"When man first set woman on two wheels with a pair of pedals, did he know, I wonder, that he had rent the veil of the harem in twain? A woman on a bicycle has all the world before her where to choose; she can go where she will, no man hindering." The Typewriter Girl, 1899.
"Every so often a bird gets up and flies some place it's drawn to. I don't suppose it could tell you why, but it does it anyway." Ian Hibell, 1934-2008
Ok, you are saying that the frame must apply driving force to front wheel to accelerate the mass of the wheel longitudinally and to accelerate the moment of inertia I to the required angular velocity. And THAT force is equal to 2 ma.
My derivation: I considered I to be based strictly on the mass at the rim, distributed in an ideal ring at a fixed and unique radius from the hub - an approximation.
The inertial torque of the wheel when accelerated is I*alpha, and moment of inertia I = (m*r^2)/2. The torque required to achieve that angular acceleration is F_I*r, where F_I is applied to the axle in order to spin up the wheel. Equating the two torques, we get
(1) F_I*r = (m*r^2*alpha)/2.
Canceling an "r" from each side we get
(2) F_I = (m*r*alpha)/2.
The angular accel alpha is directly related to the linear acceleration of the bicycle, a = r*alpha, so alpha = a/r. Inserting this into (2) and canceling both "r"s on the right, we get
(3) F_I = (m*a)/2.
From (3) the effect of imparting angular momentum to the wheel is equivalent to imparting linear velocity to half of the mass of the wheel. I think this contradicts your statement in your 6th paragraph that this equivalency is equal to m*a.
Regarding the longitudinal accel, I'm not sure why F=ma does not apply, where in this case m represents the mass of the entire bicycle with rider and gear.
I'm pretty confident in my derivation of #(3) above [famous last words, in my experience]. I'm not confident enough in my last statement (regarding F=m_total_vehicle*a) to claim you're not correct. Your argument regarding integrating around the wheel sounds good, but I have not done that integration so I don't know how exactly it would "play out" - as you say or not. But I wonder if it's an overly complex point of view. It seems to me if look at the wheel as an independent free body that gets accelerated by a force applied at its hub (or centroid), it requires a force F=ma where m is the wheel mass. The force required to increase the angular velocity is still my F = ma/2 where m is the rim mass, for a total wheel drive force of 3m_rim*a/2. This force of course is applied to the axle of the front wheel by the fork ends, and is ultimately supplied by the cyclist's pedaling. I'm not sure why integrating around the wheel is a necessary step.
Last edited by Road Fan; 12-16-13 at 07:40 PM.
To keep it simple, let's just look at a wheel with no hub -- all the weight at the edge.
If this wheel is rolling down the road, it has two components to its inertia -- rolling and moving forward.
Moving forward, it's easy -- mass * velocity.
Rolling, it's a just as easy. The bottom of the wheel is not slipping on the ground, so it's moving at the same speed as the wheel, and the entire thing is rotating at the same speed. And it has the same mass, because it's the same wheel. So the inertia from this is mass * velocity as well.
Add the two components up ... 2 * mass * velocity.
You'll get the same answer if you integrate around the wheel, calculating the speed of each part -- but it's a lot more work.
Though as Jobst says here -- 1) mass on the wheel counts twice as much as mass on the frame, but 2) the effect is so small that you can ignore it.
You are starting to confuse inertia (m or I), momentum (m*v) and inertial resistance (m*a or I * alpha). It makes it harder to follow your reasoning and confirm that it's not fuzzy. You're also not commenting on what I wrote, which I had hoped to see.
I'll allow that dynamo lights are hard to justify as an add-on on a purely economic basis. That said, my Ortliebs, Brooks, and Schmidt bike parts and accessories may be useful, but they're also luxuries as well. Some questions you might consider include the following:
When was the last time you forgot to recharge your light(s)?
What did you do when that happened?
What's the longest you've ridden in the dark you didn't plan for?
When was the last time your battery died unexpectedly?
How far from the end of your ride were you when that happened?
What did you do then?
My answers are: 2 years ago, before I installed a dynamo and associated lights. Had to drive home.
2 hours more than the 30 minutes I expected; asked directions because I was lost, but the dyno kept the light shining.
3 years ago, and 2 miles from home. Frankly scared me; I ended up practically tip-toeing down sidewalks, because it was obvious nobody in a car noticed my bright yellow jacket.
To sum up, dynamo lights are a wonderful thing to have. IMHO.
Yes, I was a bit loose with the terminology. But inertia is just resistance to changes in momentum, and rotational inertia is just resistance to rotational momentum. And since we assume that the wheel isn't slipping and all the mass is at the rim, the two are very closely related.
And this "double" relationship applies if you're looking at momentum or kinetic energy (to further fuzzy things, though of course they're closely related, so much so that we often use the terms interchangeably in casual speech.)
I've also not been looking at the forces and accelerations at all -- you can, but it's easier just to look at the end result (momentum and kinetic energy), since they are trivial to calculate when you break it down into translational and rotational components, especially since the two velocities and masses are the same (in this simplified model of a wheel where all the mass is at the rim.) I could do it the harder way, but I'd have to whip out some paper.
Here's an analysis of the kinetic energy of a rotating bicycle wheel that assumes that all weight is at the rim. (Really, this is what I should have started with, but I didn't find it before.)
You can change this to a calculation of momentum just by replacing 1/2 m*v^2 with m*v and 1/2 I*w^2 with I*w, and end up with a total momentum of 2*m*v
Last edited by dougmc; 12-17-13 at 02:02 PM.
The answer to all of the above: The Great Silver Sag-Wagon
Really, if I were truly caught without lights I'd just hop the subway with the bike. I do that almost every day without the bike anyway. No biggie. So that said, I'm still going forward with this even though I have no great reason for it. All I need now is spokes and a light, and the time to build the wheel. Actually, I'm more curious than anything else. I so rarely see dynamo lights, never mind a current version, that I don't even know their capabilities.
An even easier answer is ... a spare light. For the bikes I ride at night a lot, I typically have two head lights and two tail lights. Another option is spare batteries, though I like to have a completely spare light just in case of problems that go beyond batteries. And even if you prefer high performance lights, your spare can be a lot smaller/less impressive -- a 1 watt/50 lumen light will get you home, and if it takes AA batteries you're never far from even more if you need them.
I generally run both tail lights at the same time, but put fresh batteries in one and older batteries in the other so they won't need new batteries at the same time. But even so, a pair of AAAs doesn't take much space in the tool bag.
Personally, I'd have spare lights even if I had a dynamo light, just in case of a problem.
Dyno lights can't be used as a flashlight to change a flat or whatever. Also focused beam lights put light on the road, a blinking light on the bars helps with being seen. In addition, the standlight can be overrated IMHO. These are all cons to dyno lighting. Being able to see and be seen is huge when riding at night, unless you have a death wish. Again, all light systems have cons and having a backup is a good idea no matter what you go with.
I always have redundant lights when I'm riding longer distances. Calling for help still means a long delay before help arrives. On my commuter, I don't have any redundancy because I could just walk and the wait is short. I could even put my bike on the bus if I wanted
I have three bikes with dyno lights. One has a dyno hub. The others have bottle dynamos. I'm going to leave the bottle dynamo on the Raleigh Twenty, because it's good enough. I'll change the other bottle for a dyno hub.
I have made this longer than usual because I have not had time to make it shorter. --Blaise Pascal
Tom Reingold, noglider@pobox.com
Residences: West Village, New York City and High Falls, NY
Blogs: The Experienced Cyclist; noglider's ride blog
I guess I'm going to find all this out once I get my setup.