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Translate watt savings over a time trial/average power?
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I don't understand how to translate bike watts savings to a race. If I average 230 watts over 4.5 hours, say, without a "10 watt savings" what can I expect my average watts to be with said 10 watt savings (all other things being equal).
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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downtowndj wrote:
I don't understand how to translate bike watts savings to a race. If I average 230 watts over 4.5 hours, say, without a "10 watt savings" what can I expect my average watts to be with said 10 watt savings (all other things being equal).

If every single variable was the same, and you put on a bike addition or clothing adjustment that truly was "worth 10 free watts", then you could think of it two ways:

1. You could average 240 watts for a little less than 4.5 hours doing the same exact effort as before
2. You could average 230 watts over 4.5 hours with a slightly less effort

So in this case I think you would be #1. You want the same effort but you want to apply a 10 watts savings. I think you just take the 230 watts and add 10 to 240.

I could be wrong, but that is the general idea I think...

...I think

"If it costs you 30 minutes at Maryland so what" -dwreal
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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The general rule of thumb:

5 watts savings = 0.5 sec/km faster

blog
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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SteveJ is right... if you're saving 10 watts, you'll go 1sec/km faster at normal cycling speeds than if you weren't saving those watts. Over an Ironman, 10 watts is roughly 3 minutes at normal cycling speeds.

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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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downtowndj wrote:
I don't understand how to translate bike watts savings to a race. If I average 230 watts over 4.5 hours, say, without a "10 watt savings" what can I expect my average watts to be with said 10 watt savings (all other things being equal).

230 Watts

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Re: Translate watt savings over a time trial/average power? [JasoninHalifax] [ In reply to ]
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JasoninHalifax wrote:
downtowndj wrote:
I don't understand how to translate bike watts savings to a race. If I average 230 watts over 4.5 hours, say, without a "10 watt savings" what can I expect my average watts to be with said 10 watt savings (all other things being equal).


230 Watts


Ha... I get it... I should have asked the question differently.
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Re: Translate watt savings over a time trial/average power? [JasoninHalifax] [ In reply to ]
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JasoninHalifax wrote:
downtowndj wrote:
I don't understand how to translate bike watts savings to a race. If I average 230 watts over 4.5 hours, say, without a "10 watt savings" what can I expect my average watts to be with said 10 watt savings (all other things being equal).


230 Watts

Damn that's funny. Spit my coffee funny. ...and perfectly, exactly true.
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Re: Translate watt savings over a time trial/average power? [bluestacks867] [ In reply to ]
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bluestacks867 wrote:
downtowndj wrote:
I don't understand how to translate bike watts savings to a race. If I average 230 watts over 4.5 hours, say, without a "10 watt savings" what can I expect my average watts to be with said 10 watt savings (all other things being equal).


If every single variable was the same, and you put on a bike addition or clothing adjustment that truly was "worth 10 free watts", then you could think of it two ways:

1. You could average 240 watts for a little less than 4.5 hours doing the same exact effort as before
2. You could average 230 watts over 4.5 hours with a slightly less effort

So in this case I think you would be #1. You want the same effort but you want to apply a 10 watts savings. I think you just take the 230 watts and add 10 to 240.

I could be wrong, but that is the general idea I think...

...I think
You're confusing things.
You need to state where you're measuring the power for this version to make any sense.
In general what you say will be incorrect. It would only be correct if the saving was made in the drivetrain, upstream of the power measurement, and you're talking about the measurement not the actual power produced.
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Re: Translate watt savings over a time trial/average power? [JasoninHalifax] [ In reply to ]
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JasoninHalifax wrote:
downtowndj wrote:
I don't understand how to translate bike watts savings to a race. If I average 230 watts over 4.5 hours, say, without a "10 watt savings" what can I expect my average watts to be with said 10 watt savings (all other things being equal).


230 Watts
yep, quite obviously this.

All a "10 watt saving" means is that it will take 10 watts less to achieve the specified speed, typically a 40km/h airspeed if we're talking aerodynamics, a 30km/h or 40km/h road speed if we're talking rolling resistance, etc...
The known quantity is your ability to produce power - and we don't expect that to change because of what it's being used for. The improvements to be gained by "saving watts" are separate and answered by others here.
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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regarding "230 watts over 4.5 hours" what distance is that over?
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Re: Translate watt savings over a time trial/average power? [ericMPro] [ In reply to ]
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ericMPro wrote:
SteveJ is right... if you're saving 10 watts, you'll go 1sec/km faster at normal cycling speeds than if you weren't saving those watts. Over an Ironman, 10 watts is roughly 3 minutes at normal cycling speeds.

This.

Short of actually going out and riding the course a bunch of times with and without the aero saving. This is the best estimate that the OP is going to get.
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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Hello

let's take an example, with a rider of 75kg, CdA 0,22 average over the course, delivering 230w constantly. Flat course.

