Help! Cycling Physics Question...

Quoted message said:

Greetings...We have a discussion here and need some educated advice...

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1
hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...

On a flat track at constant speed: none whatsoever. If there are any hills: a ridiculously
small amount.

Weight only comes into play when accelerating or going uphill.

Quoted message said:

and how much less mass have you moved over that same hour.

Erm, 2 grams?

Andy

"TCG" <[email hidden]> wrote in message "]news:[email hidden]...

Quoted message said:

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1
hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have you
moved over that same hour.

If your cadence is 60rpm the question is entirely irrelevant ;-)

--
Guy
===

WARNING: may contain traces of irony. Contents may settle after posting.
chapmancentral.comchapmancentral.comOpen ↗

"Andy Koppe" <a n d y @ d c s . e d . a c . u k> wrote in message
"]news:[email hidden]...

Quoted message said:

Quoted message said:

Greetings...We have a discussion here and need some educated advice...

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on
1 hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...

On a flat track at constant speed: none whatsoever. If there are any

hills:

Quoted message said:

a ridiculously small amount.

Weight only comes into play when accelerating or going uphill.

Quoted message said:

and how much less mass have you moved over that same hour.

Erm, 2 grams?

Andy

Moving your shoes round in a circle *is* accelerating them.

Whazzarke said:

"Andy Koppe" <a n d y @ d c s . e d . a c . u k> wrote in message
"]news:[email hidden]...

Quoted message said:

Quoted message said:

Greetings...We have a discussion here and need some educated advice...

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on
1 hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...

On a flat track at constant speed: none whatsoever. If there are any

hills:

Quoted message said:

a ridiculously small amount.

Weight only comes into play when accelerating or going uphill.

Quoted message said:

Moving your shoes round in a circle *is* accelerating them.

I don't think so, because they keep moving at the same speed.

Moving them up and down doesn't matter either, because the weight on the one pedal balances the
weight on the other.

Andy

"Andy Koppe" <a n d y @ d c s . e d . a c . u k> wrote in message
"]news:[email hidden]...

Quoted message said:

Whazzarke said:

"Andy Koppe" <a n d y @ d c s . e d . a c . u k> wrote in message
"]news:[email hidden]...

Quoted message said:

> Greetings...We have a discussion here and need some educated

advice...

Quoted message said:

Quoted message said:

Quoted message said:

>
> If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based
> on 1 hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...

On a flat track at constant speed: none whatsoever. If there are any

hills:

Quoted message said:

a ridiculously small amount.

Weight only comes into play when accelerating or going uphill.

Quoted message said:

Moving your shoes round in a circle *is* accelerating them.

I don't think so, because they keep moving at the same speed.

Moving them up and down doesn't matter either, because the weight on the

one

Quoted message said:

pedal balances the weight on the other.

Andy

If I may refer the Honourable Gentleman to the first paragraph of:

physicsclassroom.comU6L1b.htmlOpen ↗

It's to do with change in velocity, not speed...

Whazzarke said:

"Andy Koppe" <a n d y @ d c s . e d . a c . u k> wrote in message
"]news:[email hidden]...

Quoted message said:

Whazzarke said:

"Andy Koppe" <a n d y @ d c s . e d . a c . u k> wrote in message
"]news:[email hidden]...
> > Greetings...We have a discussion here and need some educated

advice...

Quoted message said:

Quoted message said:

> >
> > If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based
> > on 1 hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...
>
> On a flat track at constant speed: none whatsoever. If there are any
hills:
> a ridiculously small amount.
>
> Weight only comes into play when accelerating or going uphill.

Quoted message said:

Moving your shoes round in a circle *is* accelerating them.

I don't think so, because they keep moving at the same speed.

Moving them up and down doesn't matter either, because the weight on the

one

Quoted message said:

pedal balances the weight on the other.

Andy

If I may refer the Honourable Gentleman to the first paragraph of:

physicsclassroom.comU6L1b.htmlOpen ↗

It's to do with change in velocity, not speed...

May I then refer the Honourable Gentleman on to the next part of the lesson concerning the
"Centripetal Force Requirement":

physicsclassroom.comU6L1c.htmlOpen ↗

In the case of bicycle pedals, the centripetal force is provided by the cranks.

Andy

In message <[email hidden]>, "Just zis Guy, you know?"
<[email hidden]> writes

Quoted message said:

"TCG" <[email hidden]> wrote in message "]news:[email hidden]...

Quoted message said:

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1
hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have
you moved over that same hour.

If your cadence is 60rpm the question is entirely irrelevant ;-)

Well, the answer may be insignificant but that doesn't make the question entirely irrelevant ;-)

Surely the weight of shoes has some effect apart from just increasing the all up weight of bike
and rider? I have recently returned from a holiday in the Alps where my wife and I did lots of
walking as well as cycling. On some days we used our bikes to rapidly gain height on roads or
tracks before walking. We tried wearing our walking boots to cycle on some days and I can assure
you that it was much harder work than carrying them in a rucksack and wearing trainers. I would
guess that a small amount of work is needed to unweight the recovering leg and that this work is
greater for heavier shoes.
--
Michael MacClancy

"Whazzarke" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

It's to do with change in velocity, not speed...

