Is body weight equivalent to bicycle weight?

On Fri, 22 Jul 2005 19:43:48 GMT, "Bruce W.1"
wrote:

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.


opinion: if you're talking about lard around your waist etc. then
I'll vote taking it off your body is going to have greater benefits.

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:.

Ron Ruff wrote:
Bruce W.1 wrote:

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?


The only time when a part of the bike weight is more significant is
when accelerating. Then the weight of rims and tires (and a fraction of
the spoke weight) is twice the effect of weight elsewhere.

I just don't see how this can be so. On what physical principles are
you basing this?

snip
-Ron



--
Paul M. Hobson
Georgia Institute of Technology
http://www.underthecouch.org
..:I go to GT, I like numbers:.

On Fri, 22 Jul 2005 19:43:48 GMT, "Bruce W.1"
wrote:

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.

Dear Bruce,

The army saying illustrates the principle.

When we walk, weight on our back moves along at a steady
pace, much like our bicycle frames doing the same thing when
we ride.

But weight on our feet must be swung forward by our legs,
somewhat like the wheels on our bikes, which must spin as
well as move forward.

With every step, the foot hits the ground, comes to a halt,
trails behind and provides a forward push, and is then
whipped forward (and upward) again at roughly twice the
walker's speed for the next step.

Add a pound or two on the end of your hind leg and the
repeated violent acceleration and raising becomes much more
difficult. Weight added to the hands would be just as
obnoxious, since the hands are constantly waving back and
forth during walking and running.

(If you're not sure about the odd business of the foot
coming to a halt with every step, even at Olympic sprinting
speeds, step through a puddle and then look back at your wet
footprints. They show that your foot never slipped when it
was touching the ground, just as the wet track left by your
rubber tire will show that your tire was not slipping when
it touched the pavement at 20 mph. If the feet and the tire
weren't slipping against the ground, then they must have
been equally motionless.)

Rotating mass like the wheel (and the crank) adds an extra
effort. Consider a bicycle sliding across the usual
imaginary frictionless physics pond with its wheels locked.

Everything--including the wheels--is going 20 mph across the
pond and therefore required X amount of force to reach that
speed from a standing start. To spin the wheels up to 20 mph
would require additional force.

But this extra force required for rotating isn't much. Flip
a bike upside down, crank one pedal in high gear, and you
can spin the heavier rear wheel up to 30 mph in a few
moments with a couple of quick heaves of one feeble arm.

The other arm would serve to spin the front wheel up to the
same speed, if we could attach a crank and chain to it.

Another way to judge the effort involved is to remember that
no rider can spin a real bike up from 0 to 30 mph in under
five seconds on the flats.

Now take this already minor effort and reduce it to just the
difference between a light and and a heavy wheelset. It
could still be the difference between winning and losing in
a sprint, where photo finishes can be required:

http://www.velonews.com/images/details/8436.11841.f.jpg

But it makes scarcely any difference to the rest of us. Much
of the claimed difference in feel between lighter and
heavier equipment is simply expectation--when something
costs a thousand dollars more, we tend to feel a difference,
whether it's there or not.

Two common scenarios are often raised with the minor
difference between light and heavy wheelsets and lighter and
heavier bikes.

First, the same force accelerates a larger mass more slowly,
so the heavier bike should be slower to reach the same
cruising speed on the level (and even slower if the extra
mass must also be spun up like a wheel or crank).

But the masses involved are trivial to start with.

Typically, riders are talking about pound or two less
weight, which sounds impressive on a sub-20 pound bicycle
because the weight is being reduced 5-10%.

However, the mass being accelerated includes a roughly 150
pound engine--heavy bike and rider together will weigh 170
pounds, while the 2-pound lighter bike will weigh 168
pounds, only 1.2% less.

On the flat, that means only that the rider will accelerate
to the same cruising speed 1.2% more slowly--and
acceleration to cruising speed is a very small part of a
typical ride.

After that, we mostly waver at a snail's pace, accelerating
and decelerating slightly.

Some people want to argue that the constant tiny
accelerations must add up, but this really isn't the
case--the constant tiny accelerations are pretty much
balanced by the constant tiny decelerations, where the
heavier bike slows down more slowly than the lighter bike
that sped up more quickly.

(Any long initial acceleration does tend to be unbalanced
because normally we throw away most of our hard-earned
momentum by touching our brakes. But that initial
acceleration was still only at a 1.2% disadvantage for about
30 seconds.)

So the acceleration difference on the flats just isn't going
to show up for typical small weight savings on ordinary
rides.

