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Re: Challenge Gran Can - Brownlee [milesthedog] [ In reply to ]
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milesthedog wrote:
if it's a technical downhill, you have a point. But, kilo for kilo, AB just won't be able to travel as fast on a downhill as someone who weighs 8-11kg more
Seriously why does everyone keep pointing this out? You dont travel faster on downhills if you weigh more! It's simple physics! The only thing that comes into play is drag and rolling resistance. Since rolling resistance is negligible it is just drag which we cannot ascertain simply from weight.

Terrible Tuesday’s Triathlon
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Re: Challenge Gran Can - Brownlee [oscaro] [ In reply to ]
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 you're failing to factor in inertia - "a more massive object has a greater tendency to resist changes in its state of motion" - which, with wind direction and speed changes being magnified on a descent due to increased speed, plays a larger role and favors "massive" riders.

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Last edited by: milesthedog: Apr 25, 17 2:31
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Re: Challenge Gran Can - Brownlee [milesthedog] [ In reply to ]
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milesthedog wrote:
you're failing to factor in inertia - "a more massive object has a greater tendency to resist changes in its state of motion" - which, with wind direction and speed changes being magnified on a descent due to increased speed, plays a larger role and favors "massive" riders.

Yes well this is a very small factor. Heres a bit more thorough explanation.

Terminal velocity is when the force of gravity equals the drag

vterminal is proportional to square root (mass / area)
vterminal is proportional to square root (r cubed / r squared)
vterminal is proportional to square root (r)

So at constant density if you double r the terminal velocity increases by 1.414
Double r is eight times the mass for only 1.414 the terminal velocity.

Drag proportional to v squared is the real drag (pun intended)

Now lets pretend you could double your mass and keep the same area
vterminal is proportional to square root (mass / area)
vterminal is proportional to square root (mass / constant)
vterminal is proportional to square root (mass)
If everything was constant (including your area and rolling resistance)
If you increased mass by 2 you would increase terminal velocity by 1.414
If you increased mass by 4 you increase terminal velocity by 2
Rolling resistance is not constant so it would be less than that 1.414 and 2

Let say 180 lb rider and add 20 lbs of lead to the frame - that is only 10% straight down hill
On a 10 grade that is only 0.846% - 40 mph versus 40.43 mph (without accounting for rolling resistance).

Even climbing bikes are designed to be light.
Basically up the hill you pay for all the weight and down the hill you only get credit for the square root of the weight.
Drag proportional to v squared is the real drag


Terrible Tuesday’s Triathlon
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Re: Challenge Gran Can - Brownlee [oscaro] [ In reply to ]
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My money is on Oscaro on this one!
Last edited by: northern monkey: Apr 25, 17 3:48
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Re: Challenge Gran Can - Brownlee [northern monkey] [ In reply to ]
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Well, your money is on a plagiarist - he stole that rant from the third to last post in this thread: https://bicycles.stackexchange.com/...e-go-downhill-faster
shameful. And he then failed to read the final post.

Greater mass has an easier time overcoming wind resistance, but as I referenced in my previous comment on inertia, wind speed and direction and rider direction relative to the wind are not constant, so there a lot of forces trying to change the rider's state of motion (a straight line) and those forces make the lighter rider slower. Toss in some technical turns, and the heavier rider is at a disadvantage. (*My own words)

And here's that final post for you - in quotations, like any good and descent human would do, especially one posing anonymously as an intellectual on a forum ;)

"As usual, I think the best way to consider this is through energy;
in moving from rest at the top of a hill (height h), the conservation of energy applies between potential energy at the top and kinetic at the bottom:
Mgh = MV^2 + losses(due to aero and rolling resistance)
therefore
V = sqrt( gh - losses/M )
as losses are not proportional to mass, then factoring them down by mass reduces their influence on speed for the heavier rider, regardless of whether terminal velocity has been reached or not"

wovebike.com | Wove on instagram
Last edited by: milesthedog: Apr 25, 17 5:22
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Re: Challenge Gran Can - Brownlee [milesthedog] [ In reply to ]
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I never claimed that the words were my my own, and thought it would be enough that the font was different so one would understand that it was copy pasted. There are a few factors that play to the advantage of the heavier rider, but they are hard to compute and they are far less important than drag.

Terrible Tuesday’s Triathlon
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Re: Challenge Gran Can - Brownlee [oscaro] [ In reply to ]
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Yeah, I saw the font difference and I hope you caught the jest - but I'll be sure to point it out now because too much is lost in writing - I was writing in jest. Also, didn't you see Dan's aero shootout comments where he compared this forum to an academic journal? I didn't get that he was writing in jest... but I kind of like those standards and the aerotest they possibly contributed to.

wovebike.com | Wove on instagram
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Re: Challenge Gran Can - Brownlee [milesthedog] [ In reply to ]
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Well yeah, I thought the entire internet was an academic journal because everything on the internet is true! /Pink
We will see in a few weeks how everything plays out, will be very interesting!

Terrible Tuesday’s Triathlon
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Re: Challenge Gran Can - Brownlee [lovegoat] [ In reply to ]
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AB will be very dominant in 70.3. I can only see Gomez or Sanders (depending on that day race dynamics) being able to compete against him.

But I don't see him racing well in an Ironman. The way he runs, in my opinion, is not sustainable in a marathon in an Ironman race.
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Re: Challenge Gran Can - Brownlee [BarcelonaGuy] [ In reply to ]
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I can't wait to see the predictions thread for this race.
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