Author Topic: Hoy leg press  (Read 3137 times)

Hoy leg press
« on: December 21, 2011, 08:01:18 pm »
You may have seen this doing the rounds:



Quote from: Chris Hoy
In response to the questions re leg press it was 631kg for 5 reps. Could have squeezed out a rep more if Ross hadn't been laughing at me!

631kg seems quite a lot!
Those wonderful norks are never far from my thoughts, oh yeah!

eck

  • Gonna ride my bike until I get home...
    • Angus Bike Chain CC
Re: Hoy leg press
« Reply #1 on: December 21, 2011, 08:08:43 pm »
631kg seems quite a lot!
It's 14 Mrs ecks.  :o
It's a bit weird, but actually quite wonderful.

mattc

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Re: Hoy leg press
« Reply #2 on: December 21, 2011, 08:14:46 pm »
Has anyone bothered to work out the force required on one's pedal
to maintain motion up gradient X
in gear Y

(as a multiple of bike+rider weight)

?!?
Has never ridden RAAM
---------
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Re: Hoy leg press
« Reply #3 on: December 21, 2011, 08:16:55 pm »
It's 14 Mrs ecks.  :o

or considerably fewer Mr ecks  :-*

andygates

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Re: Hoy leg press
« Reply #4 on: December 21, 2011, 08:18:02 pm »
Olympian leg monster is olympian and monstrous. :thumbsup:
It takes blood and guts to be this cool but I'm still just a cliché.
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eck

  • Gonna ride my bike until I get home...
    • Angus Bike Chain CC
Re: Hoy leg press
« Reply #5 on: December 21, 2011, 08:18:49 pm »
It's 14 Mrs ecks.  :o

or considerably fewer Mr ecks  :-*

And how many Noodleys?  :P
It's a bit weird, but actually quite wonderful.

Re: Hoy leg press
« Reply #6 on: December 21, 2011, 08:32:19 pm »
And how many Noodleys?  :P

Only a few  :-X

Re: Hoy leg press
« Reply #7 on: December 22, 2011, 06:38:05 pm »
I am shite at maths so I will not venture how many of me there are.

However, that does look an impressive feat by Mr Hoy.

Re: Hoy leg press
« Reply #8 on: December 23, 2011, 03:51:41 pm »
Working my way up to inferior.

Karla

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Re: Hoy leg press
« Reply #9 on: December 23, 2011, 04:20:09 pm »
Has anyone bothered to work out the force required on one's pedal
to maintain motion up gradient X
in gear Y

(as a multiple of bike+rider weight)

?!?

D = distance moved on ground for 1 pedal stroke (meters)
P = distance moved by pedal for 1 pedal stroke = 2*pi*crank length (meters)
S = bike speed on the road (meters/second)
M = mass of rider and bike combined (kilograms)
G = gradient of hill = height gained / height travelled horizontally = tan(gradient as an angle)
g = acceleration due to gravity at earth's surface = 9.8

Rate of vertical ascent R = SG

Force required to lift mass M at vertical speed R, F_lift = M x acceleration = M(R+g) = M(9.8 + SG)

Model the cyclist as applying equal force to both pedals at a uniform rate over the full pedal rotation.

Ratio of vertical distance travelled to pedal distance travelled = DG/P

DG/P is also the ratio of pedal force to lifting force F_lift, except we need to divide by 2 to account for both pedals.

Therefore the force applied to the pedals in order to climb a hill of gradient R at speed S, assuming you push just as hard on each pedal for the entire pedal stroke

F_pedal = DGM(9.8+SG)/2P

D/P = CN_f/2*pi*LN_r  where N_f = no. teeth on your chainring, N_r = no. teeth on your rear sprocket, C = wheel circumference.

Therefore F_p = (GMCN_f(9.8+SG)) / (4*pi*PLN_r)

I'm not sure whether there should be a gradient squared term in there, can anybody comment?
You could also model the rider as pushing on a single pedal at 90 degrees, using torque calculations.  However, I'm off to get a cup of tea!

Re: Hoy leg press
« Reply #10 on: December 23, 2011, 05:00:26 pm »
^^^ :thumbsup:     Enjoy that cuppa!!!

PS Wish I could still do applied maths like this.
'Something....something.... Something about racing bicycles, but really a profound metaphor about life itself.'  Tim KrabbĂ©. Possibly