Hour Record Official Discussion Thread

[SeriousSam]

Wiggins rides for WIGGINS, not Sky.

 

[Jancouver]

Re:

Wiggins rides for WIGGINS, not Sky.

and WIGGINS is sponsored by SKY

... or who exactly is paying for this show?

 

[Gaylord Figman]

Wiggo like Team Sky promises much but delivers little
Lost count of the races that they were just going to just turn up and win
They should keep their mouths shut and let the racing do the talking

 

[NUFCrichard]

Wiggo like Team Sky promises much but delivers little
Lost count of the races that they were just going to just turn up and win
They should keep their mouths shut and let the racing do the talking

Sky have won the tour twice in the last 3 years! They have won countless other stage races, plenty of stages in the big tours too. Wiggins also won the tour, is world road TT champion and has x olympic golds..

I don't like Wiggo, but promising lots and delivering little obviously does not apply to Wiggins.

 

[Bavarianrider]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

 

[Singer01]

Re:

will struggle to beat Dowsetts record
quote]

the hour record is something you pretty much know roughly how you are going to do due to the limited number of variables. if he says he is aiming for 55, no way he struggles to beat the current record.

 

[Taxus4a]

Will be historic for cyling.

 

[Jancouver]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

Not sure where you came up with the 6.8 w/kg. Thats just pure BS.

Track speed is slightly more about raw power and Wigo purposely gained some weight just to gain few extra watts. Those few extra kilos will not create much of an extra drag or extra power needed to carry the extra weight.

So for example, if he gained 5% of extra power from the additional muscle/weight gain that may equal to 2% of extra speed. Thats all he needs to do 55kph as his power was already few % superior over Dowsett before he even gained the extra weight/power and he needs 4% to crack 55kmh.

 

[Bustedknuckle]

Wiggo like Team Sky promises much but delivers little
Lost count of the races that they were just going to just turn up and win
They should keep their mouths shut and let the racing do the talking

http://en.wikipedia.org/wiki/List_of_Team_Sky_wins

 

[Gaylord Figman]

Re: Re:

will struggle to beat Dowsetts record
quote]

the hour record is something you pretty much know roughly how you are going to do due to the limited number of variables. if he says he is aiming for 55, no way he struggles to beat the current record.

In the recent motorpaced 10ml TT Wiggo rode at roughly the right pace to just beat Dowsetts record so I suggest that is about where he is at.That was also 18mins not an hour!
55kms is cloud cuckoo land!
He was going to win the Giro then it was Paris Roubaix now its the hour record
We shall know on Sunday

 

[Bavarianrider]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

Not sure where you came up with the 6.8 w/kg. Thats just pure BS.

Track speed is slightly more about raw power and Wigo purposely gained some weight just to gain few extra watts. Those few extra kilos will not create much of an extra drag or extra power needed to carry the extra weight.

So for example, if he gained 5% of extra power from the additional muscle/weight gain that may equal to 2% of extra speed. Thats all he needs to do 55kph as his power was already few % superior over Dowsett before he even gained the extra weight/power and he needs 4% to crack 55kmh.

Well that's not ***, that's mathematics. No matter what Wiggins did, he can't defy the laws of physics. 6.8Watt/Kilo is the least amount of power a male at Wiggins size/weight has roughly to produce to go 55km/h on a wooden track on those bikes.

By the way, you do realize that extra muscles need extra oxygen support, right?

 

[Richeypen]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

Not sure where you came up with the 6.8 w/kg. Thats just pure BS.

Track speed is slightly more about raw power and Wigo purposely gained some weight just to gain few extra watts. Those few extra kilos will not create much of an extra drag or extra power needed to carry the extra weight.

So for example, if he gained 5% of extra power from the additional muscle/weight gain that may equal to 2% of extra speed. Thats all he needs to do 55kph as his power was already few % superior over Dowsett before he even gained the extra weight/power and he needs 4% to crack 55kmh.

