why clincher more pricey???

[Gumpole]

surely the (incredibly small) extra energy used with heavier wheels when accellerating will be dissipated at a later stage, so no extra energy is used over the race as a whole...

 

excepting I guess for the final dash for the finish line, after which the race is over!

 

[Johan Bornman]

 

Cut cut cut cut...

 

Which I will not accept as weight from the hub has a MUCH lesser effect*
on energy than carbon or aluminium at the rim. It makes me think you
don't really understand rotational inertia and have (correctly in your
example might I add) plugged some numbers into some formulae you
have found.

My opinion - not an attack on your person just in case you feel inclined
to blow up again!!!

 

 

Yes' date=' we know all that.  It is a given. The issue is HOW MUCH. I think I made my point in that departement.

 

Thanks for thinking I just found some old formulae lying around in a cupboard somewhere and dared dust them off and try. You want me not to blow up?? I do understand rotational inertia. The calculations are complex and I need more data than what we have. To do an actual calculations I'll need thousands of data points starting at point zero (centre of hub) all the way to the outside of the tyre. And I'll have to integrate. I can't find my pencil sharpener right now so I'll give that a miss.

 

I think you miss the point of what I did. I simplified the problem by simply looking at the difference (of weight at the circumference) between two identical wheels. I wasn't interested in the actual intertia, just the difference. That's all we need to know.

 

 

 

*The effect reduces as r^2 as you approach the hub, so at the hub
it has zero rotational effect! This is because the moment of inertia
I = Integral(r^2 dm), so as r tends to zero, so does I.

Remember the value of I you used (m r^2) is an approximation for
an infinitely thin hoop of radius r with no hub, spokes or depth - not at
all like a bicycle wheel (more like the iron rim on an ox wagon wheel).
The hoop approximation is a good one for our purposes though, in
that it gives us an upper bound for the effect.

 

That's exactly what I did and I purposefully did it like that. The reason should be clear to you. For all intends and purposes I used a bicycle wheel with zero mass everywhere other than X grams on the circumference and another identical one with zero mass everywhere and X-200g of mass on the circumference of the rim.

 

The inertial effect of the inside bits of the two wheels in my example are equal and therefore cancel each other out.

 

You are blustering with a bookfull of jargon that serves no purpose. Acknowledge that my approach works or point out the flaws.

 

 

[flex]

Time to get the popcorn out again

[Johan Bornman]

JMac, I think I see your problem.

 

Look at the extract below in red.

 

 

 

Sorry - I was unclear. Your approach was correct, but I came across:

I've just shown you what a 200 gram difference makes. It doesn't matter whether that is from nipples at the hub or from carbon instead of aluminium at the perimeter. Accept it.

Which I will not accept as weight from the hub has a MUCH lesser effect*
on energy than carbon or aluminium at the rim. It makes me think you
don't really understand rotational inertia and have (correctly in your
example might I add) plugged some numbers into some formulae you
have found.

 

You assumed that I meant a reduction of weight at the hub when I meant MOVING THE NIPPLES FROM THE RIM TO THE HUB.

 

My interpretation of what that means is that it doesnt matter how you reduce weight at the circumference (whether by taking the nipples away or using a lighter weight material for the rim)' date=' [the difference is negligible'].

 

[] bracketed text is added on now for clarity. () brackes indicate a rephrase for clarity.

 

How's that?

 

And in case anyone is still reading, let me spell out my stance on this.

 

The addage that weight saved at the circumference of a wheel is X-times more beneficial than weight saved anywhere else on the bike is naive.

 

That's my stance, in black and white, as you seem to think I express myself.

 

On an industry note, you'll notice that the industry has done a U-turn on its nipples-at-the-hub madness. Shimano is back with nipples at the rim and so is someone else I saw the other day but can't remember.

 

Also, if you ever disassemble a nipple-at-the-hub wheel you'll discover great big blobs of die-cast lugs inside the rim to anchor the spokes. This is true for Spinergy, Shimano and Reynolds. I have not come across a single company that does a lightweight (less than a nipple) anchor at the rim. Engineering wise it is not possible to anchor a spoke in a rim without some sort of weigh.

 

 

 

 

 

[GoLefty!!]

I think if the wheels are spinning at 20,000 rpm there will be a significant improvement to acceleration benefits by reducing the MOI of the rotoation wheel,  much like mechanical engineers try to do in turbine blades.

 

Bicycle wheels spin so slowly and have other issues like high friction due to traction and poor road surfaces that the arguements for moving nipples away from the rim to the hub making a difference are really moot.

 

at the end of the day, the biggest effect on acceleration is the amount of torque the engine can produce and what total mass has to be accelerated.

 

To calculate the absolute effects of a few grams here and there taking translational, rotation, and aerodynamic effects into account is simply too big a problem to be solved without some sophisticated software or a  lot of time and excel.

 

The basic arguement held by JB is correct that the real world differences are moerse small and really won't win or loose you a race.

 

600calories is a pretty big savings but I'm not sure the savings are linear because that does not take rate of fatigue into account.

 

The reality is that in any race the number of significant accelerations are likley only in the order of around 10-15 for any one rider.

