Is body weight equivalent to bicycle weight?

Is body weight equivalent to bicycle weight? In other words, would
riding a bicycle that's five pounds lighter be the same as losing five
pounds off of your body weight?

There's an old Army saying; one pound on your foot (boot weight) is
equivalent to five pounds on your back. But I'm not sure what this has
to do with anything.

Thanks for your help.

there may be some difference between
rotating weight and non-rotating.

i e, wheels may matter more because they accelerate differently.

other than that, it should all be 'a pound is a pound'.

at least 1 lb more anywhere is a lb you have to move up a hill, and
accelerate from a stop.

no wait, i thought of another kind of weight - sprung [you]
vs unsprung [the bike].

if you get off the saddle for bumps, the bike will
'hit harder' if it;s heavier.

if you don;t get up, it;s all unsprung.

wle.

wle wrote:
there may be some difference between
rotating weight and non-rotating.

i e, wheels may matter more because they accelerate differently.

The rule of thumb that I've been told goes, "an ounce off the wheels
equals a pound off the frame."



other than that, it should all be 'a pound is a pound'.

at least 1 lb more anywhere is a lb you have to move up a hill, and
accelerate from a stop.

no wait, i thought of another kind of weight - sprung [you]
vs unsprung [the bike].

if you get off the saddle for bumps, the bike will
'hit harder' if it;s heavier.

if you don;t get up, it;s all unsprung.

wle.

"Eric Hill" wrote:

wle wrote:
there may be some difference between
rotating weight and non-rotating.

i e, wheels may matter more because they accelerate differently.

The rule of thumb that I've been told goes, "an ounce off the wheels
equals a pound off the frame."

Oh, it's getting even better! The old saying used to be: An ounce off the
wheels equals two ounces off the frame.

But it really isn't. For reasonable wheels and tires, and typical
accelerations, an ounce is an ounce regardless where it is. And yes, losing
5 pounds off your body is the same as getting a 5 pound lighter bike (but a
lot cheaper!).

Art Harris

Arthur Harris wrote:
"Eric Hill" wrote:


wle wrote:

there may be some difference between
rotating weight and non-rotating.

i e, wheels may matter more because they accelerate differently.

The rule of thumb that I've been told goes, "an ounce off the wheels
equals a pound off the frame."


Oh, it's getting even better! The old saying used to be: An ounce off the
wheels equals two ounces off the frame.

Hey now, it's time for the new saying to take over. It's much more
impressive!

-eric



But it really isn't. For reasonable wheels and tires, and typical
accelerations, an ounce is an ounce regardless where it is. And yes, losing
5 pounds off your body is the same as getting a 5 pound lighter bike (but a
lot cheaper!).

Art Harris

Arthur Harris wrote:

...And yes,
losing 5 pounds off your body is the same as getting a 5 pound
lighter bike (but a lot cheaper!).

Yeahbutt... The /bike/ won't gain it back!

Meats 'n Cheeses Bill

Arthur Harris wrote:
"Eric Hill" wrote:


wle wrote:

there may be some difference between
rotating weight and non-rotating.

i e, wheels may matter more because they accelerate differently.

The rule of thumb that I've been told goes, "an ounce off the wheels
equals a pound off the frame."


Oh, it's getting even better! The old saying used to be: An ounce off the
wheels equals two ounces off the frame.

But it really isn't. For reasonable wheels and tires, and typical
accelerations, an ounce is an ounce regardless where it is. And yes, losing
5 pounds off your body is the same as getting a 5 pound lighter bike (but a
lot cheaper!).

Art Harris

Art,

I was about to make a post disagreeing with you, but thinking about it,
you're more right than you give yourself credit for.

A rotating object's resistance to angular acceleration is called it's
moment of inertia (I). Whe

I=kmr²

with
k=some constant dependant on the distribution of the mass of the object
m=mass
r=radius of the object

Since the radii of wheels are pretty set now, you'll have to change the
mass distribution to make the wheels any easier to turn. Thanks for
making me revisit my freshman physics class!



--
Paul M. Hobson
Georgia Institute of Technology
http://www.underthecouch.org
..:you may want to fix my email
address before you send anything:.

Paul Hobson writes:

A rotating object's resistance to angular acceleration is called it's
moment of inertia (I). Whe

I=kmr²

with
k=some constant dependant on the distribution of the mass of the object
m=mass
r=radius of the object

Since the radii of wheels are pretty set now, you'll have to change
the mass distribution to make the wheels any easier to turn.

True, but you need to continue the analysis a bit further.

Increasing the angular velocity (w) requires a torque (T). The torque
is applied by a force at the axle, the lever arm is r. So the additional
force that the rider must apply is

f = (I*w')/r = I*(v'/r)/r = I*v'/r^2 = k*m*v'

The wheel radius completely drops out. The force required to accelerate
the wheel is m*v', so the total force is

ftot = (1+k)*m*v'

If all the mass of the wheel is at the rim (worst case), then k=1 and
we get the factor of 2 previously mentioned.

Note that small wheels don't accelerate any faster unless they weigh less.


Joe

On Fri, 22 Jul 2005 19:56:21 +0000, Eric Hill wrote:

wle wrote:
there may be some difference between
rotating weight and non-rotating.

i e, wheels may matter more because they accelerate differently.

The rule of thumb that I've been told goes, "an ounce off the wheels
equals a pound off the frame."

Yeah, I remember that, too. It's nonsense, but I do remember being told
it.

--

David L. Johnson

__o | Do not worry about your difficulties in mathematics, I can
_`\(,_ | assure you that mine are all greater. -- A. Einstein
(_)/ (_) |

Quoting Eric Hill :
The rule of thumb that I've been told goes, "an ounce off the wheels
equals a pound off the frame."

But that's obviously completely bogus. Even if all the mass of a wheel
was at the very edge, which it's not, it would be only twice as hard to
accelerate as non-wheel mass; and of course only a small proportion of
energy goes to acceleration anyway.
--
David Damerell Distortion Field!
Today is Gaiman, July - a public holiday.