I have thrown down a few public challenges to help me with my goal of slimming down and getting fit.
One is to break one hour on Mt. Diablo in Northern California (South Gate).
It seems people analyzing pro climbing performances have used a formula to estimate watts/kg.
Relative power (W/kg) = VAM (m/hour) / (Gradient factor x 100)
Gradient factor = 2+(percentage grade/10), so 6% would be 2.6
Since Mt. Diablo is over an hour from where I live, I was hoping to use this to guesstimate what watts/kg I will need (and I am intersted what folks think of the formula, which could be useful for estimating requirements for hills outside your area).
So, if the climb is 11.3 miles at 5.8% average, it is 981.5m to the top. If the goal is just under an hour, I need to climb at least 982m/h.
According to the above formula:
982/((2+0.58)*100)= ~3.8w/kg. Honestly that does not seem too hard, though I am not sure I can do that for an hour at the moment.
I have not gone all out recently (trying to get lots of SST in), but I did do a recent effort on a 3.3 mile, 7.3% hill (Old la Honda) in a group so I went a bit harder than normal and did 21 minutes, so more like:
VAM:393/(21/60)= 1123
1123/273= 4.11 watts/kg
My feeling is that breaking one hour on Diablo is more of a challenge than this or even breaking 20 minutes on Old la Honda (I did just under 20 a few years ago when I was slimmer). If this math is correct, though, that should not be the case and my goal should be attainable now?
Any views on this formula or the watts/kg required for Diablo?
Thanks!
One is to break one hour on Mt. Diablo in Northern California (South Gate).
It seems people analyzing pro climbing performances have used a formula to estimate watts/kg.
Relative power (W/kg) = VAM (m/hour) / (Gradient factor x 100)
Gradient factor = 2+(percentage grade/10), so 6% would be 2.6
Since Mt. Diablo is over an hour from where I live, I was hoping to use this to guesstimate what watts/kg I will need (and I am intersted what folks think of the formula, which could be useful for estimating requirements for hills outside your area).
So, if the climb is 11.3 miles at 5.8% average, it is 981.5m to the top. If the goal is just under an hour, I need to climb at least 982m/h.
According to the above formula:
982/((2+0.58)*100)= ~3.8w/kg. Honestly that does not seem too hard, though I am not sure I can do that for an hour at the moment.
I have not gone all out recently (trying to get lots of SST in), but I did do a recent effort on a 3.3 mile, 7.3% hill (Old la Honda) in a group so I went a bit harder than normal and did 21 minutes, so more like:
VAM:393/(21/60)= 1123
1123/273= 4.11 watts/kg
My feeling is that breaking one hour on Diablo is more of a challenge than this or even breaking 20 minutes on Old la Honda (I did just under 20 a few years ago when I was slimmer). If this math is correct, though, that should not be the case and my goal should be attainable now?
Any views on this formula or the watts/kg required for Diablo?
Thanks!