Well that's not ***, that's mathematics. No matter what Wiggins did, he can't defy the laws of physics. 6.8Watt/Kilo is the least amount of power a male at Wiggins size/weight has roughly to produce to go 55km/h on a wooden track on those bikes.

By the way, you do realize that extra muscles need extra oxygen support, right?

Evidence???

 

[Bavarianrider]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

Not sure where you came up with the 6.8 w/kg. Thats just pure BS.

Track speed is slightly more about raw power and Wigo purposely gained some weight just to gain few extra watts. Those few extra kilos will not create much of an extra drag or extra power needed to carry the extra weight.

So for example, if he gained 5% of extra power from the additional muscle/weight gain that may equal to 2% of extra speed. Thats all he needs to do 55kph as his power was already few % superior over Dowsett before he even gained the extra weight/power and he needs 4% to crack 55kmh.

Well that's not ***, that's mathematics. No matter what Wiggins did, he can't defy the laws of physics. 6.8Watt/Kilo is the least amount of power a male at Wiggins size/weight has roughly to produce to go 55km/h on a wooden track on those bikes.

By the way, you do realize that extra muscles need extra oxygen support, right?

Evidence???

Well check your math


The physics affecting cycling at constant speed

This web page uses physical models of forces on a cyclist to help you estimate the relationship between power P (watts) and velocity V (kph or mph) of a cyclist. To do this, you need to estimate several parameters; reasonable defaults are given.

There are three primary forces that you, as a cyclist, must overcome in order to move forward:

Gravity: If you're cycling uphill, you're fighting against gravity,but if you're cycling downhill, gravity works for you. This page measures the steepness of a hill in terms of percentage grade G: rise divided by run, multiplied by 100. The heavier you and your bike are, the more energy you must spend to overcome gravity. The combined weight of you (the cyclist) and your bike is W (kg). The gravitational force constant g is 9.8067 (m/s2).
The formula for gravitational force acting on a cyclist, in metric units, is:

Fgravity (Newtons) = 9.8067 (m/s2) · sin(arctan(G/100)) · W (kg)
Rolling resistance: Friction between your tires and the road surface slows you down. The bumpier the road, the more friction you'll experience; the higher quality your tires and tube, the less friction you'll experience. As well, the heavier you and your bike are, the more friction you'll experience. There is a dimensionless parameter, called the coefficient of rolling resistance, or Crr, that captures the bumpiness of the road and the quality of your tires.
The formula for the rolling resistance acting on a cyclist, in metric units, is:

Frolling (Newtons) = 9.8067 (m/s2) · cos(arctan(G/100)) · W (kg) · Crr
Aerodynamic drag: As you cycle through the air, your bike and body need to push the air around you, similar to how a snowplow pushes snow out of the way. Because of this, the air exerts a force against you as you ride. There are a few things that dictate how much force the air exerts against you. The faster you ride, velocity V (m/s), the more force the air pushes against you. As well, you and your bike present a certain frontal area A (m2) to the air. The larger this frontal area, the more air you have to displace, and the larger the force the air pushes against you. This is why cyclists and bike manufacturers try hard to minimize frontal area in an aerodynamic position. The air density Rho (kg/m3) is also important; the more dense the air, the more force it exerts on you.
Finally, there are other effects, like the slipperyness of your clothing and the degree to which air flows laminarly rather than turbulently around you and your bike. Optimizing your aerodynamic positions also help with this. These other effects are captured in another dimensionless parameter called the drag coefficient, or Cd. Sometimes you will see people talking about "Cd · A", or CdA. This is just the drag coefficient Cd multiplied by the frontal area A. Unless you have access to a wind tunnel, it is hard to measure Cd and A separately; instead, people often just measure or infer Cd · A as a combined number.

