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I'm willing to be convinced otherwise, but...Carl Henrik wrote: A noticeable "die" at the end of short sprints is also not as bad as during longer distances, it could even be considered good. The die actually gives back of the energy you previously invested in flywheel speed. This energy you did not get paid for in dissipation/pace until the die. And since a larger part of total work is returned in short rows than in long ones the strategy will be relatively more efficient over short distances.
I don't see recovering the energy for more then a stroke or two. The fly wheel slows down too fast even at low drag to get much benefit from this. Seems to me that if you die with 50m to go and gut it out, you guarantee that you went all out--but there is not enough strokes to loose much time, so you still end up with a good time and know you couldn't have gone much faster
- Carl Henrik
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I hope you like math and physics:
The flywheel energy is proportional to the square of the the angular velocity of the flywheel. The angular velocity of the flywheel is propotional to the diplayed dissipation wattage ^(1/3). So the energy in the flywheel is proportional to displayed wattage ^(2/3).
If I do a die from pace A to pace B the fraction of energy left will be:
wattageB^(2/3) / wattageA^(2/3) = (wattageB/wattageA)^(2/3).
Lets say A = 1:24 (591w) and B = 1:28 (514w)
The fraction of energy left in flywheel after the die is (591/514)^(2/3) = ~ 90%
According to my estimates roughly 5% of total energy is in the flywheel for a 500m. If i died like above this would only be 4.5% and I would have gained 0.5% of total power accounted for which shows in average wattage.
At close to 600 watts a half percent is 3 watts.
The "noneven pace penalty" of doing a race at 1:24 for 76 seconds and then 1:28 for 9 seconds to get a 1:24.5 instead of just doing 1:24.5 all the way is only 0.3 watt.
So all in all I will gain .2 secs with a fly and die just because more energy is accounted for. This is a very big reason why fly and die is good for 500m (and also all other distances if implemented as an inevidable die at the last 100 m or so. You don't want to be plodding along at a died pace, and you don't want to do a fake die, pulling harder is always better. As distances grow though those .2 secs will seem relatively smaller and the strategy relatively weaker)
The flywheel energy is proportional to the square of the the angular velocity of the flywheel. The angular velocity of the flywheel is propotional to the diplayed dissipation wattage ^(1/3). So the energy in the flywheel is proportional to displayed wattage ^(2/3).
If I do a die from pace A to pace B the fraction of energy left will be:
wattageB^(2/3) / wattageA^(2/3) = (wattageB/wattageA)^(2/3).
Lets say A = 1:24 (591w) and B = 1:28 (514w)
The fraction of energy left in flywheel after the die is (591/514)^(2/3) = ~ 90%
According to my estimates roughly 5% of total energy is in the flywheel for a 500m. If i died like above this would only be 4.5% and I would have gained 0.5% of total power accounted for which shows in average wattage.
At close to 600 watts a half percent is 3 watts.
The "noneven pace penalty" of doing a race at 1:24 for 76 seconds and then 1:28 for 9 seconds to get a 1:24.5 instead of just doing 1:24.5 all the way is only 0.3 watt.
So all in all I will gain .2 secs with a fly and die just because more energy is accounted for. This is a very big reason why fly and die is good for 500m (and also all other distances if implemented as an inevidable die at the last 100 m or so. You don't want to be plodding along at a died pace, and you don't want to do a fake die, pulling harder is always better. As distances grow though those .2 secs will seem relatively smaller and the strategy relatively weaker)
Carl Henrik
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
Interesting argument.Carl Henrik wrote:....
The "noneven pace penalty" of doing a race at 1:24 for 76 seconds and then 1:28 for 9 seconds to get a 1:24.5 instead of just doing 1:24.5 all the way is only 0.3 watt.
I first thought about how much harder it is to increase the pace even slightly when one is at one's limit, The additional pain involved in going a half second faster can be really significant. I don't think the additional penalty of 0.3 watts is the best way to think about it. (I calculated .6 watts)
Then I thought about the energy left in the flywheel--The variation between the max and min speed is much more then the 90% ratio you use in your example. If you want to get the most energy out of the fly wheel, you may be better off timing your stroke to be at the catch just as you hit 500m where the flywheel is at its slowest. This is more beneficial the higher the drag factor and the slower the rating.
