HOME    PHOTOS    FORUM    CALENDAR    RECORDS    HISTORY    TECH & FAQs    VIDEO    CONTACT    LINKS   
SCTA-BNI
• El Mirage
• SpeedWeek
• World Finals
USFRA
• World of Speed
ECTA
DLRA
LTA
Texas Mile
Top Speed Shootout
Bonneville Motorcycle Speed Trials
Swedish Landracing
• SpeedWeekend on Ice
El Mirage Ladies
Auxiliary
Utah Motorsports Foundation
200 MPH Club
FIA
FIM
AMA
Save The Salt




Credit cards or PayPal
It's easy !

(not tax deductible)

More Info....

Many thanks to you all!



Kudos Laser Web Site

• • •

200 MPH Club Plaques
More Info


Engraving on trophies, mugs, glassware and much more!

Kudos Laser mug

Want to know who's who when you're racing?? Ever forget who you are?

We have the solution.




More.....

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Horse Power Needed after Cd Change

One of my few good friends, yeah, Keith Turk is my friend, considered a class change for his land speed race car. The class change will require that the car be de-modified from an altered car and turned into a gas coupe. Numerous items must be added back on: for instance, the grill opening needs to be returned to production configuration, the head lights must be returned to production and bumpers must be added back. Oh, and no rear spoiler! These features all add drag back into the equation, but the question is how much to add and what does it cost in terms of horsepower.

DISCLAIMER

I have been asking for weather data from last years WOS. The reason for the request is that the altered car ran a flat out terminal speed of 225 mph using a motor that developed 500 flywheel hp at sea level with a 125 shot of nitrous oxide. So I needed to correct the sea level horsepower to the WOS conditions when it ran as well as using the climate data to figure aerodynamic drag. I never got the real data but did get a data point of barometric pressure equal to 26.5 inches of mercury and temperature of 90 degrees Fahrenheit. I don't think these are correct but since they are all that I have, I'll use them and adjust later if better information comes to me.

Assumptions and Givens
I made a few assumptions, along with the givens, that may be suspect. Those of you who have real information on how to change my assumptions, not opinions, please contact me and I will make the suggested changes.

1) I assumed that the 125 shot of Nitrous was the same at sea level as at altitude. Why? Well, it is it's own air and fuel and that is delivered the same whether or not at altitude or sea level. Right? Wrong?
2) I assumed a car weight (W) of 3400 pounds. However, in a sensitivity analysis later I show that the major effect of car weight on rolling resistance washes out in the final analyses.
3) Barometric pressure (Pr) was at 26.5 in/hg
4) Ambient air temp (T) was at 900 F
5) Terminal speed (V) was 330 ft/sec (225 mph)
6) Car frontal area (Af) is 21 square feet
7) Tire pressures were at 50 psig
8) 5 th gear mechanical efficiency is 0.97 percent (N1) Right? Wrong?
9) differential gear mechanical efficiency is 0.99 percent (N2) Right? Wrong? These numbers were taken from " Fundamental of vehicle Dynamics", 1992.

What follows was an academic exercise for me and hopefully the answer to some of the clients questions on power needed. It is not intended to be a definitive answer to anything except our collective curiosity. To that end, if you find it useful, amusing or down right distasteful,GREAT! You may find errors, wrong assumptions, or other nastiness and I will appreciate you pointing them out to me. Complaints should be backed up by real data, through either analyses of your own or documented information. I am ameniable to changing my thinking iffn there is good reason to. Opinions, by themselves, just don't cut it.

Here we GO!
Note that funny looking drag coefficients result from the analyses, but that's due to the freaky rolling drag. It does not make any difference because the analyses performed is based on delta or difference numbers.

EQUATIONS

Aero Drag = DA = 1/2 * rho * V2 * Af * Cd
Rolling Drag = DR = fr * W
fr = fo + 3.24 * fs * ( V * (60/88) /100)2.5
rho = 0.00236 * (Pr/ 29.92) * ( 519 / (460 + T))
HP = D * V / 550 = (DA + DR) * V / 550

Before getting to far along, let me define the new terms used:

rho is the air density in slugs
fr is the rolling coefficient
Cd is the dimensionless drag coefficient we are looking for and then changing

DETERMINE

How much horse power is required to go 225 mph with a normally aspirated engine given the same climatic conditions. Aerodynamic modifications are made to the car: return headlights , grill, and rear deck to production configuration.

METHOD

Step 1) Correct the given sea level horsepower data to the Bonneville conditions, include mechanical transmission and rear end losses, to get horse power at rear wheels.

Step 2) Determine the overall drag when the car made a pass at terminal velocity in the altered class configuration.

Step 3) Calculate the rolling drag based on the terminal speed, car weight and rolling coefficients from figure 4.34 of the referenced text.

Step 4) Determine the aero component of total drag

Step 5) Determine the drag coefficient

Step 6) Change the drag coefficient by adding new items (cooling, head lights, rear spoiler removal)

Step 7) Detemine new aero drag based on new drag coeficient

Step 8) Determine new total drag by adding back the rolling drag

Step 9) Determine new horsepower requirement at same terminal velocity with new drag coefficient

Step 10) Determine new flywheel horsepower

Step 11) Correct horsepower to dyno sea level

Step 12) Perform sensitivity analyses for rolling drag.