The formula for the aerodynamic drag acting on a cyclist, in metric units, is:

Fdrag (Newtons) = 0.5 · Cd · A (m2) · Rho (kg/m3) · (V (m/s))2
The total force resisting you, the cyclist, is the sum of these three forces:
Fresist (Newtons) = Fgravity + Frolling + Fdrag
For each meter that you cycle forward, you spend energy overcoming this resistive force. The total amount of energy you must expend to move a distance D (m) against this force is called the Work (Joules) that you do:
Work (Joules) = Fresist (Newtons) · D (m)
If you are moving forward at velocity V (m/s), then you must supply energy at a rate that is sufficient to do the work to move V meters each second. This rate of energy expenditure is called power, and it is measured in watts. The power Pwheel (watts) that must be provided to your bicyle's wheels to overcome the total resistive force Fresist (Newtons) while moving forward at velocity V (m/s) is:
Pwheel (watts) = Fresist (Newtons) · V (m/s)
You, the cyclist, are the engine providing this power. The power that must be provided to your bicycle's wheels comes from your legs, but not all of the power that your legs deliver make it to the wheels. Friction in the drive train (chains, gears, bearings, etc.) causes a small amount of loss, usually around 3%, assuming you have a clean and nicely lubricated drivetrain. Let's call the percentage of drivetain loss Lossdt (percent).
So, if the power that your legs provide is Plegs (watts), then the power that makes it to the wheel is:

Pwheel (watts) = (1 - (Lossdt/100)) · Plegs (watts)
Putting it all together, the equation that relates the power produced by your legs to the steady-state speed you travel is:
Plegs (watts) = (1-(Lossdt/100))-1 · (Fgravity + Frolling + Fdrag) · V (m/s)
or, more fully:

Plegs (watts) = (1-(Lossdt/100))-1 · ( ( 9.8067 (m/s2) · W (kg) · ( sin(arctan(G/100)) + Crr · cos(arctan(G/100)) ) ) + ( 0.5 · Cd · A (m2) · Rho (kg/m3) · (V (m/s))2 ) ) · V (m/s)

 

[mektronic]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

Not sure where you came up with the 6.8 w/kg. Thats just pure BS.

Track speed is slightly more about raw power and Wigo purposely gained some weight just to gain few extra watts. Those few extra kilos will not create much of an extra drag or extra power needed to carry the extra weight.

So for example, if he gained 5% of extra power from the additional muscle/weight gain that may equal to 2% of extra speed. Thats all he needs to do 55kph as his power was already few % superior over Dowsett before he even gained the extra weight/power and he needs 4% to crack 55kmh.

Well that's not ***, that's mathematics. No matter what Wiggins did, he can't defy the laws of physics. 6.8Watt/Kilo is the least amount of power a male at Wiggins size/weight has roughly to produce to go 55km/h on a wooden track on those bikes.

By the way, you do realize that extra muscles need extra oxygen support, right?

Evidence???

Well check your math


The physics affecting cycling at constant speed

This web page uses physical models of forces on a cyclist to help you estimate the relationship between power P (watts) and velocity V (kph or mph) of a cyclist. To do this, you need to estimate several parameters; reasonable defaults are given.

There are three primary forces that you, as a cyclist, must overcome in order to move forward:

Gravity: If you're cycling uphill, you're fighting against gravity,but if you're cycling downhill, gravity works for you. This page measures the steepness of a hill in terms of percentage grade G: rise divided by run, multiplied by 100. The heavier you and your bike are, the more energy you must spend to overcome gravity. The combined weight of you (the cyclist) and your bike is W (kg). The gravitational force constant g is 9.8067 (m/s2).
The formula for gravitational force acting on a cyclist, in metric units, is:

Fgravity (Newtons) = 9.8067 (m/s2) · sin(arctan(G/100)) · W (kg)
Rolling resistance: Friction between your tires and the road surface slows you down. The bumpier the road, the more friction you'll experience; the higher quality your tires and tube, the less friction you'll experience. As well, the heavier you and your bike are, the more friction you'll experience. There is a dimensionless parameter, called the coefficient of rolling resistance, or Crr, that captures the bumpiness of the road and the quality of your tires.
The formula for the rolling resistance acting on a cyclist, in metric units, is:

Frolling (Newtons) = 9.8067 (m/s2) · cos(arctan(G/100)) · W (kg) · Crr
Aerodynamic drag: As you cycle through the air, your bike and body need to push the air around you, similar to how a snowplow pushes snow out of the way. Because of this, the air exerts a force against you as you ride. There are a few things that dictate how much force the air exerts against you. The faster you ride, velocity V (m/s), the more force the air pushes against you. As well, you and your bike present a certain frontal area A (m2) to the air. The larger this frontal area, the more air you have to displace, and the larger the force the air pushes against you. This is why cyclists and bike manufacturers try hard to minimize frontal area in an aerodynamic position. The air density Rho (kg/m3) is also important; the more dense the air, the more force it exerts on you.
Finally, there are other effects, like the slipperyness of your clothing and the degree to which air flows laminarly rather than turbulently around you and your bike. Optimizing your aerodynamic positions also help with this. These other effects are captured in another dimensionless parameter called the drag coefficient, or Cd. Sometimes you will see people talking about "Cd · A", or CdA. This is just the drag coefficient Cd multiplied by the frontal area A. Unless you have access to a wind tunnel, it is hard to measure Cd and A separately; instead, people often just measure or infer Cd · A as a combined number.

The formula for the aerodynamic drag acting on a cyclist, in metric units, is:

Fdrag (Newtons) = 0.5 · Cd · A (m2) · Rho (kg/m3) · (V (m/s))2
The total force resisting you, the cyclist, is the sum of these three forces:
Fresist (Newtons) = Fgravity + Frolling + Fdrag
For each meter that you cycle forward, you spend energy overcoming this resistive force. The total amount of energy you must expend to move a distance D (m) against this force is called the Work (Joules) that you do:
Work (Joules) = Fresist (Newtons) · D (m)
If you are moving forward at velocity V (m/s), then you must supply energy at a rate that is sufficient to do the work to move V meters each second. This rate of energy expenditure is called power, and it is measured in watts. The power Pwheel (watts) that must be provided to your bicyle's wheels to overcome the total resistive force Fresist (Newtons) while moving forward at velocity V (m/s) is:
Pwheel (watts) = Fresist (Newtons) · V (m/s)
You, the cyclist, are the engine providing this power. The power that must be provided to your bicycle's wheels comes from your legs, but not all of the power that your legs deliver make it to the wheels. Friction in the drive train (chains, gears, bearings, etc.) causes a small amount of loss, usually around 3%, assuming you have a clean and nicely lubricated drivetrain. Let's call the percentage of drivetain loss Lossdt (percent).
So, if the power that your legs provide is Plegs (watts), then the power that makes it to the wheel is:

Pwheel (watts) = (1 - (Lossdt/100)) · Plegs (watts)
Putting it all together, the equation that relates the power produced by your legs to the steady-state speed you travel is:
Plegs (watts) = (1-(Lossdt/100))-1 · (Fgravity + Frolling + Fdrag) · V (m/s)
or, more fully:

Plegs (watts) = (1-(Lossdt/100))-1 · ( ( 9.8067 (m/s2) · W (kg) · ( sin(arctan(G/100)) + Crr · cos(arctan(G/100)) ) ) + ( 0.5 · Cd · A (m2) · Rho (kg/m3) · (V (m/s))2 ) ) · V (m/s)

So what is Wiggins' aerodynamic drag then

 

[damian13ster]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

Not sure where you came up with the 6.8 w/kg. Thats just pure BS.

Track speed is slightly more about raw power and Wigo purposely gained some weight just to gain few extra watts. Those few extra kilos will not create much of an extra drag or extra power needed to carry the extra weight.

So for example, if he gained 5% of extra power from the additional muscle/weight gain that may equal to 2% of extra speed. Thats all he needs to do 55kph as his power was already few % superior over Dowsett before he even gained the extra weight/power and he needs 4% to crack 55kmh.