Then I thought about it this way:
If you follow Paul's Law, i.e. you can go twice as far for a 5 second increase in 500m split time, then going out a half second fast means you die at 466m rather then 500m. {500 * 2^(-.5/5) = 466, or .5 = 5 * log_2(500/466) }.
If you can manage 1:24.5 for 500m you will suffer just as much as you would for 466m at 1:24, but at 1:24 you would still have to gut out the last 34m. Which is painful but doesn't sound much slower.
If you power falls off more quickly then Paul's law then you get further before dieing. This is likely the case if you are a good sprinter, in which case you are relatively better off with your strategy.
In any case you convinced me that blowing up with 40-70m to go in a 500m piece is not necessarily a bad thing. Knowing that could come in very handy at some point!
Just curious how did you come up with the 5% of the total energy left? Seems about right, but I need the moment of Inertia (or other data to verify it).
- Carl Henrik
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How much is it? [edit: I got this to around 80% ]Nosmo wrote:....
Then I thought about the energy left in the flywheel--The variation between the max and min speed is much more then the 90% ratio you use in your example....
The strategy of not doing the last drive, let's think of it as another type of fake die, otw is potentially good but hard to implement. On the erg this type of fake die is always bad because there is no shell speed varying around the full system speed due to body movements.
The "no last drive" strategy on the water aims to cross the finish line with a high speed of the shell, whereas on the erg we want a low speed on the flywheel.
In order to implement the "no last drive" strategy on the erg it needs to be "nonfake" and set up by carefully monitoring meters left and changing the rate accordingly the last strokes. Sounds tricky but possibly managable with practice. If you can gain .3 secs ?? it's well worth it
Yes, I believe everyone should start their finishing sprint 100m earlier now that the secret is out. I believe Ebbesen often starts his sprint with something like 400 to go which seems a lot. But if you know that is is actually good to die during the last 100 it will help you both in putting all your energy in there but also to be more effective because of this strategy.Nosmo wrote:....
Knowing that could come in very handy at some point!
....
And with approximately 50m to go you should start changing rate so that you last drive finishes with something like 5m left. It's actually a new application for hitting the zeros, race finishing.
Nosmo wrote:....
Just curious how did you come up with the 5% of the total energy left? Seems about right, but I need the moment of Inertia (or other data to verify it).
My first ideas was to either try to get the moment of inertia from C2 or disassemble the erg and try to weigh the flywheel in some fancy manner. Both sounded like they would take time so instead I did somehting quick and dirty:
I assumed that during the first 6 strokes when the phosphocreatine storage is available I'm able to deliver my max amount of power every stroke.
This max is 900w at rate 50 at pull nr 6 with an assumed drive:recovery ratio of 3:2.
So when I'm pulling I'm able to deliver 5/3 * 900w = 1500w
After my first stroke the monitor might say I did 250w but I actually probably did around 900w (around in a very loose way).
The long time of the first drive affects negatively power during drive but the low speed allows more force and longer distance with full connection which affects positively the power during the drive. On top of this the higher drive:recovery ratio for the first 5 pulls affects the stroke power average positively and made me accept 900w as actual stroke average for all of the first 6 pulls.
So for the first stroke 650w goes into the flywheel for the duration of that stroke.
For the second pull displaying maybe 460w, 440w goes into the flywheel for the duration of that pull.
Continuing along this line I'll get an idea of how much energy there is in the flywheel at a few displayed wattages (6 strokes, 6 data points). Using flywheel energy proportionality to wattage^(2/3) I can further check that my results are reasonable and do a regression.
Carl Henrik
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
With the drag factor and the moment of inertia it is an easy calculation. I've seen the moment of inertia number in some posts so it is obviously available some where. I'm sure PaulS knows it.Carl Henrik wrote: How much is it? [edit: I got this to around 80% ]
I can also get the energy from the data files from ErgMonitor. It records the instantaneous speed 6 times per revolution. I wrote a program to translate the data files from ErgMonitor a while ago (if I remember correctly it is a simple ASCII encoded base 36 time between pulses: 0-9 then a-z).
