geelhoed Posted December 11, 2006 Posted December 11, 2006 Lacking the means of measuring this correcly, an educated guess would be for my 120bhp zetec, 3.7 seconds. Is this in line with with your measurements? Quote
oldman Posted December 11, 2006 Posted December 11, 2006 Kinell geelhoed...ang on......E= MC t'power of 2.......KINELL! That would be clogging it. Quote
geelhoed Posted December 11, 2006 Author Posted December 11, 2006 Well actually I assumed a constant power P ans as the kinetic energy is porportional to the square of the velocity we'd expect the velocity to be a function of the square root of t. With boundary conditions that for t=0 u=0 and the 0-60 time of 6 seconds we can find this estimate of 3.7 seconds. Quote
Morbius Posted December 11, 2006 Posted December 11, 2006 Well actually I assumed a constant power P ans as the kinetic energy is porportional to the square of the velocity we'd expect the velocity to be a function of the square root of t. With boundary conditions that for t=0 u=0 and the 0-60 time of 6 seconds we can find this estimate of 3.7 seconds. Did you factor in the drag increasing with the cube of the velocity (through the air not over the ground) owing to the lack of smooth flow around the bodywork? Quote
Edd Posted December 11, 2006 Posted December 11, 2006 what is the drag coefficient of a westfield? Quote
pistonbroke Posted December 11, 2006 Posted December 11, 2006 Don't forget ..... heat of the beat = throb of the knob , when the angle of dangle is constant ! Quote
V8grunt Posted December 11, 2006 Posted December 11, 2006 Don't forget ..... heat of the beat = throb of the knob , when the angle of dangle is constant ! Only if you "pass the dutchy on the left hand side"! Quote
Keith@FM Posted December 12, 2006 Posted December 12, 2006 And I thought I could read English. Ouch, my brain hurts. My car was clocked by Car and Driver at 4.65 seconds for 0-60. 50-70 took 2.88 seconds. Plug that into your calculator. That's a 175 hp MX-5 donk and they wouldn't let me take the windscreen off. Quote
geelhoed Posted December 12, 2006 Author Posted December 12, 2006 No need to mock the guess, I told you it was an educated one! Aerodynamic drag in a sprint is fairly negligable the correction factor for the time for a sprint to speed V is 1+0.4*(V/Vmax)^3 where Vmax is the topspeed of your car. 4.65 to 60 gives us a velocity function of U = 60*sqrt(t/4.65) So for the time to 50 you need 3.23 sec and to 70 6.33 which gives 3.1 for 50-70. So I guess we need a 7% correction for the slower acceleration off the line than modelled! Thanks Keith for the useful info! Any others? Quote
geelhoed Posted December 12, 2006 Author Posted December 12, 2006 Oh, the Cd of a seven is around 0.7 Easily calculated(estimated) by taking the top speed (112mph) and the peak power (120bhp) P=0.5*rho*A*Cd*U^3 rho, the density of air is about 1.2kg/m^3 Cd=2/1.2kg/m^3*120bhp / (112mph)^3/1.7m^2 = 0.68 Quote
Matt Seabrook Posted December 12, 2006 Posted December 12, 2006 Easily calculated(estimated) by taking the top speed (112mph) and the peak power (120bhp) P=0.5*rho*A*Cd*U^3 rho, the density of air is about 1.2kg/m^3 Cd=2/1.2kg/m^3*120bhp / (112mph)^3/1.7m^2 = 0.68 If you say so Quote
geelhoed Posted December 12, 2006 Author Posted December 12, 2006 Sure I know that I am neglecting the linear and quadratic terms, but at topspeed that's seems justifyable. Quote
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