why clincher more pricey???

[jmaccelari]

 

 

 

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 posting a second response for this so we keep this separate from the

energy discussion to avoid confusion.

 

The effect we notice on the flat with less deceleration into the air

due to a heavier rider is due to the higher inertia of a heavier rider.

 

He takes more energy to get to the speed and thus has more energy

to free wheel. He does not get anything for free - he pays for it in the

extra energy he requires to get up to speed.

 

This assumes the drag coeffient ("size") of the cyclist is the same in both

cases. A big, fat cyclist will lose proportionally more energy to drag than

a smaller. more aerodynamic one.

 

This is the same argument as GoLefty preferring his heavier wheels. They

cost more in energy to get going, but once going have more inertia -

resistance to change in their speed - which means they roll more nicely.

 

It is also used as one of the claims for 29er superiority over 26 inchers.

But let's not go there...

jmaccelari2010-01-06 04:03:10

[SwissVan]

 

So...if you accelerate and achieve greater topspeed downhill with heavy rimmed wheels and likewise faster uphill with supposed lighter rimmed wheels, then the supposed lighter rimmed wheels theoretically win due to the greater time spent going uphill vs downhill, as we all know there is no route where more time is spent going downhill then uphill. Although the anti hubnipple brigade could argue that the amount of time and increase in speed (acceleration) whilst going uphill is insignificant for most of us.<?: prefix = o ns = "urn:schemas-microsoft-com:office:office" />

 

The conclusion imho can be safely concluded that the location of a realistic amount of weight or the lack thereof on a realistic BICYCLE wheel makes very little difference to performance, this very slight factor however can be of great psychological benefit to the "look good go good brigade". This conclusion is enforced by the occasional (very occasional luckily  ) moment when I have been overtaken by someone wearing long red socks and riding a postman?s dikwil whereby I magically found the energy to retake my position in front of ?red socks?, only to loose it when we hit the downhill?

 

[jmaccelari]

 

 

 

So...

No. As soon as we are talking hills, we are talking potential energy,

which is another discussion. I want to try and keep this one as simple

and as to the point about wheel/frame weight loss during accelerations.

 

This discussion is about kinetic energy only. Lighter wheels will provide

an advantage over a similarly lighter frame by a goodly factor. We are

talking about the RELATIVE wheel/frame energy saving - not OVERALL.

 

Losing a couple of hundred grams on a bike is a small amount, so the

overall effect is small, but cumulative (it also makes climbing easier, but

again that is a potential energy problem).

 

The results of this discussion (flat road, no wind or friction) are:

- a couple of hundred grams weight loss makes a small difference to

  energy usage;

- this energy usage is cumulative over a race and may become worthwhile

  worrying about (or may not depending on your opinion);

- losing the weight off the wheels results in greater energy saving than

  off the frame (this loss is still overall small compared to our entire

  energy usage);

- losing the weight from the rim results in a relatively MUCH larger energy

  saving  than losing the weight nearer to the hub (but it is still small

  overall).

(thanks SwissVan for the comments)

jmaccelari2010-01-06 05:47:20

[jmaccelari]

 

 

 

But the statement is not an old wives tale and is correct.

Correct.

 

 As i understand what he is saying is basically the faster the wheel spins the higher the centrifugal force becomes' date=' 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.

[/quote']

Nope. Centrifugal force does not strictly exist. We talk of it since our

frame of reference is non-inertial (it's called the "fictitious" force for this

reason). It's a counter to the "centripetal" force. In an inertial reference

frame, these "forces" disappear. I'm sorry, but I really don't know how

to explain it any better...

 

An example to try to clarify:

 

Basically if we apply a force, we need energy. That means if we rotate a

wheel in a vacuum with no friction then if we have a resultant "centrifugal"

force it will require energy and the wheel will slow down.

 

In our thought experiment (no air, no friction) it's "obvious" the wheel will

will not slow down, so there can be no resultant force. The (fictitious)

centrifugal force has been cancelled out by the (fictitious) centripetal

force, so there is no resultant force and no energy loss.

 

One could say what we gain on the uphills...

Again we have hills (potential energy) and so it's outside the scope of

this discussion.

 

Why has no on asked why am I so pedantic about excluding potential

energy?

This is because potential energy has no dependence on any rotation

of the body. It only depends on the mass and height of the body

(and the gravitational acceleration of the point in the gravity field

it is at at that point and place in time, but let's not go there ).

 

So when we are talking about energy saving due to wheel rotation

(weight saving on the wheels) there is no difference with respect to

potential energy. It does not give a rat's bum where on the bike the

weight is lost!

 

 

jmaccelari2010-01-06 04:50:07

[SwissVan]

- losing the weight from the rim results in a MUCH larger saving than
  nearer to the hub (but it is still small overall).

 

Ok so... This appears SLIGHTLY confusing contradicting:

 

MUCH larger saving [saving what? Energy?] than nearer to the hub (but it is small overall)

 

Ok so.. back to the more or less original question. Is it a worthwhile saving of (energy?) wrt some few realistic grams at the rim vs at the hub in the realistic world.

