any key to continue
========================
As you can see according to these calculations (again just theory) you need 4550 lbs on the rear wheels to use all your power, and given enough acceleration time, you should get close to 370 mph with that much power if you can hang on to it.
Not sure 70% of weight on the rear is safe, but the actual weight on the drive wheels necessary to hook that much power won't change if salt conditions allow .5 coefficient of friction (and all other assumptions are correct like amount of lift etc.
Good luck!
I will try to look you up this year.
Larry
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Yeah Larry look us up Langlo racing on trailer and we like to be on the front row to watch when we can.
The 285 mph is right on with our timing slip but as noted the car was coasting for 1100 ft or 2 sec + as seen on video
and rpm indicated 312 mph we also had a 268 mph 21/4 speed which shows that the car coasting through the lights had an increase of 17 mph from that average and 44 miles and hour from the 312 mph indicated speed.
As far as weight on rear wheels we have 3000 lbs static [but] we also have a spoiler which also increases down force, but how much we don't know but as you can see the car hooks extreamely well for conditions.
Talking to Goodyear guy at 200 mph banquet about there testing on salt coef. [they were surprised when results showed
.7 :-o] if I remember right he also said the salt seperated from itself before the tires spun.
The Bonneville Pro computer program shows a top speed of 341 but well have to see what happens in the real world.
Yeah come by and maybe you can fine tune your program with our results if we can make a full pass.
JL222 :cheers:
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Here is Larry's routine in xls. Sheets are protected but no password required. :-D
Thanks Larry!!
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Comments on Hotrod’s above calculation sheets:
1) It may be worth including the front-to-rear weight transfer due to the aero drag acting on a CP above ground level. For some configurations this would be of minor importance, but for the more upright, short wheelbase, and high horsepower vehicles such as roadsters and motorcycles, it could amount to a significant load redistribution.
2) The rolling resistance calculation is presented as proportional to velocity. More details on its origin and applicability would be interesting. Others, (Korff’s Goldenrod paper), have a quadratic relation in velocity. This may make more sense since at higher speeds, the tires and wheels are subject to a velocity squared aero drag and turbulence. Comparing the two, there can be a significant difference at higher speeds. Of course, it could be argued that the velocity squared component could be absorbed into the vehicle drag coefficient but aerodynamicists would probably prefer not to contaminate the results of their shapes, calculations, and wind tunnel data with what the wheels may be contributing, if that could even be reasonably determined short of being on the salt.
3) Some of the terminology used does not clearly convey just what the numbers really represent.
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how much different if roadster is front wheel drive ?
never thought much about that..............
6 weeks to go
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Calculating the drag-related weight transfer is relatively simple if a couple quantities are known or can be estimated. dW = (D*h)/WB where dW is the weight transfer (lb), D is the drag force (lb), h is the height of the center of pressure above the ground, and WB is the wheelbase. h and WB should have the same units, e.g. inches, feet, or whatever.
By way of illustration, assuming drag of 1600 lbs, h of 20”, and a wheelbase of 80”, the front end will get 400 lbs lighter and the rear 400 lbs heavier. Perhaps not a big effect, but at least accounting for it would get the traction calculation that much closer to reality.
As regards a front-drive roadster, this weight transfer would eventually limit the speed achieveable since traction required is increasing in order to overcome the increasing drag, but the traction available is decreasing as the drive axle weight is decreasing. Such a roadster would probably be inherently front-heavy, and the limit may be beyond the power available, but in any case, the remedy is straightforward--ballast up the front or induce some aero downforce. And, adding ballast to the front is much more friendly to aerodynamic stability than having to add it to the rear.
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Good points there, I will need to consider putting that in the calculation.
It might be more practical due to all the complicating considerations to do that as a second followup calculation as you did in the preceding post.
The only question I would have is how to properly estimate the center of pressure height.
It might be reasonable to assume the center of pressure is at the center (centroid) of the frontal area.
That would probably be reasonable for a clean well designed body with good aerodynamics like a lakester or stream liner. I suspect that assumption would break down in the case of body styles like the older pickup trucks where the "green house" of the cab might create a larger fraction of the drag compared to the front end, especially on some of the really old body styles which had near flat and near vertical windshields.
A large good scoop elevated well above the hood line could also act like a lever and if it was a high drag scoop might substantially raise the center of pressure.
It certainly is something to think about for people that have high drag and high profile body shapes, like the roadsters, older pickups and diesel trucks.
In the case of front wheel cars it is definitely something to consider as it inherently will reduce traction on the drive wheels.
Larry
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Sorry about the miscue on the weight -- I was working from home and trying to do that at the same time.
Looks like my numbers agree pretty well with your actual times (given the lack of precision for some of the input values especially).
=========================
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .3
Target speed = 305 MPH
Car weight = 4900 lbs
Weight Distribution = 61 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 percent
Using 16% drive train loss
BHP required at 305 mph ..................... = 1319.58 HP
Aero power required is ....................... = 1010.209 hp
Rolling resistance is ........................ = 98.238 hp
Total power at wheels needed at 305 mph is = 1108.447 hp
=======================
Aerodrag hp ........................................ = 1010.209 HP
Aerodrag in lbs .................................... = 1242.06 lbs
Estimated lift at 0.2 x drag ....................... = 248 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 1494 lbs
Max usable hp at speed due to traction limitations........ = 1215 HP
Thrust required to overcome rolling resistance ........... = 120 lbs
Portion of lift forces carried by the driving wheels ......= 124 lbs
Max thrust lbs deliverable at speed due to lift ...........= 1370 lbs
Max useable power at speed due to traction lost to lift ...= 1114 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= 6 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= 107 HP
Surplus traction thrust after overcoming aero drag ....... = 128 lbs
Surplus thrust avail for acceleration .................... = 7 lbs
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Press any key to continue
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Looks like you need to add some weight! ---- see below for additional weight.