 

I think we would be utilising energy more effectively if we spent more time focussing on correct bike fit and real aerodynamic gains (not wheels..)

 

[jmaccelari]

 

 

surely the (incredibly small) extra energy used with heavier wheels when accellerating will be dissipated at a later stage' date=' so no extra energy is used over the race as a whole...

 

[/quote']

You have to put in the extra energy. It does not matter how this is

dissipated. You still have to put it in.

 

It is dissipated as heat, though. Either in your brakes when braking (or

in the real world through friction, rolling resistance or air drag). Kinetic

energy is not recovered.

 

Potential energy is recoverable (but only to a degree in the real world).

jmaccelari2010-01-05 21:21:09

[jmaccelari]

 

 

 

Yes' date=' we know all that.  It is a given. The issue is HOW MUCH. I think I made my point in that departement.

[/quote']

So we agree the old wives tale is true!

 

 

The calculations are complex and I need more data than what we have. To do an actual calculations I'll need thousands of data points starting at point zero (centre of hub) all the way to the outside of the tyre. And I'll

have to integrate.

So you don't understand analytical integration. The formulae are simple.

You don't need "thousands of data points" (only if you perform a

numerical integration and the wheel can be reduced to a simple

approximational model that can be analytically solved) !

 

I can't find my pencil sharpener right now so I'll give that a miss.

You don't need it....

 

I think you miss the point of what I did. I simplified the problem by simply looking at the difference (of weight at the circumference) between two identical wheels. I wasn't interested in the actual intertia' date=' just the difference. That's all we need to know.

[/quote']

This is exactly what I did from first principles, so did you actually

read the document I put together? I read your stuff?

 

 

That's exactly what I did and I purposefully did it like that. The reason should be clear to you. For all intends and purposes I used a bicycle wheel with zero mass everywhere other than X grams on the circumference and another identical one with zero mass everywhere and X-200g of mass on the circumference of the rim.

 

The inertial effect of the inside bits of the two wheels in my example are equal and therefore cancel each other out.

Correct - so why do you say you require "thousands of data points" to

perform the integration. You have done it! Are you trying to intimidate

us by exaggerating the scale of the problem?

 

You are blustering with a bookfull of jargon that serves no purpose.

This is taken from the Johan Bornman book. You love to wax lyrical

andf then when someone comes with facts and references you accuse

them of muddying the issue. Science is science and if you can handle

the figures stay out of the lab!

 

 Acknowledge that my approach works or point out the flaws.

I have acknowledged that your approach works! Read my posts .

 

What I disagree with is your blanket statement that losing weight

on the wheel is more energy effective than on the frame is an

old wive's tale.

 

You and I have both shown (independently) that it is true - even if it

is a relatively small differernce, it is still true. You have avoided answering

me on this issue.

 

The argument is not about whether it saves energy (it does), but whether

it is worthwhile.

 

You say it is not.

 

I say it adds up slightly after a while. I do state, however, that other

effects may be more important than weight loss on the wheel.

 

You are black and white, I am open to opinion.

 

You have also avoided answering my concern that you state that it

does not matter where on the wheel the weight it lost. If you understood

integration, you would see this immediately and not make such

statements. Weight loss towards the rim has more of an effect.

(Redacted sine JB answered this in a later post)

 

Please do not cloud the issues by rehashing arguments we have

discussed and that I publically state we agree on.

jmaccelari2010-01-05 21:42:37

[jmaccelari]

 

You assumed that I meant a reduction of weight at the hub when I meant MOVING THE NIPPLES FROM THE RIM TO THE HUB.

 ...

How's that?

You are quite correct. I misinterpreted what you said. You might find

there's some aerodynamic advantage (again very small), but that's

beyond the scope of this argument.

 

So we now agree on all points except whether it's worthwhile.

 

And that's a matter of opinion!

 

[Wyatt Earp]

Jmaccelari

 

 

Deksels boet, eventually someone with nuts decided to step up to the plate.

 

I have found the discussion very interesting and have enjoyed your arguments as "simplified" as they are .

 

One thing I will say, rim weight will never worry me too much in the future.

[Spinnekop]

http://s3.amazonaws.com/advrider/lobby.gif

 

 

 

 

http://weightweenies.starbike.com/forum/images/smilies/icon_popcorn.gif

 

 

[Johan Bornman]

Jmac, thanks for the edit. You're a gentleman.

 

Spinnekop, am I the angel or the devil?  I may or may not hold this against you.  ;-)

 

 

[jmaccelari]

 

Spinnekop' date=' am I the angel or the devil?  I may or may not hold this against you.  ;-)

 [/quote']

I bags the devil!

[jmaccelari]

 

Jmac' date=' thanks for the edit. You're a gentleman.[/quote']

Pleasure, but I think you also owe Kiwi an apology for your reply to

his original solution:

 

One place where you should not take physics lessons is from bicycle company websites and from the back of bicycle product boxes.

His equations from the back of his Pronutro box were spot on (ours

reduce to his with one simple step)...

 

 

His maths was correct, but once again I state you are free to make your

own assumptions and conclusions as to the effect in the real world.