Not so sure it is the best strategy, but it is not as bad as I originally assumed. The strategy of sprinting as if you 50m less to go and gutting out the rest is probably better then building the speed all the way to the end. At least that way one is guaranteed to empty the gas tank, and knowing that dieing at 50m to go is OK could be a big psychological boost.Carl Henrik wrote:Yes, I believe everyone should start their finishing sprint 100m earlier now that the secret is out. I believe Ebbesen often starts his sprint with something like 400 to go which seems a lot. But if you know that is is actually good to die during the last 100 it will help you both in putting all your energy in there but also to be more effective because of this strategy.Nosmo wrote:....
Knowing that could come in very handy at some point!
....
I wouldn't want to try to do any of these strategies on the water. Dieing with 50m to go on the water can really kill the boat speed if technique falls apart. Also one does not always know where the finish line is--you can't see it. For the races my club puts on, and I assume other courses as well, the finish line buoys are just past the actual line. This is so everyone rows across it.
Carl Henrik wrote:Nosmo wrote:....
Just curious how did you come up with the 5% of the total energy left? Seems about right, but I need the moment of Inertia (or other data to verify it).
My first ideas was to either try to get the moment of inertia from C2 or disassemble the erg and try to weigh the flywheel in some fancy manner. Both sounded like they would take time so instead I did something quick and dirty:
....
I'm not so sure your assumptions of how much energy you can put in to the flywheel are accurate, since the power you can produce will probably vary with hand speed a fair amount. Start up and steady state conditions may be too different. But I'm guessing here.
Of course this is also easily checked with the erg monitor data files (assuming one has my program).
- Carl Henrik
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I've had second thoughts about the idea of catching the zero for the finish. The alternative is to get more energy in there by realeasing the zero.
If you time rate for an extra drive the flywheel will have had a higher average speed over the last meters and hence higher dissipation as well, even though the return is less for the energy put in. Releasing a zero will give lesser fraction of total work accounted for but still more in absolute terms. The best way to finish is to up rate the last 2 strokes or so so that you can do 2.5 strokes instead and get the extra drive in there. I checked this under the condition that drive energy was proportional to recovery time.
If you fail to release a zero otoh you know you at least got some energy "for free".
On the PM2 it also seems to me like the average watt may jump up too easy by doing a very strong last pull, not sure though.
If you time rate for an extra drive the flywheel will have had a higher average speed over the last meters and hence higher dissipation as well, even though the return is less for the energy put in. Releasing a zero will give lesser fraction of total work accounted for but still more in absolute terms. The best way to finish is to up rate the last 2 strokes or so so that you can do 2.5 strokes instead and get the extra drive in there. I checked this under the condition that drive energy was proportional to recovery time.
If you fail to release a zero otoh you know you at least got some energy "for free".
On the PM2 it also seems to me like the average watt may jump up too easy by doing a very strong last pull, not sure though.
Carl Henrik
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
- johnlvs2run
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I always try to fit in the last drive with about 4 or 5 meters to go.
This way the end of the piece will come between the end of the last drive and midpoint of recovery.
Otherwise if you drive with 8 meters to go, then the last 4 meters are recovery and time lost.
This way the end of the piece will come between the end of the last drive and midpoint of recovery.
Otherwise if you drive with 8 meters to go, then the last 4 meters are recovery and time lost.
bikeerg 75 5'8" 155# - 18.5 - 51.9 - 568 - 1:52.7 - 8:03.8 - 20:13.1 - 14620 - 40:58.7 - 28855 - 1:23:48.0
rowerg 56-58 5'8.5" 143# - 1:39.6 - 3:35.6 - 7:24.0 - 18:57.4 - 22:49.9 - 7793 - 38:44.7 - 1:22:48.9 - 2:58:46.2
rowerg 56-58 5'8.5" 143# - 1:39.6 - 3:35.6 - 7:24.0 - 18:57.4 - 22:49.9 - 7793 - 38:44.7 - 1:22:48.9 - 2:58:46.2
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Seems all this effort to time the last stroke or 2 could just be put into focusing on giving it your all. Plus whatever you learn here you'll have to unlearn in the boat, since the boat moves the fastest during the first part of the recovery so you actually don't want to be taking a stroke as you cross the finish line.
