First Generation Coker weight difference?

Unicycles and Equipment
42

Both :wink:

Seriously, when I upgraded my wheel it made a very noticeable difference to the ride. Well worth it in my opinion.

Yes, the 36" tyres are ridiculously heavy. A lighter one would be nice, but on the other hand the current ones are pretty much bullet proof - never have to worry about punctures.

Rob

43

Thanks Sask, so the rolling weight really makes a difference?

I only want it to improve manuverability and the make it easier for the long hills here in Colorado.

So is there a secret to putting on a 700 tube, I tried last night and blew it out. I really tried to keep it from twisting but it twisted anyway then blew. I reset it twice and still blew the tube. Any hints to help would be great.

44

Fitting a 29er tube in a 36 is much easier if you stretch it first. Pump the tube up (outside the tyre) bigger than normal and leave it overnight to stretch, then let it down and fit it like any other tube. You have to be a bit more careful than normal to avoid pinching it under the tyre bead as you’re fitting the tyre. You probably already know if you’ve fitted bike tyres, but just in case: rub some talcum or chalk powder on the tube to stop it sticking to the tyre, and put a small bit of air (only a tiny bit) inside the tube while fitting to help avoid pinching.

Rob

EDIT: And get the fattest 29er tube you can find (700x50c or something) The Schwalbe ones are popular.

45

I just ordered my new wheel set from Coker with some 125mm alum cranks for $150 shipped! Very stoked… so my total for the old coker and new wheel set is $325…pretty good price for a Coker.

I will have to report on the difference between the steel and new Alum wheel.

46

It’s weird to have an exact multiplication factor. But if you use it, then after the conversion you’d rather say you lost 7 pounds of static weight. If you still call it rotating weight, someone might do the conversion again :slight_smile:

47

I spoke to my engineering friend, it is not a set # for the rotating weight, it increases the further away from the centeral hub you get and the speed in which the wheel spins changes the moving weight. He didn’t have the formula but it would be cool to know that.

Yal All have a great weekend!

48

a bit o’ physics

I’m sure there is an explanation of the basic physics of a wheel somewhere in this forum, but I’ll post one anyhow. I’m going to consider the most basic dynamics possible; in particular, I’ll assume that the rider’s mass plus the cycle’s mass is basically constant – the unicycle’s mass is insignificant with regards to linear motion. (This assumption doesn’t quite work for those of us on the light side, but what the heck.)

Here are the main results, and then I’ll provide explanation below:

  1. Regarding the factor of two that’s been mentioned: Consider three possibilities: (1) your wheel is a disc whose mass is uniformly spread; (2) your wheel is a hoop whose mass is concentrated in its circumference; (3) your wheel is magical so that all its mass is concentrated in its (tiny) hub. The second wheel is the kind most of us ride. Suppose that the first two wheels have the same radius. Ignoring that a tire has some thickness, the inertia of wheel (2) is twice that of (1). Inertia is what you feel, so you might say that you feel mass in the rim and tire twice as much as mass in, say, the spokes. But the inertia of the first two wheels is r^2 (radius squared) times greater than wheel (3). So adding mass to the hub is pretty much free (except that in practice, a hub has non-zero radius).

  2. What you really care about when you accelerate your unicycle is linear acceleration – that is, your acceleration in the direction of motion. So how does a wheel’s radius and mass contribute to the amount of torque (basically force) that you need to provide for a given linear acceleration? That is, say you want to accelerate by 1 meter per second-squared. Is it easier on a small wheel or a big one? On a light wheel or a heavy one? And how does the effort relate to the wheel’s radius and mass? The answer is: the required torque is linearly proportional to mass and radius. If you go from a 24" to a 36", expect to put in 50% more effort to accelerate by the same amount (if both have the same mass). If you halve your wheel weight, expect to halve your effort. (Except that your legs and shoes have some mass, don’t they? So you won’t gain quite that much.)

  3. How much momentum does mass in the rim/tire contribute? How much does the radius contribute? Both contribute linearly to momentum for a fixed linear velocity.

As a reasonable example, let’s suppose that your 36er’s mass is twice that of your 24er’s (1.5 times the circumference, plus a bulkier tire). Then accelerating the 36er by the same amount as the 24er takes (ideally) 1.5 x 2 = 3 times the effort. And for a given speed, your 36er has 3 times the angular momentum as your 24er.

Explanation: The inertia of a wheel is given by its mass times its radius squared (mr^2), whereas the inertia of a disc is given by half that ((mr^2)/2).

The torque that you must provide to spin the wheel is equal to the wheel’s inertia times the angular acceleration you wish to impart: torque = inertia x angular acceleration = mr^2 x AA. So you’d think that torque would increase linearly with mass and quadratically with radius. And it does, if what you care about is angular acceleration. But the wheel is obviously just a means for us to move linearly in the world. So we’re interested in linear acceleration. Linear acceleration is related to angular acceleration by the wheel’s radius: LA = AA x r. So AA = LA/r, and torque = mr^2 x AA = mr^2 x LA/r = mr x LA = mass x radius x linear acceleration. So to impart a given linear acceleration requires a torque that is linear in both the wheel’s mass and its radius.

The angular momentum of a wheel is equal to its inertia times its angular velocity, which might suggest a quadratic relationship between radius and velocity. But again, we care about linear – not angular – velocity, which is the angular velocity times the radius: AV = LV/r. Once again, we end up with a linear relationship between radius, mass, and linear velocity (your speed). This makes sense: for a fixed speed, your 24er’s wheel is spinning faster than your 36er’s, so its momentum-disadvantage (because of its lower inertia) is partly compensated for by its higher angular velocity.