Well that's not ***, that's mathematics. No matter what Wiggins did, he can't defy the laws of physics. 6.8Watt/Kilo is the least amount of power a male at Wiggins size/weight has roughly to produce to go 55km/h on a wooden track on those bikes.

By the way, you do realize that extra muscles need extra oxygen support, right?

Evidence???

Well check your math


The physics affecting cycling at constant speed

This web page uses physical models of forces on a cyclist to help you estimate the relationship between power P (watts) and velocity V (kph or mph) of a cyclist. To do this, you need to estimate several parameters; reasonable defaults are given.

There are three primary forces that you, as a cyclist, must overcome in order to move forward:

Gravity: If you're cycling uphill, you're fighting against gravity,but if you're cycling downhill, gravity works for you. This page measures the steepness of a hill in terms of percentage grade G: rise divided by run, multiplied by 100. The heavier you and your bike are, the more energy you must spend to overcome gravity. The combined weight of you (the cyclist) and your bike is W (kg). The gravitational force constant g is 9.8067 (m/s2).
The formula for gravitational force acting on a cyclist, in metric units, is:

Fgravity (Newtons) = 9.8067 (m/s2) · sin(arctan(G/100)) · W (kg)
Rolling resistance: Friction between your tires and the road surface slows you down. The bumpier the road, the more friction you'll experience; the higher quality your tires and tube, the less friction you'll experience. As well, the heavier you and your bike are, the more friction you'll experience. There is a dimensionless parameter, called the coefficient of rolling resistance, or Crr, that captures the bumpiness of the road and the quality of your tires.
The formula for the rolling resistance acting on a cyclist, in metric units, is:

Frolling (Newtons) = 9.8067 (m/s2) · cos(arctan(G/100)) · W (kg) · Crr
Aerodynamic drag: As you cycle through the air, your bike and body need to push the air around you, similar to how a snowplow pushes snow out of the way. Because of this, the air exerts a force against you as you ride. There are a few things that dictate how much force the air exerts against you. The faster you ride, velocity V (m/s), the more force the air pushes against you. As well, you and your bike present a certain frontal area A (m2) to the air. The larger this frontal area, the more air you have to displace, and the larger the force the air pushes against you. This is why cyclists and bike manufacturers try hard to minimize frontal area in an aerodynamic position. The air density Rho (kg/m3) is also important; the more dense the air, the more force it exerts on you.
Finally, there are other effects, like the slipperyness of your clothing and the degree to which air flows laminarly rather than turbulently around you and your bike. Optimizing your aerodynamic positions also help with this. These other effects are captured in another dimensionless parameter called the drag coefficient, or Cd. Sometimes you will see people talking about "Cd · A", or CdA. This is just the drag coefficient Cd multiplied by the frontal area A. Unless you have access to a wind tunnel, it is hard to measure Cd and A separately; instead, people often just measure or infer Cd · A as a combined number.

The formula for the aerodynamic drag acting on a cyclist, in metric units, is:

Fdrag (Newtons) = 0.5 · Cd · A (m2) · Rho (kg/m3) · (V (m/s))2
The total force resisting you, the cyclist, is the sum of these three forces:
Fresist (Newtons) = Fgravity + Frolling + Fdrag
For each meter that you cycle forward, you spend energy overcoming this resistive force. The total amount of energy you must expend to move a distance D (m) against this force is called the Work (Joules) that you do:
Work (Joules) = Fresist (Newtons) · D (m)
If you are moving forward at velocity V (m/s), then you must supply energy at a rate that is sufficient to do the work to move V meters each second. This rate of energy expenditure is called power, and it is measured in watts. The power Pwheel (watts) that must be provided to your bicyle's wheels to overcome the total resistive force Fresist (Newtons) while moving forward at velocity V (m/s) is:
Pwheel (watts) = Fresist (Newtons) · V (m/s)
You, the cyclist, are the engine providing this power. The power that must be provided to your bicycle's wheels comes from your legs, but not all of the power that your legs deliver make it to the wheels. Friction in the drive train (chains, gears, bearings, etc.) causes a small amount of loss, usually around 3%, assuming you have a clean and nicely lubricated drivetrain. Let's call the percentage of drivetain loss Lossdt (percent).
So, if the power that your legs provide is Plegs (watts), then the power that makes it to the wheel is:

Pwheel (watts) = (1 - (Lossdt/100)) · Plegs (watts)
Putting it all together, the equation that relates the power produced by your legs to the steady-state speed you travel is:
Plegs (watts) = (1-(Lossdt/100))-1 · (Fgravity + Frolling + Fdrag) · V (m/s)
or, more fully:

Plegs (watts) = (1-(Lossdt/100))-1 · ( ( 9.8067 (m/s2) · W (kg) · ( sin(arctan(G/100)) + Crr · cos(arctan(G/100)) ) ) + ( 0.5 · Cd · A (m2) · Rho (kg/m3) · (V (m/s))2 ) ) · V (m/s)

That's a lot of work. Looks extremely fancy but for those who don't want to read everything here it is in simpler words:
3 resistance forces on a cyclist.
Gravity (function of mass, not applicable here since there is no elevation change)
Friction (function of mass)
Drag (wind/air resistance) which is not a function of mass

Power is needed to overcome the resistance and travel at certain speed. But since not all resistant forces are function of mass, power/mass does not increase in 1:1 ratio to the velocity.

 

[Richeypen]

Lol, so your evidence consists of copying and pasting something from a website. None of which actually corroborates or backs up your assertion, and is completely meaningless anyway without knowing athlete and atmospheric specific data. So I ask again, where is your evidence that to ride 55km Wiggins will need to produce 6.8 w/kg?

Plus its 'Maths'

 

[Marco Pantani]

Wiggins better take the record before Ulle goes for it.

 

[Gaylord Figman]

Re:

Lol, so your evidence consists of copying and pasting something from a website. None of which actually corroborates or backs up your assertion, and is completely meaningless anyway without knowing athlete and atmospheric specific data. So I ask again, where is your evidence that to ride 55km Wiggins will need to produce 6.8 w/kg?

Plus its 'Maths'

Maths is irrelevant it's what's in Wiggos head on the day which will really count

 

[Alex Simmons/RST]

55km/h would roughly mean 6.8 Watt/Kilo for an entire 60 minutes. Clinic would explode.
And that's a rather conservative calculation.

Not sure where you came up with the 6.8 w/kg. Thats just pure BS.

Track speed is slightly more about raw power and Wigo purposely gained some weight just to gain few extra watts. Those few extra kilos will not create much of an extra drag or extra power needed to carry the extra weight.

So for example, if he gained 5% of extra power from the additional muscle/weight gain that may equal to 2% of extra speed. Thats all he needs to do 55kph as his power was already few % superior over Dowsett before he even gained the extra weight/power and he needs 4% to crack 55kmh.

Well that's not ***, that's mathematics. No matter what Wiggins did, he can't defy the laws of physics. 6.8Watt/Kilo is the least amount of power a male at Wiggins size/weight has roughly to produce to go 55km/h on a wooden track on those bikes.

By the way, you do realize that extra muscles need extra oxygen support, right?

Speed on a track is primarily a function of a rider's ratio of power to aerodynamic drag area, power / CdA in units W/m^2.

In Wiggins's case, to do 55km on a high barometric pressure day (say 1025hPa), he'd need in the vicinity of 2300-2400W/m^2.

One cannot know the power demand without also knowing his coefficient of drag area.

e.g. if he has a CdA of 0.200m^2, then the power demand would be around 460-480W. That too high for him.

But if he had a CdA of less than that, say 0.19m^2, then the power demand is correspondingly lower at 440-455W, which is closer to his capable range, and ~6.0W/kg if he is say 75kg.