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I briefly looked at one ergmonitor file this morning. Rowing with a 3:1 ratio at 24 spm, 115 DF, 118cm stroke length and a 1:57 pace, the flywheel looses ~57% of its energy from release to catch.
If one rowed with a ratio of 1:1 and at a higher rating then the amount of energy lost would be a good deal less, however, rowing faster and with more drag results in the flywheel slowing down relatively quicker. I'm too lazy to do the math at the moment.
I'm pretty skeptical of the original explanation of finishing with a slower average pace as an effective way to improve overall time.
If someone wants to send me an ergmonitor file with a much faster pace and higher DF and SPM, I'll take a look at it.
I'd also be happy to send anyone my translation program for the Ergmonitor files if someone wants to play with the data themselves.
BTW I was wrong about the format of the files, the time between pulses is encoded base 32, '0-9' & 'a-v'. Obviously (to a programmer at least) a better way of doing it.
If one rowed with a ratio of 1:1 and at a higher rating then the amount of energy lost would be a good deal less, however, rowing faster and with more drag results in the flywheel slowing down relatively quicker. I'm too lazy to do the math at the moment.
I'm pretty skeptical of the original explanation of finishing with a slower average pace as an effective way to improve overall time.
If someone wants to send me an ergmonitor file with a much faster pace and higher DF and SPM, I'll take a look at it.
I'd also be happy to send anyone my translation program for the Ergmonitor files if someone wants to play with the data themselves.
BTW I was wrong about the format of the files, the time between pulses is encoded base 32, '0-9' & 'a-v'. Obviously (to a programmer at least) a better way of doing it.
- Carl Henrik
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If I understand it correctly there seems to be somewhat less energy then in the flywheel than my estimate, not by magnitudes though.
1:57 <=> 219w
(3:1 ratio @ 24 spm ) => (2/3 * 60/24 = 5/3 = 1.67 secs recovery)
so approximately: 1.67 x 219 = 0.57 x Energy in flywheel
<=>
Energy in flywheel at this pace and drag = 640 joule
So for the same drag a, let's say, 1:50 (263w) would have E2 in the flywheel:
E2 = 640 * (263/219) ^(2/3) = 723 joule
723 - 640 = 83 joule released by a die from 1:50 to 1:57.
Not much.
83J / 110s = 0.75 watt "for free" on average watt.
If I do a 500 as 100 secs of 268w, and 10s of 219w for a 1:50 time and 263w average the noneven punishment is 0.27w.
0.75w - 0.27w = 0.48w
This free half watt is 0.07secs at 1:50 pace so with 70% chance rower of this 500m will get a whole tenth of a second for free on his 500 due to dieing. It's worth it And, at least, you don't need to fear that these types of fly and die are wasting energy
1:57 <=> 219w
(3:1 ratio @ 24 spm ) => (2/3 * 60/24 = 5/3 = 1.67 secs recovery)
so approximately: 1.67 x 219 = 0.57 x Energy in flywheel
<=>
Energy in flywheel at this pace and drag = 640 joule
So for the same drag a, let's say, 1:50 (263w) would have E2 in the flywheel:
E2 = 640 * (263/219) ^(2/3) = 723 joule
723 - 640 = 83 joule released by a die from 1:50 to 1:57.
Not much.
83J / 110s = 0.75 watt "for free" on average watt.
If I do a 500 as 100 secs of 268w, and 10s of 219w for a 1:50 time and 263w average the noneven punishment is 0.27w.
0.75w - 0.27w = 0.48w
This free half watt is 0.07secs at 1:50 pace so with 70% chance rower of this 500m will get a whole tenth of a second for free on his 500 due to dieing. It's worth it And, at least, you don't need to fear that these types of fly and die are wasting energy
Carl Henrik
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
Carl,Carl Henrik wrote:If I understand it correctly there seems to be somewhat less energy then in the flywheel than my estimate, not by magnitudes though.
1:57 <=> 219w
(3:1 ratio @ 24 spm ) => (2/3 * 60/24 = 5/3 = 1.67 secs recovery)
Don't think you are thinking about this correctly. The power reading is the average power over the entire drive.
24 spm at a 1:57 or 219W, means you are averaging 219W over 60/24=2.5 s. Each strokes, adds 219W * 2.5s = 547.5J to the flywheel, and the same amount is lost on the recovery. Using the 57% figure for energy left, this means the energy in the flywheel changes from 725J to 1273J during the stroke.
