your simple aglebra shows that drag has more of an effect.
so how can both answers be correct with different outcomes. Help me out here tortoise
If your algabra equation isn't showing whether drag or power is a more significant factor to top speed then what is it showing ?
Remember earlier when I mentioned you had to be very careful about definitions?
The problem here is actually the word problem not the math. You asked what on the surface appeared to be a very simple question.
You basically asked "if you make equal changes to drag and power" is one more effective than the other?
The word problem that taper is holding to is actually a flawed statement with respect to your intent I believe.
A 25% reduction in drag is not equivalent to a 25% increase in power because of the way fractions and percentages work.
It suddenly dawned on me when I recalled tortises comment early in the thread regarding percentages of change and when I used a 50% change it was very obvious logically what was going on.
Lets go to the 50% change because it is a bit more intuitively obvious.
If you cut drag by 50% you are reducing drag to 1/2 of its original value. Increasing power 50%
IS NOT an equivalent change, the equivalent change is doubling power. That becomes obvious when you look at the fractions instead of as percentages. Drag reduction to 1/2 its original value if you invert that fraction it becomes 2/1 ---------
that is the correct "equivalent" change in power, not a 50% increase.
1.5 x power is not the reciprocal of 0.5
If you use whole number fractions instead you can easily see that a 50% increase in power is really 3/2 of the original power, where the true reciprocal is 2/1 or 4/2.
Likewise the 25% reduction in drag is really 3/4 of the original so the reciprocal change to power is 4/3 = 1.33333 not 125%.
The word problem as stated does not actually do what you intended which was to determine what equivalent changes in power and drag rate with each other. It is a subtle problem with percentages that bites people all the time!
In very small percentage changes the error is so small it can be pretty much ignored, for example a 2% reduction in one thing vs a 2% increase in another.
The 2% reduction if reduced to a common fraction is 49/50, so its reciprocal is 50/49 which is 1.020408. The error being so small it gets lost in most cases. If however you make big changes however that error quickly grows.
Taper is correct
if we stick to the word problem exactly as stated.
If how ever you restate the word problem to match what I believe was your intent, and how I saw your first query is "if you make equivalent changes to drag and power do they produce the same results?".
Restated that way and doing the proper math to find the reciprocal of the drag reduction they are exactly equal. BUT you will notice that your power increase as a percentage will always need to be larger than the equivalent reduction in drag.
In the real world land racing cars are always running at the limits of the technology and the reductions in drag that are possible within the rules and the increases in power available in the rules are usually small. Drag reduction gets very much harder as you drop the drag, where power is a bit more open to increase if your willing to change engine displacement, add a blower or switch to a fuel class etc.
This is why my first calculations where the terminal speed was held constant checked out exactly but when we changed the form of the calculation to vary drag and power by a percentage and see how the terminal speed changed we started to see the error.
Larry
