hotrod


« Reply #15 on: May 02, 2010, 01:58:14 PM » 

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 recoding 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


« Last Edit: May 03, 2010, 08:40:57 AM by hotrod »

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interested bystander


« Reply #16 on: May 02, 2010, 04:43:15 PM » 

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 allencompassing and Bonnevile specific.


« Last Edit: May 02, 2010, 04:52:30 PM by interested bystander »

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5 mph in pit area (clothed)



jl222


« Reply #17 on: May 02, 2010, 06:46:22 PM » 

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'' JL222


« Last Edit: May 02, 2010, 06:53:47 PM by jl222 »

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hotrod


« Reply #18 on: May 02, 2010, 10:47:30 PM » 

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


« Last Edit: May 02, 2010, 10:55:14 PM by hotrod »

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mkilger
Sr. Member
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Age: 48
Location: Riverside ca
Posts: 997


« Reply #19 on: May 02, 2010, 10:52:28 PM » 

what?? more power= more ballast



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jl222


« Reply #20 on: May 03, 2010, 10:07:27 AM » 

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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jimmy six


« Reply #21 on: May 03, 2010, 04:18:48 PM » 

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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First GMC 6 powered Fuel roadster over 200, with 2 red hats. Pit crew for Patrick Tone's Super Stock #49 Camaro



hotrod


« Reply #22 on: May 03, 2010, 08:16:30 PM » 

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


« Last Edit: May 03, 2010, 08:24:16 PM by hotrod »

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jl222


« Reply #23 on: May 04, 2010, 12:47:58 PM » 

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 ] 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



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WOODY@DDLLC


« Reply #24 on: June 18, 2010, 03:08:58 PM » 

Here is Larry's routine in xls. Sheets are protected but no password required. Thanks Larry!!



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Interested Observer


« Reply #25 on: July 06, 2010, 12:11:35 AM » 

Comments on Hotrod’s above calculation sheets:
1) It may be worth including the fronttorear 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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mstrdinan


« Reply #26 on: July 06, 2010, 08:31:29 PM » 

how much different if roadster is front wheel drive ? never thought much about that.............. 6 weeks to go



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Interested Observer


« Reply #27 on: July 07, 2010, 08:14:03 PM » 

Calculating the dragrelated 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 frontdrive 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 frontheavy, and the limit may be beyond the power available, but in any case, the remedy is straightforwardballast 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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hotrod


« Reply #28 on: July 07, 2010, 09:38:42 PM » 

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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BALS aSALT
Full Member
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Age: 56
Location: Belmont, California
Posts: 299
das bullet


« Reply #29 on: October 16, 2010, 04:58:39 PM » 

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
who says hotrodding is'nt rocket science.?



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it is better to live one day as a lion, than it is to live one thousand days as a lamb. or maybe it is better to burn out than fade away.



