why clincher more pricey???

[NotSoBigBen]

snip snip snip

 

 Why cut up a perfectly good wheel?

 

For flip sake now you tell me .....

 

Now what must I times by what again ........

BigBen2010-01-08 13:11:08

[Slowbee]

and just a thought that occured while free wheeling today .. is it better to get better hubs and bearings, or change the weight on the rim ?

[Johan Bornman]

and just a thought that occured while free wheeling today .. is it better to get better hubs and bearings' date=' or change the weight on the rim ? [/quote']

 

 

Better? Define better.

 

Better for performance or better for maintenance?

 

Besides, if you were freewheeling, you were either going downhill or flat; hubs with lower drag would have been more advantageous and weight at the rim neither here nor there.

 

Better hubs in my world are hubs with flanges designed to minimise spoke breakages, with bearings that are easy and cheap to service and, uses standard components.

 

Better to someone else is a hub that has chicks throwing panties at him.

 

Better is in the eye of the beholder.

 

 

[Slowbee]

 

and just a thought that occured while free wheeling today .. is it better to get better hubs and bearings' date=' or change the weight on the rim ? [/quote']

 

 

Better? Define better.

 

Better for performance or better for maintenance?

 

Besides, if you were freewheeling, you were either going downhill or flat; hubs with lower drag would have been more advantageous and weight at the rim neither here nor there.

 

Better hubs in my world are hubs with flanges designed to minimise spoke breakages, with bearings that are easy and cheap to service and, uses standard components.

 

Better to someone else is a hub that has chicks throwing panties at him.

 

Better is in the eye of the beholder.

 

 

 

jislaaikie, I need to buy you some more beer JB!

 

By better I meant an upgrade in quality - or is this also a marketing ploy that makes us believe that some hubs are better than others ?

 

[Guest Agteros]

 

Better to someone else is a hub that has chicks throwing panties at him.

 

I would love to see that when the wheel goes at 70km/h. Definitely don't want chicks like that... Their lingerie account will blow my budget!!!

 

[jmaccelari]

 

Jmac' date=' I see you are back to your patronising self. Stop it.

[/quote']

Yes, Mr Pot.

 

 

I asked you a simple question and you sidestepped.

No' date=' I did not. I told you the question is irrelevant. I am not interested in

answering it as it will be an absolute answer for one wheel and will

not add anything new to the argument. The problem is that you have

a hassle admitting your are fallible and shot your mouth off without

thinking and are now desperately trying to salvage some face.

 

 

 

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.

I gave you the equations for the components. As I told you and you

obviously have problems understanding: measure the components

yourself and then add up the bits. You think you are a mathematical,

brain - then stretch yourself by doing the simple sums.

 

As I stated, we have proved that for ANY wheel that the energy saving on

rotational mass on the wheel is greater than the energy saving on the

frame per gram.

 

I am not being patronising. I am just tired of you harping on about

irrelevancies either because you refuse to admit you are wrong or

because you really do not understand the fundamentals.

 

My opinion is that it is the first option.

 

If it's the second send me:

 

The mass and outer radius of the hub.

The mass and length of one spoke.

The mass and inner and outer radius of the rim.

The mass and inner and outer radius of the tyre (with tube).

The mass and radial distance of a nipple.

 

I will then compare the MOI to the ideal wheel we have been

discussing and you'll see there is a difference in the absolute

figures, but not the final conclusion. The more accurately modelled

wheel will be the equivalent of an ideal wheel as we have been

discussing, but of a smaller radius (see if you can figure that out!).

 

If you do this, please remember you are going to be further

embarrassed as I'll will show how easy it is. And you'll be supply

the ammunition to shoot yourself down with again...

 

[GoLefty!!]

If it's the second send me:

 

The mass and outer radius of the hub.

The mass and length of one spoke.

The mass and inner and outer radius of the rim.

The mass and inner and outer radius of the tyre (with tube).

The mass and radial distance of a nipple.

 

 

 

You will need the dimension "r" to the centre of mass of the various components, not to the outer surfaces. All forces and properties act through the centre of mass. For a hub we will need to slice it half along its axle and calculate the centre of mass above the centre line. Then add the whole spoke, whole nipple and section of the rim between spokes.