SOLUTION

Step 1) HP at rear wheels

HP = ((HPsea level dyno ) * (Pr / 29.92) * (519 / (460 + T)) + HPnitrous) * N1 * N2
HP = ((500) * (26.5 / 29.92) * (519 / (460 + 90)) + 125) * 0.97 * 0.99
HP = 521 at rear wheels at Bonneville in altered car configuration

Step 2) Determine over all drag (D)

HP = D * V / 550
D = HP * 550 / V D = 521 * 550 / 330 D = 868.9 lbs

Step 3) Calculate rolling drag (DR)

DR = W * (fo + 3.24 * fs * ( V * (60/88) /100)2.5
DR = 3400 * (0.08 + 3.24 * 0.0015 * (330 * (60/88) /100)2.5
DR = 3400 * 0.1169
DR = 397.5 lbs
This seems very high! Do sensitivity analysis later to determine effect.

Step 4) Determine aero drag (DA)

DA = D - DR
DA = 868.9 - 397.5
DA = 471.4 lbs

Step 5) Determine altered car drag coefficient (Cd)

Aero Drag = DA = 1/2 * rho * V2 * Af * Cd
DA = 1/2 * [0.00236 *(Pr / 29.92) * (519 / (460 + T))] * V2 * Af * Cd
471.4 = 1/2 * [0.00236 *(26.5 / 29.92) * (519 / (460 + 90))] * 3302 * 21 * Cd
471.4 = 1/2 * [0.00197] * 3302 * 21 * Cd
471.4 = 2255.4 * Cd
Cd = 0.2090
This seems low due to the high rolling drag! I think it is ok though.

Step 6) Determine new drag coefficient (Cd - new)

Cd - new = Cd + Cd - cooling system + Cd - headlights + Cd - spoiler delete
Cd - new = 0.2090 + 0.0124 + 0.005 + 0.01
The cooling system, headlights and spoiler delete are my estimates based on text values
Cd - new = 0.2364

Step 7) Determine new aero drag (DA - new)

DA - new = 2255.4 * Cd - newDA - new = 2255.4 * 0.2364DA - new = 533.3 lbs

Step 8) Determine new total drag (Dtotal - new)

Dtotal - new = DA - new + DR
Dtotal - new = 533.3 + 397.5
Dtotal - new = 930.8 lbs

Step 9) Determine new rear wheel horsepower for Bonneville conditions (HPnew)

HPnew = Dtotal - new * V / 550
HPnew = 930.8 * 330 / 550
HPnew = 558.46

Step 10) Determine new flywheel HP (HPnew - flywheel)

HPnew - flywheel = HPnew / (N1 * N2)HPnew - flywheel = HPnew / (N1 * N2)HPnew - flywheel = 558.46 / (0.97 * 0.99)
HPnew - flywheel = 581.55

Step 11) Determine new dyno (corrected to sea level) horse power (HPnew - flywheel - sea level)

HPnew - flywheel - sea level = HPnew - flywheel * (rho0 / rhoalt)
HPnew - flywheel - sea level = 581.55 * (0.00236 / 0.00197)
HPnew - flywheel - sea level = 697 hp

Step 12) Perform sensitivity analyses for rolling drag
.

let DR = 200 lbs (instead of the 397.5 lbs calculated) then DA = 868.9 - 200 = 668.9 lbs (instead of the 471.4 lbs calculated) and 668.9 = 2255.4 * Cd hence Cd = 668.9 / 2255.4 = 0.29623 therefore Cd - new = 0.29623 + 0.0124 + 0.005 + 0.01 = 0.3236 also DA - new = Cd - new * 2255.4 = 0.3236 * 2255.4 = 729.8 lbs so DT - new = DA - new + DR so DT - new = 729.8 + 200 = 929.8 lbs The previously calculated new total drag was 930.8 lbs using the very high rolling drag vs 929.8 lbs using an assumed rolling drag 50% less. In this case, there is insensitivity to rolling drag. I believe this vindicates some of the assumptions I have made in other analyses also.

This was a serious effort to try and determine how much a competitor was going to have to increase his normally aspirated horsepower to meet a stated speed goal after changing his car to meet a different class and given similar climatic conditions at Bonneville. It was fo rme fun, but maybe not for him. If you use a similar technique or the same methodology, just remember, your milage may vary :^}.

Copyright (C) 1998 - 2004, all dates inclusive, L.E. Mayfield - All Rights Reserved


2023 Calendar (in progress)

Audio Streaming & Archives
Next Live Event...
Speedweek starts
Friday Aug. 5

Bonneville SaltCam

New Chat Room! Non-Java!
Tues. 7PM MT
Register, it's easy....
(just no spaces in your name)
....and give it a try!

Bonneville Lodging

Save the Salt






Landracing.com
Thanks YOU!




Aussie Invader Site
(Monthly Newsletter)



Follow the Team...

Introduction

Episode 1) Overview Video

Episode 2) The Transmission

Episode 3) The Engine

Episode 4) The Drive Train

Episode 5) Body and Paint

Episode 6) Dyno Run




Wendover, UT