It Doesn't matters, It's not the bike/kit you have...

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Ming the Merciless

There is no mercy
Photo Winner
Location
Inside my skull
OK, include power input. Both twins (one on a 7kg bike, one on 15kg bike, ceteris paribus) start from rest (zero momentum) and accelerate (same power) to 30kph (8.3m/s). Relative acceleration will be inversely proportional to mass (say 80kg v 88kg) and the time taken to get to 30kph will be proportional - shall we say the power each twin generates allows them to get to 30kph in 10 seconds and 11 seconds respectively. As an aside please note the rider of the heavier bike now has more momentum: 11% more. Each twin cycles for an hour at the same power that, with equal aero drag, etc, etc (only difference is mass) achieves about 30kph. 30km later the twin on the lighter bike has to wait for about 30 seconds for her brother (who weighs the same as her btw). The second or two extra taken to accelerate to 30kph is negligible.
HTH (and @bpsmith - go to a dentist that pulls teeth with an anaesthetic)

Continue with your assumptions and apply them for relative deceleration and what do we end up with? Given that the aero drag and rolling resistance are decelerating both riders all the time. Plus aero drag is not increasing linearly.
 
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mjr

Comfy armchair to one person & a plank to the next
I hope he wasn't riding his supperbike at dinner time. That would be a terrible faux pas.
Indeed. The faux pas should be danced between dinner and supper.


View: https://www.youtube.com/watch?v=bVIsPm2KH58
 

Ajax Bay

Guru
Location
East Devon
Continue with your assumptions and apply them for relative deceleration and what do we end up with? Given that the aero drag and rolling resistance are decelerating both riders all the time. Plus aero drag is not increasing linearly.
I'm not sure I know what you mean by 'relative deceleration' - do you mean in order to stop at the end of 30km? In this model neither aero drag nor rolling resistance are 'decelerating' the riders: their speed increases from 0 to 29.5/30kph and stays there. Those two forces are countered by the drive (equal power = equal drive force) provided by the riders, overmatched during the initial acceleration and then matched, exactly. Aero drag increases exponentially but 1) it's the same (in this model) for both riders and 2) the speed is constant so the aero drag is constant. Perhaps you could "continue with [my] assumptions and apply them" yourself; and share that with us? What do you end up with?:snowball:
Interestingly the OP @itaa hasn’t been back to comment :rolleyes:
Well, not that interesting, but I know what you mean (at least I think I do :laugh:).
 

Ajax Bay

Guru
Location
East Devon
Worth looking at the OP's post history to understand where the road bike v MTB rant (kernel of truth/fact nevertheless) came from. 9 year old MTB with new tyres and a 'below entry level' road bike (bought for £26 - a bargain btw) which he is considering converting to a flat bar.
 

Ming the Merciless

There is no mercy
Photo Winner
Location
Inside my skull
In your model they accelerate to 30km/h a second apart. The heavier bike / person combination is now in the slip stream of the lighter / bike combination. They are putting out the same power and due to reduced drag the heavier rider / bike combo will now accelerate faster than the lighter rider and be able reach a higher speed (for the same power input) than the one in front. In fact they will be able to draw level as much as matters. So the riders will finish together.
 
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