 

In reality any quality hub from a reputable manufacturer has an extremely miniscule influence on the MOI of the wheel because it's radius is so small and the bearings can be considered frictionless.

 

[jmaccelari]

 

 

 

 

You will need the dimension "r" to the centre of mass of the various components' date=' not to the outer surfaces. All forces and properties act through the centre of mass. For a hub we will need to slice it half along its axle and calculate the centre of mass above the centre line. Then add the whole spoke, whole nipple and section of the rim between spokes.

[/quote']

No. We do not deal with the centre of mass. The COM is for potential

and translational kinetic energy modelling (the cartesian reference frame).

So forget about all the COM stuff here.

 

The concepts are the same, except we need to look at kinetic effect in the

rotational reference frame. We will reduce the wheel geometry to an

ideal wheel (infinitely thin hoop) of a specific radius, which will be the

equivalent of a COM in our cartesian frame.

 

So we replace the "centre of mass" of the body with the moment arm of

inertia - the radius of the ideal wheel.

 

To illustrate the difference:

For the wheel, the COM is in the centre of the wheel - in the middle of the

axle. The effective moment arm for the rotational "COM" is the radius of

the ideal wheel. This is where all the mass effectively sits in the angular

reference frame.

 

Your concepts are correct, just not correctly framed in terms of a

angular reference frame.

 

If the body is a hollow cylinder (like a rim), we need the inner and outer

radii to calculate the "moment arm" (analogous to the centre of

mass you are talking about). We can then take this radius and the mass

of the wheel and model the real wheel as an idealised perfect hoop

of a different (smaller) radius that is the exact analogy of the real wheel.

 

This is the concept Mr Bornman cannot grasp. Every real wheel can

be reduced to an ideal wheel (perfect hoop) and then we use the

old formulae to model it. So if we prove a hypothesis is true for an

idealised wheel, then we can deduce it is true for all real wheels.

 

This is why I don't care what stupid wheel Mr Bornman is holding in his

hand. Using the moment of inertia (COM analogy), we can reduce any

flipping wheel in either of his hands to an idealised one...

 

In reality any quality hub from a reputable manufacturer has an extremely miniscule influence on the MOI of the wheel because it's radius is so small and the bearings can be considered frictionless.

 

The bearings have nothing to do with the argument since they do not

affect the rotational energy requirements of the wheel (they are part

of the frictional component we are ignoring for the comparison).

 

You are quite correct that the hub has a smaller effect because of the

smaller radius. This is the exact reason. The radius (the "moment arm")

is proportional to the all important MOI. The smaller the radius, the smaller

the MOI and the smaller the effect the component has on the system.

 

That is why nipples on the hubs are better than ones on the rim. BUT:

the effect in this case is so small you won't notice it. I would assume the

aerodynamic saving is much more significant in this case...

 

jmaccelari2010-01-09 04:49:00

[jmaccelari]

 

and just a thought that occured while free wheeling today .. is it better to get better hubs and bearings' date=' or change the weight on the rim ? [/quote']

 

 

Better? Define better.

 

Better for performance or better for maintenance?

 

Besides, if you were freewheeling, you were either going downhill or flat; hubs with lower drag would have been more advantageous and weight at the rim neither here nor there.

 

Better hubs in my world are hubs with flanges designed to minimise spoke breakages, with bearings that are easy and cheap to service and, uses standard components.

 

Better to someone else is a hub that has chicks throwing panties at him.

 

Better is in the eye of the beholder.

 

Good lord! I concur!

 

[GoLefty!!]

ah I got ya! Indeed I was looking at it from the cartesian perspective

[Johan Bornman]

jislaaikie' date=' I need to buy you some more beer JB!

By better I meant an upgrade in quality - or is this also a marketing ploy that makes us believe that some hubs are better than others ?
[/quote']

 

That wasn't a facetious question. Better is not easy to answer for hubs.

 

The best hubs are forged, not CNC machined. They have good bearings with adjustable pre-load and they have two seals, one contact, one labyrinth.

