i know one answer is mtfu and ride hills but ignoring that - from stationary bikes in gym i pretty much know what my HR is for a given steady(ish) watts output any rule of thumb or on line table/calculator to work out what output has to be for say a constant 1in10 gradient?
Bike Forum
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sort of darkside - watts versus gradient rule of thumb?
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Posted 1 year ago #
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Depends how heavy you are.
Posted 1 year ago # -
and it depends how fast you want to ride.
and how accurate the gym bikes are
and how strong the wind is blowing
and what pressure you have in your tyres
and whether you have free running bearings and chain
etc etc
Posted 1 year ago # -
PE=mgh
hthPosted 1 year ago # -
and how accurate the gym bikes are
^this will be your downfall; they are wildly different.
Posted 1 year ago # -
The physics of cycling is complicated:
PTOT = (PAT + PKE + PRR + PWB + PPE)/Ec
PTOT = (0.5ρVa2Vg(CdA + Fw) + 0.5(mt + I/r2)(Vgf2 - Vgi2)/(tf - ti) +
VgCrrmtgCOS(TAN-1(Gr)) + Vg(0.091+0.0087Vg) + VgmtgSIN(TAN-1(Gr)))/EcWhere:
PTOT = total power required (W) mt= total mass of bike+rider system (kg)
PAT = power required to overcome total aerodynamic drag (W) I = moment of inertia of wheels (kgm2)
PKE = power required to change kinetic energy (W) r = outside radius of tire (m)
PRR = power required to overcome rolling resistance (W) Vgf = final ground velocity (m/s)
PWB = power required to overcome drag of wheel bearings (W) Vgi = initial ground velocity (m/s)
PPE = power required to change potential energy (W) tf = final time (s)
ρ = air density (kg/m3) ti = initial (s)
Va = air velocity (relative to direction of travel) (m/s) Crr = coefficient of rolling resistance (unitless)
Vg = ground velocity (m/s) g = acceleration due to gravity (9.81 m/s2)
Cd = coefficient of drag (dependent on wind direction) (unitless) Gr = road gradient (unitless)
A = frontal area of bike+rider system (m2) Ec = efficiency of chain drive system (unitless)
FW = wheel rotation factor (expressed as incremental frontal area) (m2)
Mt = total mass of the bike and rider system (kg)
I = moment of inertia of the wheels (kgm2)
r = outside radius of the tyre (m)
Vgf = final ground velocity (m/s)
Vgi - initial ground velocity (m/s)
tf = final time (s)
ti = initial (s)
Crr = coefficient of rolling resistence (unitless)
g = acceleration due to gravity (9.81 m/s)
Gr = road gradient (unitless)
Ec = drivetrain efficiency (unitless)Posted 1 year ago # -
and it depends how fast you want to ride
not sure about that - think maybe the effort (watts) is the same just over a longer period of time?i know gym bikes aren't great but i've think got some sort of idea and yes i guess weight would be an issue but say an average weight rider on a bike that doesn't have shonky bearings and assume no head or tailwind or excess barfing
Posted 1 year ago # -
and it depends how fast you want to ride
not sure about that - think maybe the effort (watts) is the same just over a longer period of time?
Watt is a unit of power, i.e. energy over time.
Total energy may be the same (although probably not as I doubt your body is equally efficient at all power output levels), but power is directly proportional to speed.
Posted 1 year ago # -
Watt is a unit of power, i.e. energy over time
True
but power is directly proportional to speed.
Not really - coasting downhill gives you speed for no effort.
Posted 1 year ago # -
Assuming negligable resistnce from anything else (not a bad assumption on a steep-ish hill).
Then for a given hill,
E=mgh
P=E/tP=mgh/t
P = power output as usefull energy (moving you up the hill)
m = your bodymass in kg
g = 9.81
h = vertical height gained
t = time takenAs long as the hill is steep enough to keep you slow enough to ignore wind resistance, on a day with no wind it'll be fine.
And anyway, its just a number you can use to compare against yourself, who cares if its not accurate to more than +/-20W.
Posted 1 year ago # -
thanks labmonkey - what i was wondering was if that sort of calculation was available as a chart anywhere - with a few of the variables fixed as a lot relate to the bike which for timetrials is probably important but was loooking for "fattish old bloke" on "bog standard bike" at "no were near as fast as anyone else"
Posted 1 year ago # -
side note; the same equatios apply going downhill.
Sit in yoru normal climbing position, measure speed on a hill you know the gradient of and with some simple maths you can work out the energy lost through wind resistance.
Graph that against speed (your goign to need a lot of hills of different gradients to ride down), find the resistance at the speed you were climbing at and add that to the first power you calculated.
Posted 1 year ago # -
Assuming negligable resistnce from anything else (not a bad assumption on a steep-ish hill).
Then for a given hill,
E=mgh
P=E/tP=mgh/t
P = power output as usefull energy (moving you up the hill)
m = your bodymass in kg
g = 9.81
h = vertical height gained
t = time takenAs long as the hill is steep enough to keep you slow enough to ignore wind resistance, on a day with no wind it'll be fine.
And anyway, its just a number you can use to compare against yourself, who cares if its not accurate to more than +/-20W.
THIS!
Posted 1 year ago # -
thanks thisisnotaspoon looks simple enough (other than checking how a % road gradient is calculated!)
thanks GlenH for correcting me on watts = energy over time - total energy will be same over given distance but time to cover it is dependent on power output (without getting into efficiency)
Posted 1 year ago # -
get bicycling science by wilson has lots of graphs
power requirment versus speed and slope
8m/s 2% slope 220w
6% slope 550w
10% slope 750wPosted 1 year ago #
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