Virtual / Effective bike weight

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Bianco
Posts: 3
Joined: Thu Sep 23, 2004 2:15 am
Location: Pasadena, CA

by Bianco

Has there ever been a discussion on virtual or effective bike weight? (At least, that's what I'm calling it). They say rotational weight is everything. So, let's say you have two bikes and they both weigh exactly 6kg. But, they are built-up with different components. One of the bikes has a wheelset which is 1500 g, and the other has a wheelset of 1100 g.

Wouldn't the bike with the lighter wheelset be "effectively" lighter than the other one, even though both of the bikes have the same complete weight of 6000 g??? If thats the case, any math wizards try to figure out a formula which can determine the effective weight of the two bikes in question?

by Weenie


Bozopelli
Posts: 164
Joined: Tue Sep 21, 2004 10:26 am
Location: age

by Bozopelli

Well, it wouldn´t be lighter, as teh scale shows the same weight, but it would feel or drive lighter in the way that you would need less Power to accellerate the bike and keep it rolling at the same speed (equal transmission preassumed). Scientificially speaking of translated and rotaded masses. In this case, everything but the wheeels is translated only (unless you perform a wheelie or nosewheelie :lol: ). The wheels are translated and rotated at the same time. You will need power to move the bike and aditional power to rotate the wheels. For the rotational forces, you will also have to add the friction the hubs, BB, chain, pedal axels, .... have. Therefore it is nearly impossible to find out an exact formula. For someone more capable of math than me, it could be possible to postulate a formula which is near to th effective bike "efficiancy".

As for the Physical basics, you´ll have to imagine, that if you speed up your bike, you´ll need the most power. If it´s on the desired speed on a horizontal surface, you need only power to compensate the external forces like wind and friction (that´s why a bike rolls quite a llong time without pedalling (best on downhills :lol: )).

If you add the bike´s weight to the complete weight of the system (bike + driver +accessories) you can see, that a light bike is about 10 % of the systems weight (for a light driver) and 6 % for a heavy biker. A heavy bike would be about 15 % (light driver) and 10 % for a eavy biker (bikeweigts 6 Kg light and 9 kg heavy, rider with 60 and 95 kg).
The biker, which is the biggest part of the System in any case, has no rotational masses at all (hopefully).

But not only teh wheels are rotated but also the crankarms and Pedals and BB. But they are not so far away from the rotational axels, where you can se, that it is more important to have lighter Rims and Tires than to have light hubs/cassettes (and for the drivetrain part: light pedals before light crankset/before light BB).

You can explain this with the "Laws of the levers" (Hebelgesetze in german, i don´t know if tht expression is correct in english). Same mass in double distance to the point to the joint (axle in this case) means double force (momentum).

Hope i could make some things a little bit more intransparent :wink: to you.

Greets, Bozo
Quäl dich, du Sau!

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gonzo
Posts: 54
Joined: Tue Nov 02, 2004 11:01 am
Location: Sydney

by gonzo

This concept exists, it is called the moment of inertia. So for a rotating body you take the weight of the object and multipy it by the distance it is from the centre. To calculate it easily you would have to make a few assumptions such as assuming that the spokes can be represented as a point source halfway between the rim and hub but it could be done without too much difficulty.

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