Concept2 Forum

Does anyone have a table of values for power compared to 500m split/ time, e.g.<br />2:00 - 205W<br />1:59 - 210W<br />1:58 - 217W<br />Does that make sense?

<!--QuoteBegin-ajspook+Jul 31 2005, 05:47 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Jul 31 2005, 05:47 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Does anyone have a table of values for power compared to 500m split/ time, e.g.<br />2:00 - 205W<br />1:59 - 210W<br />1:58 - 217W<br />Does that make sense? <br /> </td></tr></table><br /><br />There's a partial table on the British C2 site, stepped in 25-watt increments, at <a href='http://www.concept2.co.uk/guide/guide.p ... conversion' target='_blank'>Pace to Watts Conversion Table</a>.<br /><br />The formula is Watts =( 1/((Pace/Distance)^3)) * 2.8. Here are values for 2:00 splits down to 1:30, rounded to the nearest whole watt:<br /><br />Pace Watts<br />2:00 203<br />1:59 208<br />1:58 213<br />1:57 219<br />1:56 224<br />1:55 230<br />1:54 236<br />1:53 243<br />1:52 249<br />1:51 256<br />1:50 263<br />1:49 270<br />1:48 278<br />1:47 286<br />1:46 294<br />1:45 302<br />1:44 311<br />1:43 320<br />1:42 330<br />1:41 340<br />1:40 350<br />1:39 361<br />1:38 372<br />1:37 383<br />1:36 396<br />1:35 408<br />1:34 421<br />1:33 435<br />1:32 449<br />1:31 464<br />1:30 480<br /><br />Hope this helps.

Thanks Nav, exactly what I wanted.. cheers

<!--QuoteBegin-NavigationHazard+Jul 31 2005, 11:57 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(NavigationHazard @ Jul 31 2005, 11:57 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-ajspook+Jul 31 2005, 05:47 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Jul 31 2005, 05:47 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Does anyone have a table of values for power compared to 500m split/ time, e.g.<br />2:00 - 205W<br />1:59 - 210W<br />1:58 - 217W<br />Does that make sense? <br /> </td></tr></table><br /><br />There's a partial table on the British C2 site, stepped in 25-watt increments, at <a href='http://www.concept2.co.uk/guide/guide.p ... conversion' target='_blank'>Pace to Watts Conversion Table</a>.<br /><br />The formula is Watts =( 1/((Pace/Distance)^3)) * 2.8. Here are values for 2:00 splits down to 1:30, rounded to the nearest whole watt:<br /><br />Pace Watts<br />2:00 203<br />1:59 208<br />1:58 213<br />1:57 219<br />1:56 224<br />1:55 230<br />1:54 236<br />1:53 243<br />1:52 249<br />1:51 256<br />1:50 263<br />1:49 270<br />1:48 278<br />1:47 286<br />1:46 294<br />1:45 302<br />1:44 311<br />1:43 320<br />1:42 330<br />1:41 340<br />1:40 350<br />1:39 361<br />1:38 372<br />1:37 383<br />1:36 396<br />1:35 408<br />1:34 421<br />1:33 435<br />1:32 449<br />1:31 464<br />1:30 480<br /><br />Hope this helps. <br /> </td></tr></table><br /><br /><br />So I guess that pulling an average 1:35 pace for 500 metres works out at twice the amount of work as pulling an average 2:00 pace for 500 metres? Sounds about right.<br /><br />Please educate me though - I follow that 2:00 pace is the average time to cover 500 metres, but what does the wattage mean? Is it power for a specific time? or power over the distance of 500m?<br /><br />Excuse the ignorance folks!<br />RichardT

<!--QuoteBegin-R S T+Aug 2 2005, 09:07 AM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(R S T @ Aug 2 2005, 09:07 AM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><br />Please educate me though - I follow that 2:00 pace is the average time to cover 500 metres, but what does the wattage mean? Is it power for a specific time? or power over the distance of 500m?<br /><br />Excuse the ignorance folks!<br />RichardT <br /> </td></tr></table><br /><br />The flywheel is essentially a "Joule" absorber, Watts = Joules/Second(s), and in this case the seconds are determined by stroke rate, i.e. a SR=30 would be 2 seconds/stroke, if your pace were 2:00 (203 watts) it would mean that there were 406 joules absorbed during that 2 seconds; 406J/2sec=203 watts.<br /><br />Remember though that the Joules are being absorbed at a different rate during the stroke cycle, dependent on the RPM's and DF. So in the end, you are getting "average watts" for a given stroke reflected as a pace for that stroke. This is why increasing rate produces a faster pace, even if input force is constant, the denominator gets smaller, making the result (watts) larger.<br /><br />Any help?

