How Drag Reduction Effects Velocity 1

[JCT81]

Hello,

There is an industry standard in cycling that states for 100g of drag reduction you will save 40sec over a 40km distance if you are traveling 30mph. I am trying to prove this mathematically and am having trouble. If I were to compare two scenarios, the first with a cyclist with a CdA of 0.6 and the second with a CdA of 0.4, there would be an increase in velocity for scenario #2 assuming the riding conditions were the same. Is there an equation to determine the increase in velocity when there is a reduction in CdA?

Take care,

Jon

 

[IRstuff]

http://en.wikipedia.org/wiki/Drag_equation

Drag force is proportional to the square of velocity.

TTFN
faq731-376

 

[rb1957]

as above aero. drag is proportional to V^2. see wiki, the full expression is D = (1/2*rho*V^2)*Cd*A ... rho is reasonably fixed, your problem says that Cd*A is reduced.

i think to understand your problem (sounds pretty basic to be a professional (ie non-student) problem) you need a dynamics approach.

assume you have a motive force F, aero drag is proportional to V^2, so you'll accelerate (increase velocity) untill F = drag and then continue at constant velocity (30 mph), and so your time to go 40km (love the mixed units ! ... another clue that this is non-professional). you might simplify the problem by assuming constant velocity.

now reduce your drag force by 100g (again, the units !). to understand what this means you need to assume a velocity, maybe 30 mph ? in any case, lower drag means a higher speed, shorter time. but then you have two values of Cd*A ??

since you don't know the original drag force, (actually you do if you know Cd*A and V) maybe the point is to figure out the initial drag force such that 100g less force results in 40sec less time. You know the initial time (40km/30mph) and the initial velocity (30mph). Assume an initial Cd*A which'll give you drag force at 30mph which i'd also say is the driving force applied. Then reduce the drag (at 30 mph) by 100g ... how much faster can you go if you have the same driving force ? how much time will you save over 40km ? knowing that you want to save 40sec tells you how much your speed has increased (t = 40km/30mph ... v = 40km/(t-40) ... again, watch the mixed units.

a confusing answer to a confusing question ... as i reread to OP i added things ...

 

[drawoh]

JCT81,

You need to state your problem clearly and correctly.

For starters, grams are a unit of mass, not force. You reduce your drag by one Newton, not 100g. This is nitpicking, but it will mess up any attempt at the unit balancing you need to do when you mix mph with kilometres.

If two bicycles ride 40km at 30mph, and one bicycle has 1N less drag, the bicycles will arrive at the same time. The rider of the lower drag bike will be less tired, or the other cyclist will have taken more steroids.

You need to assign a standard output power to your cyclists, assign some value of rolling resistance, and a nominal drag force. From that, you can work out the relative velocities.

--
JHG