I don't know what his current body mass is. It would certainly be higher than in his GT days. Track (and flat riding) favours higher body mass as it doesn't add so much to aero drag.

Wiggin's track CdA has never been published as far as I know, and even if it had, he's done quite a bit of aero refinement work these past months, so in reality neither you nor I know. But it is possible to say what W/m^2 is required.

On a good day with right conditions, I think 54.5km is about right. Something like 1-km to 1-mile more than Dowsett depending on environmental conditions.

That assumes he paces well and doesn't go out too hard.

 

[Alex Simmons/RST]

http://alex-cycle.blogspot.com.au/2015/06/pressure-on-hour.html

 

[Bavarianrider]

Re:

Lol, so your evidence consists of copying and pasting something from a website. None of which actually corroborates or backs up your assertion, and is completely meaningless anyway without knowing athlete and atmospheric specific data. So I ask again, where is your evidence that to ride 55km Wiggins will need to produce 6.8 w/kg?

Plus its 'Maths'

Well those are the mathematic formulas on how to calculate a riders power output at certain speeds. There are various tools on thze internet that do the calculation for you using those formulas.
Denying that those formulas aren't true is like denying that eartj circles sun.

 

[DFA123]

Re: Re:

Lol, so your evidence consists of copying and pasting something from a website. None of which actually corroborates or backs up your assertion, and is completely meaningless anyway without knowing athlete and atmospheric specific data. So I ask again, where is your evidence that to ride 55km Wiggins will need to produce 6.8 w/kg?

Plus its 'Maths'

Well those are the mathematic formulas on how to calculate a riders power output at certain speeds. There are various tools on thze internet that do the calculation for you using those formulas.
Denying that those formulas aren't true is like denying that eartj circles sun.

No-one is denying that the formulas are accurate. However, your input values are highly questionable. You don't know Wiggins' weight, his aerodynamic drag or the atmospheric conditions. Without fairly precise values for all three variables, it is absolute nonsense and guesswork to say he would need 6.8w/kg as a conservative estimate.

 

[Alex Simmons/RST]

Re: Re:

Lol, so your evidence consists of copying and pasting something from a website. None of which actually corroborates or backs up your assertion, and is completely meaningless anyway without knowing athlete and atmospheric specific data. So I ask again, where is your evidence that to ride 55km Wiggins will need to produce 6.8 w/kg?

Plus its 'Maths'

Well those are the mathematic formulas on how to calculate a riders power output at certain speeds. There are various tools on thze internet that do the calculation for you using those formulas.
Denying that those formulas aren't true is like denying that eartj circles sun.

Formula are well and good, but GIGO applies.

IOW if your inputs are nonsense, so will be the results of the formulas.

 

[Richeypen]

Re: Re:

Lol, so your evidence consists of copying and pasting something from a website. None of which actually corroborates or backs up your assertion, and is completely meaningless anyway without knowing athlete and atmospheric specific data. So I ask again, where is your evidence that to ride 55km Wiggins will need to produce 6.8 w/kg?

Plus its 'Maths'

Well those are the mathematic formulas on how to calculate a riders power output at certain speeds. There are various tools on thze internet that do the calculation for you using those formulas.
Denying that those formulas aren't true is like denying that eartj circles sun.

So I take it you cant answer the question then. Please just admit that you have no evidence and you made the number up.

 

[Bavarianrider]

Well, using an air desnity that is most likely lower than it will be at Wiggins record attempt, using values for an extreme areo frame(which Wiggins isn't allowed to use) and using a CWA value of 0.21 (0,25 is a normal triathlon position while 0.2 would be an extreme position ala Boardman for a 75 Kilo rider) which a guy Wiggins size can't hardly achieve on such a bike, we can bring the Watts down to 6.7 per kilo. Assuming that Wiigins will be able to go 55km/h with anything lowewr than 6.7 Watt/Kilo is absurd and not worth debatable.

Feel free to do your own calculations.