<!--QuoteBegin-PaulS+Aug 2 2005, 05:42 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(PaulS @ Aug 2 2005, 05:42 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-R S T+Aug 2 2005, 09:07 AM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(R S T @ Aug 2 2005, 09:07 AM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><br />Please educate me though - I follow that 2:00 pace is the average time to cover 500 metres, but what does the wattage mean? Is it power for a specific time? or power over the distance of 500m?<br /><br />Excuse the ignorance folks!<br />RichardT <br /> </td></tr></table><br /><br />The flywheel is essentially a "Joule" absorber, Watts = Joules/Second(s), and in this case the seconds are determined by stroke rate, i.e. a SR=30 would be 2 seconds/stroke, if your pace were 2:00 (203 watts) it would mean that there were 406 joules absorbed during that 2 seconds; 406J/2sec=203 watts.<br /><br />Remember though that the Joules are being absorbed at a different rate during the stroke cycle, dependent on the RPM's and DF. So in the end, you are getting "average watts" for a given stroke reflected as a pace for that stroke. This is why increasing rate produces a faster pace, even if input force is constant, the denominator gets smaller, making the result (watts) larger.<br /><br />Any help? <br /> </td></tr></table><br /><br /><br />Spot on - thanks Paul. Males perfect sense now.<br /><br />RichardT

Out of interest does any one know what power is used on the recovery, i.e. to pull your body back up to the catch?

Hello,<br /><br />Have a trawl through this site<br /><br /><a href='http://www-atm.physics.ox.ac.uk/rowing/physics/#ergo' target='_blank'>http://www-atm.physics.ox.ac.uk/rowing/ ... rgo</a><br /><br />its fascinating and you will find the answer to your last question somewhere in here - i am sure I've seen it here before but cannot find it just now.<br /><br />Bill<br /><br /><br /><br /><br /><br />

<!--QuoteBegin-ajspook+Aug 7 2005, 03:27 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Aug 7 2005, 03:27 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Out of interest does any one know what power is used on the recovery, i.e. to pull your body back up to the catch? <br /> </td></tr></table><br /><br />On the Erg, with the assistance of the bungee (return mechanism) and normal rail slope, there is little required of us, however we did have to stretch the bungee and climb the hill, so basically it's a return from earlier effort.<br /><br />In a boat it gets rather more complicated and probably the most convenient answer is, "no". <br />

<!--QuoteBegin-PaulS+Aug 8 2005, 10:04 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(PaulS @ Aug 8 2005, 10:04 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-ajspook+Aug 7 2005, 03:27 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Aug 7 2005, 03:27 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Out of interest does any one know what power is used on the recovery, i.e. to pull your body back up to the catch? <br /> </td></tr></table><br /><br />On the Erg, with the assistance of the bungee (return mechanism) and normal rail slope, there is little required of us, however we did have to stretch the bungee and climb the hill, so basically it's a return from earlier effort.<br /><br />In a boat it gets rather more complicated and probably the most convenient answer is, "no". <br /> </td></tr></table><br />I read through Bills link, a 75kg rower at 30spm uses about 37W just moving himself, so add that to the score on the screen to get your real power..

<!--QuoteBegin-ajspook+Aug 10 2005, 10:49 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Aug 10 2005, 10:49 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-PaulS+Aug 8 2005, 10:04 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(PaulS @ Aug 8 2005, 10:04 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-ajspook+Aug 7 2005, 03:27 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Aug 7 2005, 03:27 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Out of interest does any one know what power is used on the recovery, i.e. to pull your body back up to the catch? <br /> </td></tr></table><br /><br />On the Erg, with the assistance of the bungee (return mechanism) and normal rail slope, there is little required of us, however we did have to stretch the bungee and climb the hill, so basically it's a return from earlier effort.<br /><br />In a boat it gets rather more complicated and probably the most convenient answer is, "no". <br /> </td></tr></table><br />I read through Bills link, a 75kg rower at 30spm uses about 37W just moving himself, so add that to the score on the screen to get your real power.. <br /> </td></tr></table><br /><br />Ah, so the heavier person even uses more, giving lwts an 'advantage' according to the PM (which doesn't account for this)? <br /><br />And you saying that it is 37 watts/stroke, or is it per 30 strokes?<br /><br />I'll stick with my earlier "no".