 

Man i cant weight wait for someone to say ask how much faster ceramic wheel bearings are vs hi quality steel bearings (lets ignore the weight saving  benefits).

 

 

 

 

 

 

 

SwissVan2010-01-06 04:53:53

[Johan Bornman]

 

 
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' date=' when you both should both be designing the new space shuttle

 

[/quote']

 

I don't know whether you remember your lecturer with fondness or not, so I'll assume some history between us to be in the mix. I'll keep my comment short.

 

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' date=' 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.

cut cut cut cut cut

[/quote']

 

You are confusing centripital force with inertia.

[jmaccelari]

 

 

 

- losing the weight from the rim results in a MUCH larger saving than

  nearer to the hub (but it is still small overall).

 

Ok so... This appears SLIGHTLY confusing contradicting:

 

MUCH larger saving [saving what? Energy?] than nearer to the hub (but it is small overall)

What I mean is that at the rim we might save (for example) 16J of

rotational energy - which is a small overall saving. At the hub' date=' we

will save 0J due to rotational energy (there is no rotation exactly

on the axle so no rotational energy is saved there). So if we compare

0J to 16J the relative saving is much more (you can't get much more

than infinite!), but the overall saving is still small - around 1% for our

example.

 

In the real world we never will get exactly 16J (all exactly on the rim)

or 0J (all exactly in the axle), so our real figures will fall within this

range.

 

The important statement is that the further out, the more energy we

save.

 

Ok so.. back to the more or less original question. Is it a worthwhile saving of (energy?) wrt some few realistic grams at the rim vs at the hub in the realistic world.

There is an energy saving, but it is small. Is it worth it is up to

you. How much are you prepared to pay to lose 400g (for example)?

How much will it cost you to save off the frame and how much

on the wheel?

 

What we do know is that the biggest bang per gram for energy saving

is to lose weight off the rim.

 

Will you notice it? You tell me!!!

 

Man i cant weight wait for someone to say ask how much faster ceramic wheel bearings are vs hi quality steel bearings (lets ignore the weight saving  benefits).

Are you trolling here?

 

The rotational energy and kinetic energy savings in a bicycle bearing

are MUCH smaller than the small savings for a wheel (they are lighter

and smaller, so less kinetic, rotational and potential energy in the

system and thus to lose).

 

My guess is that we might be able to lose 1g on a bearing and this will

save us 0.1J in the best case for our example. I'll side with JB you'll never

notice that!! This is about 0.0025% of the overall energy budget.

Remember the wheels best case is about 1%.

 

Ceramic bearing have other advantages discussed on other threads

that I will not go into now...

jmaccelari2010-01-06 05:42:33

[SwissVan]

Will you notice it? You tell me!!!

Man i cant weight wait for someone to say ask how much faster ceramic wheel bearings are vs hi quality steel bearings (lets ignore the weight saving  benefits).

Are you trolling here?

 

Will you notice it? You tell me!!!
Doubt it, imo overall weight of the wheel is more important and realisticaly the real amount you save by having nipples on the hub side is negligble for the average rider racer.

 

Are you trolling here?
Nooit, just seeking a conclusion to all the theory and math.

 

 

 

 

[jmaccelari]

 

 

Are you trolling here?

Nooit' date=' just seeking a conclusion to all the theory and math.

 [/quote']

You must be unique on the Hub!!!

[NotSoBigBen]

So why was a clincher more pricey (or not) again?

[jmaccelari]

 

So why was a clincher more pricey (or not) again?

 

**** knows - I never got that far!!!

 

My guess is that a tubular sits in a low cavity/dip around the rim which is

easy to manufacture. The dip has no structural requirement apart from

providing a surface to glue the tubular on.

 

A clincher will need a strong side wall on the rim to hold the clincher

tyre bead in. This, because of the high pressures road tyres are under,

would require a lot of design and manufacturing effort to make light

but still strong enough to hold the tyre.

 

My opinion. No maths involved.

 

[GoLefty!!]

yeah J thats about right. The clincher requires a more expensive mould than does a tubbie rim so the cost is passed on to the end user.

Theres also slightly more complex engineering involved which also adds to the cost.

But the biggest reason is that Clincher rims are by far the dominant design, therefore have higher volume =  higher demand = higher price.

 

[Johan Bornman]

Jmac, I'd like to return to this one point if I may.

 

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....
 and that I publically state we agree on.

 

 

   ***************

 

You say I don't need thousands of data points (OK, dozens then) to calculate the MOI of a wheel rotating at a given speed. You also say I don't need to integrate.

 

How on earth are you going to avoid that? It is not an ideal wheel but a real one with spokes, tyre, nipples, rim and the works.

 

Please calculate the MOI for me for the wheel I have spinning in my hand.  I'll supply all the data, you ask the questions.

 

 

[jmaccelari]

 

 

 

The calculations are complex

No they are very simple. And by "very" simple' date=' I mean "very,

very simple".

 

Moment of inertia for a rotating object is defined by:

I = integral(m * r^2)

 

It doesn't get much simpler than that. Cut your wheel up into bits.

