Quote:
Originally Posted by dcb
I saw a coastdown calculator and wanted some help understanding it before I try and integrate it into the mpguino.
Basically you (or your computer) times a coastdown at high speed (say 55mph to 50mph) and another at low speed (say 7mph to 2mph) and put the speeds and times and weight in and it tells you your cDA and rolling resistance. But I'd like to understand what it is doing a little better. I appreciate that they have approximated things like pressure and temperature for the sake of simplification, but can anyone determine what the "6" on the cDA line represents or the "28.2" on the rr line actually is?
The crux of it is:
a1=(va1-vb1)/t1;
a2=(va2-vb2)/t2;
v1=(va1/2)+(vb1/2);
v2=(va2/2)+(vb2/2);
cDa=((6*m)*(a1-a2))/(v1*v1-v2*v2);
rr=(28.2*(a2*v1*v1-a1*v2*v2))/(1000*(v1*v1-v2*v2));
abbreviations:
a1: high speed deceleration
a2: low speed deceleration
v1: high speed average velocity KPH
v2: low speed average velocity KPH
cDA: coefficient of drag * area
rr: Rolling resistance
va1: high speed start KPH
vb1: high speed end KPH
va2: low speed start KPH
vb2: low speed end KPH
m: mass in Kilograms
Other conversion factors
1 mile = 1.609344 KM
1 pound = 0.45359237 kilograms |
Well, the numbers that you have quoted are for 'metric' units. Read on for explanations...
In standard metric units (mks - metres/kilograms/seconds) there should be 2/1.2 in stead of "6" and 1/9.8 instead of "(28.2.../(1000...".
So, re-written:
cDa=((2*m)*(a1-a2))/((1.2)*(v1*v1-v2*v2))
rr=(a2*v1*v1-a1*v2*v2)/(9.8*(v1*v1-v2*v2))
From the (hand-derived) math:
- the "2/1.2" is the reciprocal of the standard 0.5*'rho' ['rho' being the density of air @ 1.2 kg/m^3]
- the 1/9.8 is the reciprocal of acceleration due to gravity [9.8m/s^2]
As you have stated the speed units as km/h, there are factors of 3.6 (1 km/h = 1000 m / 3600 s = 1m/3.6s). So, used appropriately, one will achieve the original equations with the "6" and "28.2".
And, as you have already suspected, the results will be an indication of real values but not absolute. I don't feel like doing the calculus to get the exact results - too time consuming and the external influences when taking the readings will not justify its precision!
Also note that there is another component that is often overlooked: a 'viscous' drag - a velocity component in the equation:
F = 'rho'/2*cDA*v^2 + cV*v + m*g*cF
BTW = this is bringing back memories of my master's thesis...