The second scenario is the dreadful task of hauling the
extra weight uphill. Again, the problem is that people think
of the extra weight in terms of the bike, not the total mass
that must be pedalled up the hill--the difference is a 168
pound mass versus a 170 pound mass.

Here's a calculator for speed:

http://www.kreuzotter.de/english/espeed.htm

Use hands-on-tops, 150-lb rider, 20-lb bike, 7% slope, and
10 mph. It predicts 287 watts are needed.

Change to an 18-lb bike, blank the watts, re-calculate, and
the prediction is that 284 watts are needed.

Change to a 15-lb bike, blank the watts, do it again for 10
mph, and prediction is 279 watts.

(Remember, after the initial acceleration to cruising speed,
it doesn't matter whether the weight was lost from the
rotating wheels or from the linear-motion-only frame.)

Not only is the slow-down effect of the extra 2 pounds small
(you're just pushing a slightly heavier weight up the same
surprisingly gentle 7% slope), but it's somewhat balanced by
the speed gained when you descend:

http://www.kreuzotter.de/english/espeed.htm

Put the same 150-lb rider on a 20-lb bike, put his hands on
the drops, set cadence and watts to 0, and roll him down a
-7% slope.

The prediction is 39.5 mph.

Less mass fighting the same wind drag lowers terminal
velocity, so it drops to 39.3 mph for an 18-lb bike.

And to 38.9 mph for a 15-lb bike.

So when you roll back down the mountain, you gain back some
of what you lost raising that extra pound or two.

In short, typical bicycle weight differences matter only for
accelerations, not cruising speeds on the flats. And the
differences are so small that they don't matter much even if
magnified by rotation.

http://www.kreuzotter.de/english/espeed.htm

Use hands-on-tops, set the power to 200 watts, the rider to
150 lbs and the bike to 20 lbs, and send him up a 7% grade
for 10 miles:

http://www.kreuzotter.de/english/espeed.htm

The prediction is 83 minutes, 20.0 seconds at 7.2 mph.

Reduce the bike weight by 2 pounds, and the time for a
10-mile 7% climb drops to 82 minutes, 11.5 seconds, a full
68.5 seconds faster at 7.3 mph.

A 68.5 second advantage in an hour and twenty minutes
matters to a racer, but the rest of us aren't likely to
notice it. At 7.3 mph, it works out to about a 730-foot lead
after ten miles uphill.

For fun, recalculate with a 1 mph headwind--the 2-lb lighter
bike is right back to 83 minutes, 20.0 seconds.

Carl Fogel

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

Paul Hobson wrote:
Ron Ruff wrote:

The only time when a part of the bike weight is more significant is
when accelerating. Then the weight of rims and tires (and a fraction of
the spoke weight) is twice the effect of weight elsewhere.

I just don't see how this can be so. On what physical principles are
you basing this?

The wheels must be accelerated rotationally, as well as linearly (along
with the rest of the bike/rider). The extra force required to "spin up"
the tire and rim is the same as the force required to accelerate it
linearly, since the radius times the angular speed is equal to the
linear speed. So for acceleration, the rim and tire are double the
importance of weight elsewhere.

Extra energy is needed to "spin up" all rotating parts on a bike, but
angular speed x radius x mass of other parts (like hubs and pedals) is
small compared to the rims and tires.

-Ron

"Ron Ruff" wrote:

Paul Hobson wrote:
Ron Ruff wrote:

The only time when a part of the bike weight is more significant is
when accelerating. Then the weight of rims and tires (and a fraction of
the spoke weight) is twice the effect of weight elsewhere.

I just don't see how this can be so. On what physical principles are
you basing this?

The wheels must be accelerated rotationally, as well as linearly (along
with the rest of the bike/rider). The extra force required to "spin up"
the tire and rim is the same as the force required to accelerate it
linearly, since the radius times the angular speed is equal to the
linear speed. So for acceleration, the rim and tire are double the
importance of weight elsewhere.

Thing is, since you can give a loose wheel a spin with a finger and
easily get it up to "sprinting speed", it's pretty clear that this
"acceleration tax" is minimal.

Not only that, but unless you come to a stop, you really don't pay the
"tax" again, since any energy you put into the wheel via acceleration
is there to prevent DEceleration (think of it as a flywheel and it
becomes more obvious). This is certainly the case when you're JRA.

Mark Hickey
Habanero Cycles
http://www.habcycles.com
Home of the $695 ti frame

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
(_)/ (_) |

Joe Riel wrote:
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

I'm 98% sure you just took me to school...but I'm gonna have to think
about it tomorrow morning on my weekly ride to the bank.

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