[madmarc]

 

 

 

 

A point to remember is that weight on the edge of the wheel, eg the rim and tire, increases at the square of the speed. So a 100 grams saved at the rim will have a much bigger effect than a lighter frame for instance.

 

 

This old old-wive's tale refuses to die. If you do the maths, and believe me, it is not difficult stuff, you'll see that this story is nonsense. The effect is negligible, so please stop perpetuating this myth.

 

Most metals in daily use are alloys. The exceptions being aluminium kitchen foil which is almost pure aluminium and copper in electrical wire.

 

Therefore, when we refer to an alloy of sorts, we call it by the name of its biggest component. Most aluminium alloys have less than 15% other metals in there and we then simply call it aluminium. Unless of course the fact that it is alloyed is of importance. The practice of calling aluminium alloy leaves all the other alloyed metals out in the cold. What do we call them?

 

The exception to the rule is steel and that is simply because iron alloy has a very specific name - steel.

 

The gold on your wedding finger is gold - not gold alloy. The platinum in your car's exhause is called....platinum and so on and so on.

 

Clarity of speech helps us all understand what you're trying to say.

 

 

 

 

Holy crap boringman you remind me of my old N6 Mech Design teacher. All that data from Kiwi's simple statement above. Once again we bow down in the shadow of your wisdom.

 

I often wonder why my teacher was a lecturer and you build wheels, when you both should both be designing the new space shuttle

 

But the statement is not an old wives tale and is correct. As i understand what he is saying is basically the faster the wheel spins the higher the centrifugal force becomes, the CF increases by the square of the RPM of the wheel. so a heavier wheel rim will have a higher CF. But because we ride at different speeds (RPM) there is no constant CF and the CF will only become a factor at higher speeds. So when climbing a hill the CF will be neglible, but when hooking down a hill the CF will drastically increase, giving more inertia and getting to top speed quicker than a lighter rim.

 

One could say what we gain on the uphills in a lighter bike for easier or faster climbing (lighter rims) we lose on the downhills with a heavier bike (heavy rims) as the increased weight will give us more inertia enabling the bike to get to max speed quicker.

 

So a cyclist wanting more weight for downhills to increase the inertia for more speed would be very clever if he used heavier wheels instead of a heavier frame (say 2 kg more), as he would need to lug the extra 2kg heavier frame up the hills to have the gain for the downhills. With heavier wheels of say 200 grams he could extract the same benefit on the downhill due to the extra 200 grams having the same effect as the 2kg on the frame as a result of Kiwi's statement, but he would not have to lug the difference of 1.8kg up the hills.

 

 

 

[Gumpole]

 

 

surely the (incredibly small) extra energy used with heavier wheels when accellerating will be dissipated at a later stage' date=' so no extra energy is used over the race as a whole...

 

[/quote']

You have to put in the extra energy. It does not matter how this is

dissipated. You still have to put it in.

 

It is dissipated as heat, though. Either in your brakes when braking (or

in the real world through friction, rolling resistance or air drag). Kinetic

energy is not recovered.

 

Potential energy is recoverable (but only to a degree in the real world).

 

I think I'm talking about potential energy , ie at the top of the hill with the same aero influences heavier bike goes very slightly faster/further, or after the big accelleration on the flat the rider will have to work less (very slightly) hard when he is decelerating* which will save the energy that was put into the system in the first place.

 

* = I'm assuming the real world where there is wind resistance, so rider is still work even when gently decelerating.

 

all the examples that you mention (braking, rolling resistance, wind) would effect the lighter wheeled bike too.

 

[jmaccelari]

 

 

I think I'm talking about potential energy ' date=' ie at the top of the hill with the same aero influences heavier bike goes very slightly faster/further,

[/quote']

Correct - this is potential energy and is recoverable perfectly in an

ideal world on the downhill, but imperfectly in the real world as there

will be some losses due to friction and air drag. This potential energy

is converted into kinetic energy (motion) and then dissipated as heat.

 

 or after the big accelleration on the flat the rider will have to work less (very slightly) hard when he is decelerating* which will save the energy that was put into the system in the first place.

No. On a flat the energy we generate is converted solely into kinetic energy. You are talking here about "recovering" kinetic energy.

 

For potential energy we have:

dE(potential) = m * g * dh

where dE (delta Energy) is the energy we need to put in to climb dh

metres (dh = delta height). Since dh is zero (flat) dE is zero.

 

What you see here is the energy you put in to accelerate being used

to overcome friction and air resistance. The energy is not saved, but

expended in overcoming these factors. We have specifically excluded

wind and friction losses in our discussion to keep to the point of weight

savings on the wheel.

 

In our ideal example you would continue coasting at the speed you

accelerated to until you clapped on brakes at the end of the race

(sounds lovely, doesn't it?). This is Newton's First Law of Motion you

would have learned from school.

 

To regain kinetic energy, we would require a KERS system similar to

those on Formula 1 cars. As cyclists in the real worls all we can do is lose

this kinetic energy to heat (moving the air, overcoming friction and braking)

and then keep on putting it in so we keep on going. This is why we have

to keep on pedalling...

jmaccelari2010-01-06 03:51:25