With some other hypothesis fixed (bike friction, tyre loss, road efficiency, air pressure, ....), average speed is 37.7 km/h.

On a 180km IM, this lead to 4,77h time (approx. 4h 46n12s).

If you go CdA 0,21 average through new helmet + position change + new wheel + ... whatever, all other hypothesis staying unchanged, you save 7 watts.

If you keep pushing 230w average, then speed goes up to 38.2 (not going to far in precision).

Gain is 0,5 km/h for 7 watt aero gain.

So, kind of 0,07 km/h per watt. This value change if wattage, CdA, or delta gain, or other parameters... change.

Over 180km, time go down to 4,71 ( approx. 4h 42mn 36s)

So IM gain is 3mn36s, for 7 watts.

Kind of 31s per watt gain for an IM.
So, 15,5s gain per watt for a 70.3, at this power/CdA/ + other hypothesis.

All this being non linear, this value per watt do not apply on a linear way, so can only be applied for limited gains (5 to 10 watts), with equivalent hypothesis.

Philippe
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Re: Translate watt savings over a time trial/average power? [Ai_1] [ In reply to ]
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Ai_1 wrote:
You're confusing things.
You need to state where you're measuring the power for this version to make any sense.

I don't know... I mean if I ride a max effort at 290 watts and a magical fairy turns back time and gives me 10 free watts, I feel like I could ride max effort at 300 watts.

...or ride a little bit below max effort at 290 watts again.


hmm....

"If it costs you 30 minutes at Maryland so what" -dwreal
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Re: Translate watt savings over a time trial/average power? [stevej] [ In reply to ]
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This seems simple enough for me to understand. It seems like the description of whatever savings one gets in terms of watts should be based on some time/distance metric... like watts/hour something? Science and math are NOT my strong suit (obviously).


stevej wrote:
The general rule of thumb:

5 watts savings = 0.5 sec/km faster
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Re: Translate watt savings over a time trial/average power? [Pyrenean Wolf] [ In reply to ]
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Pyrenean Wolf wrote:
Hello

let's take an example, with a rider of 75kg, CdA 0,22 average over the course, delivering 230w constantly. Flat course.

With some other hypothesis fixed (bike friction, tyre loss, road efficiency, air pressure, ....), average speed is 37.7 km/h.

On a 180km IM, this lead to 4,77h time (approx. 4h 46n12s).

If you go CdA 0,21 average through new helmet + position change + new wheel + ... whatever, all other hypothesis staying unchanged, you save 7 watts.

If you keep pushing 230w average, then speed goes up to 38.2 (not going to far in precision).

Gain is 0,5 km/h for 7 watt aero gain.

So, kind of 0,07 km/h per watt. This value change if wattage, CdA, or delta gain, or other parameters... change.

Over 180km, time go down to 4,71 ( approx. 4h 42mn 36s)

So IM gain is 3mn36s, for 7 watts.

Kind of 31s per watt gain for an IM.
So, 15,5s gain per watt for a 70.3, at this power/CdA/ + other hypothesis.

All this being non linear, this value per watt do not apply on a linear way, so can only be applied for limited gains (5 to 10 watts), with equivalent hypothesis.

Philippe
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Re: Translate watt savings over a time trial/average power? [bluestacks867] [ In reply to ]
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bluestacks867 wrote:
Ai_1 wrote:

You're confusing things.
You need to state where you're measuring the power for this version to make any sense.


I don't know... I mean if I ride a max effort at 290 watts and a magical fairy turns back time and gives me 10 free watts, I feel like I could ride max effort at 300 watts.

...or ride a little bit below max effort at 290 watts again.


hmm....
Wattage is your output. You aren't changing, you will put out 290w wearing a skin suit, you will put out 290w wearing jeans and a sweatshirt. It all matter what forces are you overcoming with those 290w. If you get "10 free watts", it's just taking fewer watts to overcome those forces, ergo you put out 280w to overcome them. The extra 10w (since you are a constant 290w) goes towards increased speed.
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Re: Translate watt savings over a time trial/average power? [ripple] [ In reply to ]
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Your output doesn’t change. But the measurement of your output changes depending on where the measuring device is located.

If you gain 10w by reducing chain and pulley friction, then you won’t see any difference in your crank or pedal based PM, but your Powertap hub would read 10w higher.