Now say, you are rowing at a 1:50 (263W) at 24 spm, then you are adding
657J to the flywheel, but you could imagine a slightly higher ratio higher so the flywheel has more time to slow down, then the final flywheel speed is the same as above: the energy in the flywheel would change from 725J to 1382J
The difference between the max flywheel energy in these two cases is much smaller then the variation during a stroke. So one could easily argue that timing the final stroke is more of a benefit then dieing late [edited: originally said early].
Looking at it another way:
Just to pick convient numbers: Imagine you are at a drag factor such that you are rowing at 30 rating, 2:1 ratio, and a 1:40 pace and energy in the flywheel at the catch is also 57% of the max. The speed of the flywheel changes by sqrt(.57)=75%. A 1:40 pace is 5 m/s. The pace will change speed will change from roughly 4.5 to 6 (quick estimating here--it is not a linear average). So if you finish just after the peak in speed vs just after the valley, the difference is speed may be close to a full m/s, or at this pace close to 0.2 seconds.
Also Back to Paul's law:
The human body is capable of producing more energy in a given amount of time if its power output does not vary. Obviously it is much easier to row 1000m at a 1:45 then it is to do the first 500m in 60 seconds and the second in 2:30. The real question becomes, how fast can you go after dieing. If you die with 50 meters to go, Every second in pace/500 you slow down, takes 0.1 seconds longer to finish.
This is all kind of pedantic! I still think, no matter what one does--timing the final stroke, dieing with50 m to go, doing a steady pace--knowing what one can do and choosing the correct pace is much more important then any of the strategies we have been discussing.
Last edited by Nosmo on May 2nd, 2008, 8:11 pm, edited 1 time in total.
- Carl Henrik
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Having been a fairly regular visitor to the C2 forums for 5 years or so I too feel this is not the core of erg strategizing, but I don't think these strategies have been analyzed to this detail before and it's nice to learn some new stuff.This is all kind of pedantic! I still think, no matter what one does--timing the final stroke, dieing with50 m to go, doing a steady pace--knowing what one can do and choosing the correct pace is much more important then any of the strategies we have been discussing.
First off, which of these do you mean?:
1)
2)the flywheel looses ~57% of its energy from release to catch
I used your first statement having not heard the second, if it's the second with more energy in the flywheel then certainly 1 tenth of a second would be "free" due to dieing.Using the 57% figure for energy left
[edited]Your precise way to get the energy dissipation I'm not convinded by[/edited]
I interpret you mean dieing late. One could argue so but I don't know how successfully since the timing for the last stroke should be such that you cross the finish at the release of the last stroke, not at the catch (of the stroke that will not be).So one could easily argue that timing the final stroke is more of a benefit then dieing early.
Managing the rate for the last strokes it is more rewarding to opt for putting more enegy in by sneaking in an extra drive than it is to get a high percentage accounted for from less energy. As I discussed in previous post. How much more rewarding it is I did not determine though.
The "late die" strategy at 50m left has already guaranteed as good as max energy put in but it succeeds in giving a higher erg internal efficiency too.
Last edited by Carl Henrik on May 4th, 2008, 5:39 am, edited 1 time in total.
Carl Henrik
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
- Carl Henrik
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Again I was first convinced but now have second thoughts of a method.Each strokes, adds 219W * 2.5s = 547.5J to the flywheel, and the same amount is lost on the recovery.
All though you can precisely determine how much work is performed each drive, all that energy will not be saved in the flywheel, waiting to start dissipating until the recovery. Much of the 547.5J will already have dissipated when the recovery starts. Although how much precisely is difficult to say.
We do know the flywheel speed is fairly similar during drive and recovery though and averaging a corresponding 219w. So recovery time of 1.67 times 219 is a reasonable approximation for the recovery dissipation at least.
Maybe this is to off topic or to deep in topic to continue on this thread though
Carl Henrik
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min
M27lwt, 181cm
1:13@lowpull, 15.6@100m, 48.9@300m, (1:24.4)/(1:24.5)@500m, 6:35@2k, 36:27.2@10k, 16151m@60min