Conservation of angular momentum.

I must admit your argument looks more than reasonable, and conservation of angular momentum
duly noted

However, Plan 36B: If we are on about a bike on a flat track, at constant speed, exactly the same
for the two cases - lighter and heavier shoes, then why not take the bike out of it altogether and
think about me sat sitting on front of the television moving my feet in a circular fashion watching
Le Tour. The lighter shoes have gotter be easier.

Michael MacClancy said:

Quoted message said:

If your cadence is 60rpm the question is entirely irrelevant ;-)

Quoted message said:

Well, the answer may be insignificant but that doesn't make the question entirely irrelevant ;-)

Nah - roadies never spin at less than 90.

Guy
===
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TCG said:

Greetings...We have a discussion here and need some educated advice...

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1
hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have you
moved over that same hour.

It's been slow at the office lately...

Work Done (W) = Force (F) x Distance Moved (D)

but

F= mass (m) x acceleration (a)

Substituting,

W = maD - or at least you probably are, keeping up that cadence for an hour!

Simplifying greatly,

Ignore friction (sigma), gravity (g), and the fact that the direction of linear acceleration is
always changing:

Linear acceleration is always towards the centre of the crank. Assume this is constant. For a radius
of 172.5mm:

Linear Velocity (V) = Radius (r) x Angular Velocity (Omega)

Omega = 60rpm = 120 Pi Radians per minute = 2 Pi Radians per second

Substituting

V = 0.1725 x 2 x Pi metres per second = 1.083

Acceleration = rate of change of distance with respect to time

or

a = v^2 / r

so

a = 1.083*1.083 / 0.1725 = 6.81 m/s^2

So,

W = maD

W = 0.002 x 6.81 x D

Linear Distance moved in one hour (D) = circumference x rpm x 60

Circumference = pi X Diameter (d)

So

W = 0.002 x 6.81 x pi x 2 x .1725 x 60 x 60

W = 53 Joules

So, the energy needed to move (and thus saved, if you don't have to move it) 2g through one hour of
revolutions at 60 rpm on 172.5mm cranks is 53 Joules - the eqivalent of moving a mass of 53kg (the
mass of a small adult) through a distance of one metre in one second. That's also how much less mass
you've moved over the hour.

Comparing like for like, if two identical riders, of mass 53kg, save for the missing 2g, rode
identical bikes in ideal conditions at the same cadence over the same route, one would finish the
course in 1 hour - the lighter rider would finish the course one second faster.

You can substitute the values for the other sized crank yourself...

--
MatSav

"MatSav" <matthew D O T savage A T felthamscouts DOT org D O T uk> wrote in message
"]news:[email hidden]...

Quoted message said:

TCG said:

Greetings...We have a discussion here and need some educated advice...

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1
hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have
you moved over that same hour.

It's been slow at the office lately...

Work Done (W) = Force (F) x Distance Moved (D)

but

F= mass (m) x acceleration (a)

Substituting,

W = maD - or at least you probably are, keeping up that cadence for an hour!

Simplifying greatly,

Ignore friction (sigma), gravity (g), and the fact that the direction of linear acceleration is
always changing:

Linear acceleration is always towards the centre of the crank. Assume this is constant. For a
radius of 172.5mm:

Linear Velocity (V) = Radius (r) x Angular Velocity (Omega)

No velocity is taken in the direction of the force applied, in this case, as you state the force is
radial, hence we are interested in radial velocity which is 0 (obviously the shoes move throughtout
a rotation so its not, but lets simplify and assume point mass at the end of the crank).

Work done is independent of mass for constant angular velocity withough friction.

Think flywheel, planets etc, conservation of angular momentum.

MatSav said:

Work Done (W) = Force (F) x Distance Moved (D)

but

and loads more bollocks

Quoted message said:

Comparing like for like, if two identical riders, of mass 53kg, save for the missing 2g, rode
identical bikes in ideal conditions at the same cadence over the same route, one would finish the
course in 1 hour - the lighter rider would finish the course one second faster.

You can substitute the values for the other sized crank yourself...

No No No, the mass of 2g has just gone round in circles, once its up to speed no more work
is required.

By your thinking the world would stop spinning if it wernt for the big magic cyclist in the sky.

Finish you O levels before replying.

--
Andy Morris

AndyAtJinkasDotFreeserve.Co.UK

Love this: Put an end to Outlook Express's messy quotes
home.in.tum.deoe quotefixOpen ↗

Frank X said:

"MatSav" <matthew D O T savage A T felthamscouts DOT org D O T uk> wrote in message
"]news:[email hidden]...

Quoted message said:

TCG said:

Greetings...We have a discussion here and need some educated advice...

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on 1
hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have
you moved over that same hour.

It's been slow at the office lately...

Work Done (W) = Force (F) x Distance Moved (D)

but

F= mass (m) x acceleration (a)

Substituting,

W = maD - or at least you probably are, keeping up that cadence for an hour!

