Chris K Posted March 14, 2003 Share Posted March 14, 2003 Im guessing that there is a clever equation calculating; how many BHPs are needeto push a car with drag co-efficent through the air at a given speed.. Im trying to work out what gear ratios I should have and I think this should be a important factor (waits to be corrected)... so that the gearing is not soo low that the engine cannot generate the required BHP to push you throught the air at 70mph in 5th when you are only doing (example) 3000rpm or am I barking up the wrong tree.. Chris Now, Beakie, we'll just flip this switch and 60,000 refreshing volts of electricity will surge through your body. Ready?.. See my CAR here Link to comment Share on other sites More sharing options...
Super_Rich_Bernie Posted March 14, 2003 Share Posted March 14, 2003 You need to know frontal area too. Power to overcome air drag = fA x Cd x 0.00256 x mph cubed / 375 Remember to factor in the rolling resistance. Jonathan Link to comment Share on other sites More sharing options...
Chris K Posted March 14, 2003 Author Share Posted March 14, 2003 -------------------------- Power to overcome air drag = fA x Cd x 0.00256 x mph cubed / 375 Remember to factor in the rolling resistance. -------------------------- ok.. that looks like what aI was looking for.... but now how does it work.. 😬 Cd being drag co-efficent???? what is the frontal area of a 7 with cycle wings + full screen??? how do i factor in rolling resistance??? /numpty mode Chris Now, Beakie, we'll just flip this switch and 60,000 refreshing volts of electricity will surge through your body. Ready?.. See my CAR here Link to comment Share on other sites More sharing options...
CHRIS CLARK Posted March 14, 2003 Share Posted March 14, 2003 Have a look in the Tony Weale book "Lotus Seven" Restoration/preparation/maintenance. Page 150 is dedicated to this sort of info + a graph for non techies to be able to see the visual impact of BHP v speed/drag. Link to comment Share on other sites More sharing options...
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