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* The following is one man's opinion. It is not gosphel, nor is it *
* to be taken as law. If you disagree, then so be it. If there are *
* other factors involved or experience with real hard data then I *
* be interested in receiving the data. *
* *
* Warranty is now void. Your milage may vary. *
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I'll throw my dog into the fight! My source data is "Fundamental of Vehicle
Dynamics" by Thomas D. Gillespie, copyright 1992, ISBN 1-56091-199-9.
There has been a question of drive train efficiencies and how much power is
lost in the process of driving over the salt, concrete or the dry lake bed.
First, efficiency of the drive train.
A typical vehicle is used as an example problem in the referenced text.
Here is the data as presented:
Tranny - gears 1 2 3 4 5
inertia(in-lb-sec^2) 1.3 0.9 0.7 0.5 0.3
ratios 4.28 2.79 1.83 1.36 1.00
efficiency 0.966 0.967 0.972 0.973 0.970
Final Drive (rear end) inertia 1.2 in-lb-sec^2
ratio 2.92
efficiency 0.99
For those running the Halibrand type rear ends where you have an additional
set of spur gears, and additional efficiency must be included. For spur
gears the mechanical efficiency is around 0.97.Because there are additional
bearings these should also be factored in.
This is obviously a manual transmission but notice that the total
efficiency from wheel touching the ground to the flywheel, in the final
gear is 0.9603 or a loss of only 4%. With a quickchange rear the
efficiency is 0.9315 for a loss of 6.85%. So adding the quick change
requires an additional 2.85% horsepower. This loss of horsepower to drive
he rear end has to be factored against the ability to vary the overall gear
ratio quickly to suit track conditions.
In an automatic the results are similar iff a lockup converter is used. The
converter slips otherwise and results in heat losses in stirring the fluid.
But not in a lock up at speed. Transmissions with high milage may suffer
some slippage in the clutch plates, but we all keep ours at top efficiency,
right?
Driving on the various surfaces is a different story. Just reasoning it out
would say that the dry lake bed (dirt) is the worst surface to drive on
because the tire has to "plow" it's way. This causes some high drag at the
wheel. Concrete is probably the best, short of the drag strip (which has
traction modifiers added to get coefficient of friction greater than 1).
Salt is somewhere in between.
>From the same reference several simple equations are offered. One such is
Fr = Fo + 3.24 Fs *(v/100)^2.5 where Fr is rolling resistance coefficient.
(Basic Coefficient + 3.24 times Speed Effect Coefficient times
Speed(mph) divided
by 100, raised to the 2.5 power)
This is for a generally good road, asphalt.
For tires which have had time to warm up and are inslated to at least 45
psig, Fo = 0.008 while Fs = 0.002
Sloving at 200 mph...
Fr= .008 + 3.24 * 0.002 * (200/100)^2.5 = 0.04466 for asphalt
For concrete in good shape Fr = 0.02
For a "medium hard" surface Fr = 0.08
For sand Fr = 0.3
Rolling load is = Fr * vehicle weight (say 2000 lbs) = 0.04466 * 2000 =
89.31 pounds (asphalt)
= Fr * vehicle weight = 0.02 * 2000 = 40 pounds (concrete)
= Fr * vehicle weight = 0.08 * 2000 = 160 pounds (medium
hard surface)
= Fr * vehicle weight = 0.3 * 2000 = 600 pounds for sand
I don't believe the dry lakes should be classified as sand, but more likea
medium hard surface when the vehicle is at speed. I think that the tires
almost hydroplane at speed because the silt surface acts as a fluid at
those speeds. I think the salt is worse than concrete but better than a
medium hard surface, say Fr = 0.06.
= Fr * vehicle weight = 0.06 * 2000 = 120 pounds (salt)
= Fr * vehicle weight = 0.10 * 2000 = 200 pounds ( dry lake
bed)
If we use my Sunbeam as an example of horsepower required on asphalt, then
Cd * Frontal Area = 8.13
So horsepower for aero drag = (1/2)* rho * (V ft/sec)^2 * Cd * A
= (0.5) * 0.002377 * (200 *1.467)^2 * 8.13
= 831 pounds drag
So, if you take a percentage increaae required due to rolling loads then
Power increase % on asphalt is = (89/ 831) * 100 = 10.7 %
Power increase % on concrete is =(40 / 831) * 100 = 4.8 %
Power Increase % on "salt" is = (120 / 831) * 100 =14.4 %
Power increase % on dry lake bed = (200 / 831) * 100 = 24.06 %
Of course, as speeds go up, drag goes up. Converse is true for slower
speeds. And while the numbers are different than those spefied in the
original post, they do show that the surface plays a tremendous role in
ascertaining the horsepower needed.
mayf( can't resist a math problem....)
At 05:49 PM 1/12/00 -0800, Skip Higginbotham wrote:
>Hi Team,
>Over the past few weeks the subject of "how much HP" to go "how fast" has
>received much attention and Mayf's chart report is an excellent vehicle to
>get to the required horsepower....at the rear wheels on the ground. Thank
>you Mayf you have really helped me.
>If I have misinterpreted the charts......please somebody correct me!
>
>Now how about sharing some ideas and opinions (data?) on how much Hp is
>lost between the crankshaft flange and the ground at, say Bonneville with
>dry salt (more of less) and at El Mirage on a good day and asphalt (as a
>reference).
>
>I think that the loss is 65% total at B-ville, 75% at El Mirage and 40-45%
>on asphalt.
>Any takers?
>
>Skip Higginbotham, Bastrop, Tx (where it is now 80 degrees) at 5:43 PMCST
>Air conditioning in January???????
>
>
L.E. Mayfield
124 Maximillion Drive
Madison, Al. 35758-8171
1-256-837-1051
old >>>>>>>> http://www.hsv.tis.net/~mayfield
new >>>>>>>> http://home.hiwaay.net/~lemayf
DrMayf@AOL.com
lemay@hiwaay.net <<<<preferred
Sunbeam Tiger, B9471136
Sunbeam Alpine Bonneville Land Speed Racer,
'66 Hydroplane Drag Boat (390 FE)
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