<!--QuoteBegin-PaulS+Aug 11 2005, 11:50 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(PaulS @ Aug 11 2005, 11:50 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-ajspook+Aug 10 2005, 10:49 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Aug 10 2005, 10:49 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-PaulS+Aug 8 2005, 10:04 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(PaulS @ Aug 8 2005, 10:04 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin--><!--QuoteBegin-ajspook+Aug 7 2005, 03:27 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Aug 7 2005, 03:27 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Out of interest does any one know what power is used on the recovery, i.e. to pull your body back up to the catch? <br /> </td></tr></table><br /><br />On the Erg, with the assistance of the bungee (return mechanism) and normal rail slope, there is little required of us, however we did have to stretch the bungee and climb the hill, so basically it's a return from earlier effort.<br /><br />In a boat it gets rather more complicated and probably the most convenient answer is, "no". <br /> </td></tr></table><br />I read through Bills link, a 75kg rower at 30spm uses about 37W just moving himself, so add that to the score on the screen to get your real power.. <br /> </td></tr></table><br /><br />Ah, so the heavier person even uses more, giving lwts an 'advantage' according to the PM (which doesn't account for this)? <br /><br />And you saying that it is 37 watts/stroke, or is it per 30 strokes?<br /><br />I'll stick with my earlier "no". <br /> </td></tr></table><br />Paul, I don't think you've still got your head around the power concept properly.<br />The 37W, is per stroke, per minutes, per 100m, what ever you want to make of it.<br />Power is simply energy expended per unit of time. So for example if a strole takes 2 seconds, and during the pull phase you 'pull' 400J (this is the energy required to lift a 10kg weight from the ground to 4m in the air - roughly), your average power for the whole stroke is 400/2, or 200W. You power for the actual pull would be 400/1=400W if the pull takes 1 sec. So the power reading you se on the ergo screen is simply a running average of energy in divided by time. Did that make sense?

In which case 37watts is a nonsense number. This means it takes your 75kg rower 37 joules of energy per second to move his body back down the slide on the recovery, right? So if a nice controlled recovery takes 2 seconds, say, he will use 74 joules of energy going back down, whereas if he rushes the recovery and does it in 1 second he'll use only 37joules? Can't be right.<br /><br />Sit yourself at the finish, with the handle in your hands, and take your feet off the foot plates - what happens? You slide down the rail to the front. Did you use any energy doing this? Put the handle down, try the same again, you still just roll down the slide to the front.<br /><br />Or are we talking about on the drive, and the energy used to push your body weight up the slight slope of the slide? I personally think this is completely negligible, because of the change in position of your CoG during the drive, up, and then down again.<br /><br />Pete

<!--QuoteBegin-ajspook+Aug 7 2005, 10:27 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(ajspook @ Aug 7 2005, 10:27 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Out of interest does any one know what power is used on the recovery, i.e. to pull your body back up to the catch?<br /> </td></tr></table><br /><br />But then it was the recovery that you asked about, which prompted Paul's reply. So I'll agree with Paul's answer. You don't pull your body back up the catch, you slide back down to the catch.<br /><br />Pete

Well the physics says it's 37W.. sounds fair to me.. you can't tell me that if you had no handle to pull and you swished back and forth on the slide it would use no energy, 37W isn't much after all.. obviously the 37W will vary if recovery is at different speeds.. Newtonian physics is a pretty well established field, and the article bill pointed to looked reasonable.<br />Part of that 37W would also be on the drive as well, and slowing the recovery towrds the catch position.<br />Also as a cyclist I know I can do 300W for about 20 minutes, I can do 240W for 30 on the ergo, so that helps explain the difference as well.<br />I'd bet the house that the figure is right, I don't understand how you can argue against it using hear say..