View Full Version : Amusing intersection of physics and cycling

MattTuck

04-16-2015, 07:08 PM

Potential Energy = Mass x Gravity x Height

I've set two new personal records for strava segments this year, both have been downhill segments.

Since the hills are the same height, and I'm pretty sure gravity has remained constant... we have identified the culprit. :help:

Louis

04-16-2015, 07:17 PM

You've started to use a fairing?

http://www.green-racer.com/wp-content/uploads/2009/12/boat-tail-geo-metro1.jpg

http://i47.photobucket.com/albums/f152/uprightmike/CervelowRAAMfairing2.jpg

Cornfed

04-16-2015, 07:28 PM

I prefer to blame climate change-based evolution ... the longer winters where you are has necessitated an increase in the number of fat cells in your body as it fights to survive in lower and lower temperatures.

In the future, only the fat survive. ;)

montag

04-16-2015, 07:32 PM

You've started to use a fairing?

SO AERO! :banana:

mhespenheide

04-16-2015, 09:07 PM

PE=mgh, but KE=(1/2)mv^2

mass cancels.

Clearly you've gained skill in descending!

Louis

04-16-2015, 09:30 PM

PE=mgh, but KE=(1/2)mv^2

mass cancels.

Unless his mass has resulted in a rounder, more aero, shape and therefore less drag...

MattTuck

04-16-2015, 10:26 PM

I was wearing some rapha shorts today. Apparently my descending got more epic.

cloudguy

04-17-2015, 01:58 AM

PE=mgh, but KE=(1/2)mv^2

mass cancels.

terminal velocity = sqrt(2mg/(rho_air*Cross_Sectional_Area*Drag_coefficient))

No escaping increased mass, unless drag is proportionally reduced, as Louis pointed out.

Treat the cyclist has a sphere: Cross_Sectional_Area=pi*r^2; m = (4/3)*rho_cyclist*pi*r^3.

terminal velocity = sqrt((8*rho_cyclist*g*r/(3*rho_air*Drag_coefficient))

Mass wins out, assuming rho_cyclist is constant.

In reality, of course, most mass is probably added to the front of the cyclist (beer belly), so Cross_Sectional_Area likely remains fixed. In that case, mass really wins out.

oldpotatoe

04-17-2015, 07:32 AM

terminal velocity = sqrt(2mg/(rho_air*Cross_Sectional_Area*Drag_coefficient))

No escaping increased mass, unless drag is proportionally reduced, as Louis pointed out.

Treat the cyclist has a sphere: Cross_Sectional_Area=pi*r^2; m = (4/3)*rho_cyclist*pi*r^3.

terminal velocity = sqrt((8*rho_cyclist*g*r/(3*rho_air*Drag_coefficient))

Mass wins out, assuming rho_cyclist is constant.

In reality, of course, most mass is probably added to the front of the cyclist (beer belly), so Cross_Sectional_Area likely remains fixed. In that case, mass really wins out.

Ya know, I was gonna write the same thing!!

SlackMan

04-17-2015, 08:17 AM

Ya know, I was gonna write the same thing!!

Don't be intimidated by those hard equations. They're pretty easy to derive.

http://www.cs.berkeley.edu/~luca/cs103-14/steptwo.jpg

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