Convert Watts To Pace?

[[old] Morning Wudi]

Is there a way to convert watts to 500 meter pace? I am assuming that on the Concept II, the two are directly proportional in some way. I did a perfunctory google search and on this forum but didn't come up with anything.<br /><br />Relatedly, are watts on the Concept II comparable to watts on the bicycle?<br /><br />Thanks,<br /><br />t.

[[old] Morning Wudi]

I did find one post (Stephen Seiler) about a specific formula provided by Concept II. "This formula closely approximates the expected cubic relationship between power and velocity"<br /><br />409 watts corresponds to 1:35 pace.<br />335.3 watts corresponded to 1:42 pace.<br /><br />See more on <a href='http://home.hia.no/~stephens/eliteerg.htm' target='_blank'>http://home.hia.no/~stephens/eliteerg.htm</a><br /><br />Anyone know how to extrapolate the formula from this data?<br /><br />Thanks,<br /><br />t.<br /> <br />

[[old] arakawa]

I've seen <a href='http://www.atm.ox.ac.uk/rowing/physics/ ... l#section9' target='_blank'>this discussion</a> in the The Physics of Ergometers FAQ. I've also taken the pace vs. watts data from the <a href='http://concept2.co.uk/guide/' target='_blank'>C2 training manual</a>, typed the data into a spreadsheet, and confirmed the cube rule relationship posited in the FAQ.<br /><br />To make a long story short, the pace vs. watts relationship is cubic - if you want to double the pace (e.g. go from 3:00 per 500 m to 1:30 per 500 m), you have to multiply your power output eight-fold (eight being two cubed).<br /><br />Given that the relationship is cubic and that the pace vs watts curve goes through the origin (i.e. if your speed is zero, your power output is zero), you only need one point on the curve (that is not the origin) to figure out all other pace/watt pairs. For example, assuming that the relationship 1:35 per 500m equals 409 W is true, then power = 2.8 v^3, where v is speed in meters per second, which is what the aforementioned FAQ states. The speed in meters per second is simply 500 m divided by the pace (500 m / 95 s = 5.26 m/s; 2.8 x 5.26 x 5.26 x 5.26 = 409).

[[old] Morning Wudi]

Thanks,<br /><br />That's exactly the info I was looking for. I thought I had seen it somewhere! <br /><br />T.<br /><br /><!--QuoteBegin-arakawa+Nov 10 2005, 09:46 PM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(arakawa @ Nov 10 2005, 09:46 PM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->I've seen <a href='http://www.atm.ox.ac.uk/rowing/physics/ ... l#section9' target='_blank'>this discussion</a> in the The Physics of Ergometers FAQ. I've also taken the pace vs. watts data from the <a href='http://concept2.co.uk/guide/' target='_blank'>C2 training manual</a>, typed the data into a spreadsheet, and confirmed the cube rule relationship posited in the FAQ.<br /><br />To make a long story short, the pace vs. watts relationship is cubic - if you want to double the pace (e.g. go from 3:00 per 500 m to 1:30 per 500 m), you have to multiply your power output eight-fold (eight being two cubed).<br /><br />Given that the relationship is cubic and that the pace vs watts curve goes through the origin (i.e. if your speed is zero, your power output is zero), you only need one point on the curve (that is not the origin) to figure out all other pace/watt pairs. For example, assuming that the relationship 1:35 per 500m equals 409 W is true, then power = 2.8 v^3, where v is speed in meters per second, which is what the aforementioned FAQ states. The speed in meters per second is simply 500 m divided by the pace (500 m / 95 s = 5.26 m/s; 2.8 x 5.26 x 5.26 x 5.26 = 409). <br /> </td></tr></table><br />

[[old] Ton]

I had the same discussion with a friend of me last week. he gave me the following page:<br /><a href='http://home.hccnet.nl/m.holst/ErgoDisp.html' target='_blank'>http://home.hccnet.nl/m.holst/ErgoDisp.html</a><br /><br />It might be the same information as you are just given.

[[old] Citroen]

<!--QuoteBegin-Morning Wudi+Nov 11 2005, 12:52 AM--><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><div class='genmed'><b>QUOTE(Morning Wudi @ Nov 11 2005, 12:52 AM)</b></div></td></tr><tr><td class='quote'><!--QuoteEBegin-->Is there a way to convert watts to 500 meter pace? I am assuming that on the Concept II, the two are directly proportional in some way. I did a perfunctory google search and on this forum but didn't come up with anything.<br /><br />Relatedly, are watts on the Concept II comparable to watts on the bicycle?<br /><br />Thanks,<br /><br />t. <br /> </td></tr></table><br /><br />According to <br /><a href='http://www.concept2.co.uk/guide/pdfs/tr ... pendix.pdf' target='_blank'>http://www.concept2.co.uk/guide/pdfs/tr ... pdf</a><br /><br />The formula for the C2 rower is <br /><br />Power (Watts) = 2.8 / ( (pace)^3)<br /><br />There's a really useful set of calculators at <a href='http://www.machars.net/spi.htm' target='_blank'>http://www.machars.net/spi.htm</a>

[[old] Carl Henrik]

I usually use (3.5 * 10^8) / (500m-pace)^3 since then you just need to plug in the 500m-pace given by the display. It's quite easy to remember as well if you know the speed of light: 3.0 * 10^8 m/s approxiamtely. just .5 more for the formula.