Simplifying greatly,

Ignore friction (sigma), gravity (g), and the fact that the direction of linear acceleration is
always changing:

Linear acceleration is always towards the centre of the crank. Assume this is constant. For a
radius of 172.5mm:

Linear Velocity (V) = Radius (r) x Angular Velocity (Omega)

No velocity is taken in the direction of the force applied, in this case, as you state the
force is radial, hence we are interested in radial velocity which is 0 (obviously the shoes
move throughtout a rotation so its not, but lets simplify and assume point mass at the end of
the crank).

Work done is independent of mass for constant angular velocity withough friction.

Think flywheel, planets etc, conservation of angular momentum.

I was going to write: Are you suggesting that changing two grammes at the pedals and rotating at
constant speed will actually stretch the pedals more due to the increase centripetal force? If they
don't stretch to a greater extent than without the two grammes, nothing moved so you've not done any
work. Even if they do, once the constant speed is reached, they dont move any more in their new
position, so you are still not doing any work wrt rotating the extra mass at a constant speed.

But you beat me to it.

Just my 2 grammes worth, as they say.

--
Jim Price

jimprice.dsl.pipex.comjimprice.dsl.pipex.comOpen ↗

Conscientious objection is hard work in an economic war.

Aye!.

MatSav <matthew D O T savage A T felthamscouts DOT org D O T uk> wrote in message
news:<[email hidden]>...

[much nonsense snipped]

The point you seem to have overlooked is that the acceleration (due to circular motion) is
perpendicular to the direction in which the cyclist is doing work. Therefore no work is involved.

James

Whazzarke said:

However, Plan 36B: If we are on about a bike on a flat track, at constant speed, exactly the same
for the two cases - lighter and heavier shoes, then why not take the bike out of it altogether and
think about me sat sitting on front of the television moving my feet in a circular fashion
watching Le Tour. The lighter shoes have gotter be easier.

Yes, they'll accelerate just a little quicker up to the cruising speed in the first instance from
the start at rest.

Pete.
--
Peter Clinch University of Dundee Tel 44 1382 660111 ext. 33637 Medical Physics, Ninewells Hospital
Fax 44 1382 640177 Dundee DD1 9SY Scotland UK net [email hidden]
dundee.ac.uk~pjclinchOpen ↗

"James Annan" <[email hidden]> wrote in message
"]news:[email hidden]...

Quoted message said:

MatSav <matthew D O T savage A T felthamscouts DOT org D O T uk> wrote in

message news:<[email hidden]>...

Quoted message said:

[much nonsense snipped]

The point you seem to have overlooked is that the acceleration (due to circular motion) is
perpendicular to the direction in which the cyclist is doing work. Therefore no work is involved.

Nope. There will be a difference between adding one gram at the pedal end or one gram at the heel
end. Some of the force to keep the heel going round in (near) circles will be from the cyclist.

Quoted message said:

However, Plan 36B: If we are on about a bike on a flat track, at constant speed, exactly the same
for the two cases - lighter and heavier shoes, then why not take the bike out of it altogether and
think about me sat sitting on front of the television moving my feet in a circular fashion
watching Le Tour. The lighter shoes have gotter be easier.

Yep, that's right. Trouble is, you can't just take the bicycle out of the equation, because thereby
you take out the cranks and pedals, which force your feet into a circular motion. In your couch
potato experiment of course you have to provide that force yourself.

Here's another, admittedly rather less scientific, thought on the topic. If the weight on a rider's
feet was so crucial, the pros on the Tour would surely ride barefoot. Come to think of it, the
contenders for the yellow jersey would probably have their toes amputated, in order to save that
little bit extra 🙂

Andy

Michael MacClancy said:

Quoted message said:

Quoted message said:

If you shave 1 gram off of each cycling shoe (total 2g), how much energy will you save based on
1 hour of riding at cadence of 60rpm using 172.5 or 175 crank arms...and how much less mass have
you moved over that same hour.

Quoted message said:

Surely the weight of shoes has some effect apart from just increasing the all up weight of bike
and rider?

You're right, it has some effect when accelerating, because in addition to getting the moving in the
direction of travel, you also have to get them rotating around the bottom bracket. That effect is
very small though.

The crank radius is about 170mm, which means that during one revolution the pedals travel about
2*3.14*0.17m = 1.06m. At 60 RPM, you do one revolution per second, i.e. the pedals travel at a speed
of 1.06 m/s (2.4mph).

The energy needed for accelerating your feet and shoes to such a low speed is always gonna be
negligible compared to the energy needed to accelerate bike and rider to whatever speed you
travel at.

At constant speed, the extra speed has no effect.

Quoted message said:

I have recently returned from a holiday in the Alps where my wife and I did lots of walking as
well as cycling. On some days we used our bikes to rapidly gain height on roads or tracks before
walking. We tried wearing our walking boots to cycle on some days and I can assure you that it was
much harder work than carrying them in a rucksack and wearing trainers.

I'd think that's due to the fact that walking boots restrict the movement of your ankles.

Quoted message said:

I would guess that a small amount of work is needed to unweight the recovering leg and that this
work is greater for heavier shoes.

No. The weight on the other pedal will do that for you. That's what's so clever about bikes, and why
they're more efficient than walking.

Andy