 

The flanges are slanted to better follow the spoke line and prevent bends in the spokes. The spoke holes are chamfered, not square.

 

Then, my preferences, but not everybody's, is that they use standard spokes, standard (and cheap) bearings and standard tools. This would exclude all Mavic products for instance. Yet, some Mavic owners will tell you they're very happy with their hubs.

 

 

I drink Windhoek Lite.

 

[flex]

I drink Windhoek Lite.

 

 

 

Think you may need a Windhoek Lager after all this, JB flex2010-01-09 08:24:54

[Johan Bornman]

Cut cut cut cut cut out all sorts of rude stuff........

If it's the second send me:

The mass and outer radius of the hub.
The mass and length of one spoke.
The mass and inner and outer radius of the rim.
The mass and inner and outer radius of the tyre (with tube).
The mass and radial distance of a nipple.

 

I'll risk the threatened embarrassment and carry on.

 

The hub weights 378 grams and has one high and one low flange. Their radii are 38 and 72mm respectively.

 

The spokes on the one side are 280mm and the other 284. They weigh 6 and 5.6 grams respectively.

The rim's radii are 631 and 590mm. It weighs 530 grams.

The tyre's inner radius is 664 and 616mm

The nipple is 12mm long and weighs 1 gram.

 

If you need more info, I'm here, armed with tape measure  scale and baited breath.

[100Tours]

this thread has hit 18 pages...

[flex]

this thread has hit 18 pages...

 

 

Forget the Windhoek, think we may need a bottle of Scotch!

[jmaccelari]

 

I'll risk the threatened embarrassment and carry on.

Sorry about the delay. I got my 14 year old son to do the sums

and it took him a little while and I had to check them.

 

 

The hub weights...

OK. I assumed 36 (18 of each) spokes' date=' since this was not specified. I also

assume nipples (36) on the inside of the rim. You didn't give me the

tyre/tube weight, so I measured one of my own (0.280kg).

 

I(hub) is about 4.65x10-6 k/m^2

(minimum bound is 2.73x10-6, maximum is 9.80x10-6)

You were given I=m*r^2 / 2

(Note that I don't have the exact geometry of the hub, so I calculated

the maximum and minimum values, which are both so small we can

actually ignore the hub entirely. The value given is a ratio approximation

from measurements off one of my Shimano hubs.)

 

I(spokes) is about 5.53x10-3 kg/m^2

You were given I=m*L^2 / 3

 

I(nipples) is about 12.531x10-3 kg/m^2

You were given I=m*r^2

 

I(rim) is about 197.759x10-3 kg/m^2

You were given I=0.5*m*(r1^2 + r2^2)

 

I(tyre) is about 114.849x10-3 kg/m^2

You were given I=0.5*m*(r1^2 + r2^2)

 

 

Total is about 330.677x10-3 kg/m^2. To get this we simply "integrate".

This means in plain, simple English that we add all the little bits up...

 

Now that wasn't all that "complex" was it? You didn't need to go the

numerical integration route (you would not be able to do it with a pen

and paper, either. And, I must admit, neither would I).

 

Why then the nonsense about how difficult it is?

 

So the hub is about three orders of magnitude smaller in effect than the

rest of the wheel. As GoLefty!! says we can ignore it as it is less than

the rounding error.

 

Spokes are about 1.3% of total MOI

Nipples are about 3.7% of total MOI

Rim is about 60% of total MOI

Tyre is about 35% of total MOI

 

The equivalent "ideal wheel" radius is about 480mm. So if we reduced

your "real in the hand" wheel to an ideal wheel of the same mass,

the radius we would be working with would be 480mm.

 

If course. we could keep the radius constant and reduce the mass

to 950g which would have the same effect. Or we could choose any

one of the infinite combinations...

 

So the question now is: what is your point? Or are you just playing

a little game and wasting my time?

 

As I have stated before: the actual wheel is immaterial, so this exercise

was pretty pointless unless you are still trying to convince yourself of

something...

 

If you need more info, I'm here, armed with tape measure  scale and baited breath.

Surely you mean "bated"? "Baited" could be a little smelly.