Want to make sure I got this right.
Assume 25 mph requires 300 watts for simplicity.
25 mph equals about 11.2 m / s
Now I want to solve for X
300 watts = X (9.8 m/s^2) (11.2 m/s)
X = 2.73 kg
So if such and such wheel or aero helmet saves my 50 grams of drag at 25 mph that would be a 1.8% savings in energy at 25 mph?
Off the top of my head, 50 grams of drag would be roughly 5.5 watts. So yes, 5.5 watts is ~1.8% of the 300w.
-Physiojoe
what is all this “math” stuff? can’t you just FEEL the power?
c’mon Luke!
what is all this “math” stuff? can’t you just FEEL the power?
c’mon Luke!
I felt it yesterday!
powertap confirmed it though =)
HOPE I FEEL IT TODAY! CAT 4 HEEEROOOOOO
the math works in two ways; right and wrong.
i LOVE it when people seem to think (i’m not saying the OP thinks this, just an observation) if the aero helmet “saves” 5.5 watts, then that means instead of me only being able to put OUT 300 watts i can NOW put out 305.5watts! dude, SCORE!
you can still “only” put out 300watts… just now you will be going a little bit faster (insert official math for that here)
so, how much SPEED does an aero helmet get you over 25mph on 300watts?
To quote Barbie, Math is hard, let’s go shopping!
the math works in two ways; right and wrong.
i LOVE it when people seem to think (i’m not saying the OP thinks this, just an observation) if the aero helmet “saves” 5.5 watts, then that means instead of me only being able to put OUT 300 watts i can NOW put out 305.5watts! dude, SCORE!
you can still “only” put out 300watts… just now you will be going a little bit faster (insert official math for that here)
so, how much SPEED does an aero helmet get you over 25mph on 300watts?
If you ask instead, “how much faster with an aero helmet is a rider generating 300W, who rides 25 mph with a regular helmet?”, then:
0.2 to 0.4 mph faster, and
As long as the power is the same in both cases, the extra speed doesn’t depend on the riders power. It doesn’t matter if it’s 300W or 0W.
yeah, 0.2mph or ~30 seconds per 40k is about right
assuming helmet fits well and is worn properly.
ok so…i took the original math and worked it backwards a little bit.
x='d 2.73kg the NEW x is 2.725 thanks to the 50g savings with the helmet
so
y=new speed
300w=2.725k(9.8m/s^2)(Y)
300/9.8/2.725=Ym/s=11.2338
11.2338m/s=25.129294918mph
so… .129mph gain with aero lid
Forgive my ignorance. “Grams” of drag? Why are grams used to describe force? I tried to look this up on the internet without success. Plus drag is a function of velocity, density, and viscosity, so it seems like any reported value should contain at least the air temperature.
I think it’s grams times 9.8 m/s^2 and not just grams. I think.
That’s correct. And as for fluid drag on bicycle riders–it’s generally assumed to be inviscid (ie skin drag is neglected). The other factors are accounted for in density, which is part of the CdA equation.
300w=2.725k(9.8m/s^2)(Y)
300/9.8/2.725=Ym/s=11.2338
11.2338m/s=25.129294918mph
so… .129mph gain with aero lid
You might be able to get all your answers with less assumptions using the software on http://www.analyticcycling.com
Edit - thanks for the clarification
I assume 9.8 m/s^2 because I think the left axis should be in newtons when you see those graphs about wheels and bikes. Since they use grams instead of newtons,** I assume** 9.8 m/s^2 for acceleration. The could have used lbs instead of grams, that would make more since. And when people use lbs and grams as if they measure the same thing, often 9.8 m/s^2 is used to compare them.
And to quote Doc Cog:
"Memorize this (I did):
0.1 lbs (50 g) of drag (at 30 mph) = 0.5 s/km = 5 W = 0.005 m^2 CdA = 0.0005 Crr
The above rule-of-thumb (which refers to changes) is based on various assumptions/rounding everything to a 1 or a 5, but is fairly accurate and hence quite handy when doing in-your-head (or back-of-the-envelope) calculations."
.
300w=2.725k(9.8m/s^2)(Y)
300/9.8/2.725=Ym/s=11.2338
11.2338m/s=25.129294918mph
so… .129mph gain with aero lid
Why 9.8? Gravity doesn’t really come into it. Assuming zero wind shouldn’t it be 11.23^3 ?
You might be able to get all your answers with less assumptions using the software on http://www.analyticcycling.com
1 gram-force equals 9.80665 millinewtons. It’s the weight due to gravity of a mass of 1 gram.