Measure r for each bit (the distance from the hub), weigh it and

add (m(bit) * r(bit)^2) up for each bit. The smaller the pieces

the better the approximation.

 

It may be tedious, but the calculations are primary school level.

 

"Complex" my arse! I have to laugh at that one! Don't try to

intimidate people by using the word "complex"!

 

You say I don't need thousands of data points (OK, dozens then) to calculate the MOI of a wheel rotating at a given speed.

What I stated was that you don't need to do a numerical integration. An

analytical approximation will do. Why cut up a perfectly good wheel?

 

You also say I don't need to integrate.

I state you do not need to do a numerical integration. You do need

to do an analytical one. Without an integration of some form you

do not have a moment of inertia to start with! The numerical integration

 is a lot of work and will not provide much improvement over an analytical

one.

 

You do understand the difference between a numerical and an analytical

integration' date=' don't you?

 

How on earth are you going to avoid that?

By an analytical approximation. Duh. I'm repeating myself again...

 

It is not an ideal wheel but a real one with spokes' date=' tyre, nipples, rim and the works.

[/quote']

The actual geometry of the wheel is not important to the original

argument. Don't change the goal posts, please!

 

This is why we have dealt with an ideal wheel.

 

We needed to prove that taking mass off a rotating wheel saves more

energy than taking if off the frame. The ideal wheel provides an upper

limit to the saving, telling us that the maximum we can save is twice

the energy saved from the frame weight saving.

 

The exact amount is not important as we are not talking about exact

values. We are just setting the limits to show that saving rotational mass

(away form the hub) will save more energy than pure translational mass.

By keeping on about your "thousands of data points" you are avoiding the

real issue by stating it is too difficult a problem. It's p*ss easy.

 

Please calculate the MOI for me for the wheel I have spinning in my hand.  I'll supply all the data' date=' you ask the questions.

[/quote']

Again, I don't need to since the argument is not over the absolute

value of the savings, rather the relative saving.

 

The question was a comparison between loss of weight on the frame

versus loss of weight on the wheel. It has nothing to do with the

real wheel in your hand - this will only give us an absolute figure.

 

You're harping on the wrong issue again. I think you need to go back to

page one and read the comment again because you really do not

understand the fundamental concept of the question!

 

To reiterate (so that hopefully you get it this time!):

 

Whether it is 16J or 8J or 1J makes no difference. The important thing is

that more energy is saved by removing rotational mass (i.e. mass off

the wheel) than by removing pure translational mass (i.e. mass off

the frame).

 

Do you get it now?

For your interest, here are the components of inertia for the wheel

components:

I(rim) = 0.5 * m(rim) * (r(inner)^2 + r(outer)^2)

I(tube+tyre) = 0.5 * m(tube + tyre) * (r(inner)^2 + r(outer)^2)

I(spoke) = m(spoke) * l(spoke)^2  /  3.0

I(nipple) = m(nipple) * r(nipple)^2

I(hub) = m(hub) * r(hub)^2

 

So if you measure and weigh your wheel components (rim, tyre+tube,

spoke (times the number of spokes), nipples (times number of nipples)

and hub) and add up (technically known as "integration") you will get

a pretty good (within a couple of percent) approximation to the exact

inertia of the wheel.

 

In fact, you can get a pretty (in fact a very) good approximation to the

moment of inertia for any wheel! Ain't science wonderful?

 

Note that numerical integration (your "thousands of data points") will

only improve on this approximation to a small degree and will not

change the conclusion (against your will and wishes) that taking

rotational mass off the wheel saves more energy than taking mass

off the frame. Get used to it...

 

BTW, the only way to get an exact measure, is to spin the wheel up and

measure the force required for a specific angular acceleration (or energy

expended if we integrate this with respect to time to get it up to a certain

speed). Then work backwards.

 

We would have a little loss due to friction and air resistance, so we'd

have to do this in a vacuum (or near vacuum) and with ceramic

bearings to reduce friction - but then you reckon ceramic bearings

are snake oil, too! 

 

The improvement of the numerical integration technique over the

analytical approximation is not worth the work, especially seeing as

we are not interested in the absolute values. Also, the difference

will not be measurable in real life under non-laboratory conditions.

 

We would be looking at less than a Joule in our example between

an analytical model and a numerical model.

 

Can I be any clearer?

 

If you still don't understand what I am talking about, please post again...

jmaccelari2010-01-08 10:56:29

[GoLefty!!]

So before we end up with another 84 pages of debate, in summary.

 

"an ounce off the wheels is worth 2 off the frame" holds true mathematically.

 

 

"it makes feckall difference in the real world".

 

 

 

Glad to clear that up 

 

 

 

BTW, to anyone still interesting in reading this, Better wheels is still the cheapest upgrade you can make to your bike for the largest single weight loss not on your body.

 

Have midrange bike, get lekker wheels. Make sure they look pretty and or fast to go pretty fast.

 

 

 

[Johan Bornman]

Jmac, I see you are back to your patronising self. Stop it.

 

I asked you a simple question and you sidestepped.

 

I have a spinning wheel in my hand and I want to calculate the MOI. You say it is a simple exercise. Ask me the relevant questions and give me an answer.