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .3
Target speed = 350 MPH
Car weight = 5900 lbs
Weight Distribution = 70 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 percent
Using 16% drive train loss
BHP required at 350 mph ..................... = 1987.786 HP
Aero power required is ....................... = 1526.567 hp
Rolling resistance is ........................ = 143.173 hp
Total power at wheels needed at 350 mph is = 1669.74 hp
=======================
Aerodrag hp ........................................ = 1526.567 HP
Aerodrag in lbs .................................... = 1635.608 lbs
Estimated lift at 0.2 x drag ....................... = 327 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 2065 lbs
Max usable hp at speed due to traction limitations........ = 1927 HP
Thrust required to overcome rolling resistance ........... = 153 lbs
Portion of lift forces carried by the driving wheels ......= 163 lbs
Max thrust lbs deliverable at speed due to lift ...........= 1901 lbs
Max useable power at speed due to traction lost to lift ...= 1774 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= 104 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= 257 HP
Surplus traction thrust after overcoming aero drag ....... = 265 lbs
Surplus thrust avail for acceleration .................... = 112 lbs
******************************************************************
Press any key to continue
================
Compute Aerodrag and Rolling drag
Compute traction available, and max usable power
Frontal area = 21.5 ft^2 Cd = .3
Target speed = 370 MPH
Car weight = 6500 lbs
Weight Distribution = 70 % on drive wheels
Coefficient of Lift = .2 ratio of lift to drag
Lift applied to drive wheels = 50 percent
Using 16% drive train loss
BHP required at 370 mph ..................... = 2350.117 HP
Aero power required is ....................... = 1803.503 hp
Rolling resistance is ........................ = 170.594 hp
Total power at wheels needed at 370 mph is = 1974.098 hp
=======================
Aerodrag hp ........................................ = 1803.503 HP
Aerodrag in lbs .................................... = 1827.875 lbs
Estimated lift at 0.2 x drag ....................... = 365 lbs
Percent of lift applied to drive wheels ............ = 50 %
=======================
Compute traction available for acceleration
Assumption Traction Coef on salt .5
======================
Max tractive thrust which can be delivered (static) ...... = 2275 lbs
Max usable hp at speed due to traction limitations........ = 2244 HP
Thrust required to overcome rolling resistance ........... = 172 lbs
Portion of lift forces carried by the driving wheels ......= 182 lbs
Max thrust lbs deliverable at speed due to lift ...........= 2092 lbs
Max useable power at speed due to traction lost to lift ...= 2064 HP
Reserve power available (difference between needed power to reach
target speed, and power which can be transmitted due to traction limits)
******************************************************************
THESE VALUES MUST BE POSITIVE
POWER RESERVE (static).....................= 90 HP
(how much traction limited power exceeds needed power)
Surplus transferable power at speed (dynamic)..............= 270 HP
Surplus traction thrust after overcoming aero drag ....... = 264 lbs
Surplus thrust avail for acceleration .................... = 91 lbs
******************************************************************
Press any key to continue
========================
As you can see according to these calculations (again just theory) you need 4550 lbs on the rear wheels to use all your power, and given enough acceleration time, you should get close to 370 mph with that much power if you can hang on to it.
Not sure 70% of weight on the rear is safe, but the actual weight on the drive wheels necessary to hook that much power won't change if salt conditions allow .5 coefficient of friction (and all other assumptions are correct like amount of lift etc.
Good luck!
I will try to look you up this year.
Larry
who says hotrodding is'nt rocket science.?
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Hotrod... You should have reminded me about this equation when I bought that nice foto from you.
As far as traction goes we hat a 21/4 time of 280 with a 294 3 mile time when the piston blew right at end of the 3rd mile with an indicated speed of 318 mph at 7000 rpm, This took 50 sec from 1st applying throttle.
More weight would help at the start because in low gear tps was 31% max.. 2nd gear 44%... 3rd 50%
4th gear 100% with no tire slip but, instead of more weight I would rather have a variable spoiler [like the Chaperrals had 40 yrs ago] :-o
JL222
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This whole thread was very enlightening. I am very impressed and humbled by all the brainiacs out there. I am so glad that I did not experience floating mountains….highway…..starting line………during my maiden voyage this year at Speedweek. :cheers:
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Remind me: is fixed ballast required, or is "restrained" (can move, but not freely) acceptable if safety is OK?
I'm sure some see where I'm going...
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Since ballast changes CG I thought I would add this here. (Also posted in parachute thread.)
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Ok, i am now preparing for Speedweek 2011 and have to report on years one and two. During the maiden voyage I had a theoretical value for ballast and was unable to install the desired amount. At 172mph I experienced Cd = traction. I still had plenty of gear and engine. The post speedweek analysis using various models, some provided on this thread (thanks you all), and pone that I developed on my own, was within 6% of the value I experienced when I held my actual Weight and Balance constant.
We subsequently added the appropriate ballast for speedweek 2011 and achieved 203 mph. This year we are 66 days out as of this post and ballast is fixed....on to other variables.