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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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downtowndj wrote:
This seems simple enough for me to understand. It seems like the description of whatever savings one gets in terms of watts should be based on some time/distance metric... like watts/hour something? Science and math are NOT my strong suit (obviously).


stevej wrote:
The general rule of thumb:

5 watts savings = 0.5 sec/km faster

:-)

with the hypothesis(s) I took, it comes out at :

5 watts saving = 0,86 sec/km faster

0,5 sec/km certainly valid for significantly higher watt value ?
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Re: Translate watt savings over a time trial/average power? [ripple] [ In reply to ]
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Yes, this is kind of my point I think. To translate the marketing claims into something tangible is a bit convoluted. Unless that is just me and everyone else sees "5 watts per tire" and goes "Oh! I'll shave 2 minutes off of my Ironman bike at the same output!".


ripple wrote:
bluestacks867 wrote:
Ai_1 wrote:

You're confusing things.
You need to state where you're measuring the power for this version to make any sense.


I don't know... I mean if I ride a max effort at 290 watts and a magical fairy turns back time and gives me 10 free watts, I feel like I could ride max effort at 300 watts.

...or ride a little bit below max effort at 290 watts again.


hmm....

Wattage is your output. You aren't changing, you will put out 290w wearing a skin suit, you will put out 290w wearing jeans and a sweatshirt. It all matter what forces are you overcoming with those 290w. If you get "10 free watts", it's just taking fewer watts to overcome those forces, ergo you put out 280w to overcome them. The extra 10w (since you are a constant 290w) goes towards increased speed.
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Re: Translate watt savings over a time trial/average power? [JasoninHalifax] [ In reply to ]
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JasoninHalifax wrote:
Your output doesn’t change. But the measurement of your output changes depending on where the measuring device is located.

If you gain 10w by reducing chain and pulley friction, then you won’t see any difference in your crank or pedal based PM, but your Powertap hub would read 10w higher.
I realize now we're splitting hairs but that's an issue with measurement. Without gains in fitness, the rider himself is not magically putting out 10w more. If the measurement now reads 10w higher, he was actually already at 300w. (I know that you know where I'm coming from.)
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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Speaking from my POV, yes.. that's exactly what they see.
downtowndj wrote:
Yes, this is kind of my point I think. To translate the marketing claims into something tangible is a bit convoluted. Unless that is just me and everyone else sees "5 watts per tire" and goes "Oh! I'll shave 2 minutes off of my Ironman bike at the same output!".


ripple wrote:
bluestacks867 wrote:
Ai_1 wrote:

You're confusing things.
You need to state where you're measuring the power for this version to make any sense.


I don't know... I mean if I ride a max effort at 290 watts and a magical fairy turns back time and gives me 10 free watts, I feel like I could ride max effort at 300 watts.

...or ride a little bit below max effort at 290 watts again.


hmm....

Wattage is your output. You aren't changing, you will put out 290w wearing a skin suit, you will put out 290w wearing jeans and a sweatshirt. It all matter what forces are you overcoming with those 290w. If you get "10 free watts", it's just taking fewer watts to overcome those forces, ergo you put out 280w to overcome them. The extra 10w (since you are a constant 290w) goes towards increased speed.
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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my brain hurts. good points everyone. following.

"If it costs you 30 minutes at Maryland so what" -dwreal
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Re: Translate watt savings over a time trial/average power? [downtowndj] [ In reply to ]
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downtowndj wrote:
Yes, this is kind of my point I think. To translate the marketing claims into something tangible is a bit convoluted. Unless that is just me and everyone else sees "5 watts per tire" and goes "Oh! I'll shave 2 minutes off of my Ironman bike at the same output!"
The problem is that people and companies are trying to communicate technical information to a non-technical audience. It's always tricky.
As it stands the standard means for conveying the benefit from various changes in equipment, position, etc, is to provide a "watts saved" figure based on a declared speed. That's already a distortion of the actual data because that's not really what they're measuring. IMO providing it as seconds over a given race distance would be even worse. It might make it easily digestible but that's at the expense of adding further assumptions and goes further down the road of degrading the data's accuracy and legitimacy.
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Re: Translate watt savings over a time trial/average power? [Ai_1] [ In reply to ]
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I just keep chuckling to myself every time a cycling product website has an advertisement for the $1200 ceramic speed jockey wheels. LOL! This topic made me think of that for some reason.


Also, while this stuff is entirely relevant, and it is people's own money..........I feel like a lot of money is wasted on bling when instead people aren't even using the time they've already got on hand to be faster.


So many people would save so much more time and energy getting fit properly and doing personal testing than just buying the bling. Or just bothering to sweat a little during a workout instead of thinking endless Z1/Z2 is the answer.


They can't be bothered to feel the pain to improve but will believe anything posted on a product website.


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Re: Translate watt savings over a time trial/average power? [burnthesheep] [ In reply to ]
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Watt production : Z2 is only 80% of the answer, Z3/Z4/Z5 are the other 20%

Watt consumption : body being 70% or 80% of drag, the best way to save watts is to improve position, which can be free of charge
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