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After reading the very heated clinchers vs. tubulars debate with a sub debate on rotational weight perceptions and realities I came up with a couple of my own ideas. I am very good at math but I am not a physics major, so I would welcome any comments.
Wouldn't riders who do not spin very well (like myself) benefit more from a reduced rotational weight due to the micro acceleration differences derived from the alternating 2 major pedal strokes of your revolution? In essense having two accelerating and two decelerating periods with every revolution of your crank. Although your actual velocity might not vary more than a few tenths or hundreths wouldn't more than 1/2 of your pedal stroke be spent on true acceleration.
I have wrapped my head around the idea that at a "constant" speed, rotational weight makes very minimal difference, but who is pedalling at a truely constant speed without acceleration? As far as "engine" smoothness goes, I would think a dual piston engine (insert rider here) would not be a very constant propelling force. Combine that with the increased inertia? required to complete a revolution on a much less smooth uphill climb, where many riders have to get out of the saddle and 1 2 mash their way to the crest. I wonder how long it would take to develop a good smooth spin technique while out of the saddle.
edited to add.....I would think the periods of acceleration and deceleration climbing a hill would be much more pronounced with less (shorter) "maintainence" energy. For instance... and this is anything but scentific....
Let's say you are pedalling a slow (but mathmatically easy) cadence of 60 with each full revolution of your crank taking 1 full second.
Say the downstoke of your drive side crank arm takes 4/10 a second as does the downstroke of your non-drive side crank arm.
The "lull" or transitional period where the stroke is not as smooth (which I am sure would not be near as pronounced with a pro rider) is a tenth of a second on each side.
So here we are with 8/10 of a second spent on downstrokes of each piston and 2/10 of a second on the "transitional" time.
Assume the 3 of the 4 tenths on each downstroke is spent accelerating and 1 of the 4 tenths on the downstroke and the tenth of transitional time is spent decelerating. (there would be two much smaller fractions of a second where there would be neither acceleration or deceleration)
Wouldn't that mean we have spent 60% of our revolution as acceleration and 40% as deceleration? Surely at least on a micro level?
Does this then mean that rotational weight could be more important than formulas that use "constant" speed prove?
Wouldn't riders who do not spin very well (like myself) benefit more from a reduced rotational weight due to the micro acceleration differences derived from the alternating 2 major pedal strokes of your revolution? In essense having two accelerating and two decelerating periods with every revolution of your crank. Although your actual velocity might not vary more than a few tenths or hundreths wouldn't more than 1/2 of your pedal stroke be spent on true acceleration.
I have wrapped my head around the idea that at a "constant" speed, rotational weight makes very minimal difference, but who is pedalling at a truely constant speed without acceleration? As far as "engine" smoothness goes, I would think a dual piston engine (insert rider here) would not be a very constant propelling force. Combine that with the increased inertia? required to complete a revolution on a much less smooth uphill climb, where many riders have to get out of the saddle and 1 2 mash their way to the crest. I wonder how long it would take to develop a good smooth spin technique while out of the saddle.
edited to add.....I would think the periods of acceleration and deceleration climbing a hill would be much more pronounced with less (shorter) "maintainence" energy. For instance... and this is anything but scentific....
Let's say you are pedalling a slow (but mathmatically easy) cadence of 60 with each full revolution of your crank taking 1 full second.
Say the downstoke of your drive side crank arm takes 4/10 a second as does the downstroke of your non-drive side crank arm.
The "lull" or transitional period where the stroke is not as smooth (which I am sure would not be near as pronounced with a pro rider) is a tenth of a second on each side.
So here we are with 8/10 of a second spent on downstrokes of each piston and 2/10 of a second on the "transitional" time.
Assume the 3 of the 4 tenths on each downstroke is spent accelerating and 1 of the 4 tenths on the downstroke and the tenth of transitional time is spent decelerating. (there would be two much smaller fractions of a second where there would be neither acceleration or deceleration)
Wouldn't that mean we have spent 60% of our revolution as acceleration and 40% as deceleration? Surely at least on a micro level?
Does this then mean that rotational weight could be more important than formulas that use "constant" speed prove?