Watts / kg - explain please

dsoutar

Firstly excuse my ignorance here

This is the definition of a Watt:

One watt is the rate at which work is done when an object's velocity is held constant at one meter per second against constant opposing force of one newton.

W = Joule / sec = Nm / sec = kg*m^2 / sec^3

Now we already have kg in this equation so if we now have Watt's per kg we're taking it back out leaving m^2 / sec^3.

So why is it that people constantly talk about Watts / kg especially in relation to climbing ?

Also when people say "I'd have to lose a stone to generate that sort of power surely this is also flawed 'cos part of the equation is related to their weight in the first place and hence would affect the calculation of the power they are outputing (i.e, it would reduce it) ?

Edited to correct s^3 as per below

amaferanga

Watts/kg allows you to compare riders of different kg. So the fact kg is no longer in the equation (which you've got wrong anyway - s^3 on the bottom of the expression on the right) is exactly what you want.

You can achieve a similar result by losing weight or increasing your power output (or through a bit of both).

dsoutar

Thanks. Yes - I sort of get that, but m^2 / sec^3 seems such a bizarre concept I can't get my head around it

alihisgreat

dsoutar wrote:

Thanks. Yes - I sort of get that, but m^2 / sec^3 seems such a bizarre concept I can't get my head around it

W/KG is a power to weight ratio.. what's so hard to understand about that?

whyamihere

dsoutar wrote:

Thanks. Yes - I sort of get that, but m^2 / sec^3 seems such a bizarre concept I can't get my head around it

Think of it just in terms of the ratio, ignore the units. Say rider A weighs 50kg and can produce 250W for an hour. Rider B weighs 40 kg and can generate 200W for an hour. Both riders have a power to weight ratio of 5W/kg for an hour. That means that if both riders climbed for an hour then, all else being equal (aerodynamics etc), they would reach the top at the same time.

Alex_Simmons/RST

When climbing steeper gradients, your speed is directly proportional to your power output divided by your mass (you and bike etc).

e.g. for any given mass of bike and rider going up a steep hill a higher power output will travel faster, and for the same power, a lighter rider will travel faster.

W/kg is a way of normalising a measure of performance for riders of different size, and W/kg over various durations (e.g. 5-min, 20-min, 1-hour) is a very good indicator of performance potential (especially for climbing - on flatter terrain, air resistance dominates, so the relationship is less simple).

A rider that can sustain 5W/kg can climb faster than one that can hold 4W/kg.

Top Grand Tour riders are capable of sustaining ~ 6W/kg for about an hour.

dsoutar

Thanks all but I understand perfectly the concept of Watts / kg (or power to weight) and how lighter, more powerful cyclists will succeed over heavier or less powerful. I've just not been eloquent enough to state my issue, namely the rationalisation of the units. I know whyamihere said ignore the units but it does leave one with a very strange mathematical concept once the Watts / kg is broken down to the lowest denominator.

Sorry, ALIHISGREAT but it's about the maths of power to weight. As power itself is dependant on weight it's leaves you with a rather strange outcome in terms of the formula, i.e. m^2 / sec^3.

I got an A in A-level maths (admittedly in the Cretaceous period) so I thought I was performing this correctly, but if I am then the mathematical result makes no sense despite the fact that the overall concept does.

dw300

dsoutar wrote:

Thanks all but I understand perfectly the concept of Watts / kg (or power to weight) and how lighter, more powerful cyclists will succeed over heavier or less powerful. I've just not been eloquent enough to state my issue, namely the rationalisation of the units. I know whyamihere said ignore the units but it does leave one with a very strange mathematical concept once the Watts / kg is broken down to the lowest denominator.

Sorry, ALIHISGREAT but it's about the maths of power to weight. As power itself is dependant on weight it's leaves you with a rather strange outcome in terms of the formula, i.e. m^2 / sec^3.

I got an A in A-level maths (admittedly in the Cretaceous period) so I thought I was performing this correctly, but if I am then the mathematical result makes no sense despite the fact that the overall concept does.

The point is that a less powerful rider can go faster, or a more powerful one go slower.

And didn't someone mention that the point was to remove the weight aspect from the equation?

Tom Dean

dsoutar wrote:

Sorry, ALIHISGREAT but it's about the maths of power to weight. As power itself is dependant on weight it's leaves you with a rather strange outcome in terms of the formula, i.e. m^2 / sec^3.

You can rearrange all sorts of equations and come up with units that are hard to make sense of in words. Work it back mathematically and it's simple - joules per second per kg body weight.

This is why we have SI derived units

pst88

If you can lose mass whilst maintaining the same power output you will go faster. The formula you state is the amount of power a person is generating at that particular mass and velocity. Imagine a person has a maximum power output... if you are already producing your maximum power you can't go any faster unless you either decrease your mass, in which case your maximum power is going further, or increase your maximum power.

Herbsman

:roll:

thegibdog

The OP has hit on an important point and this is why energy saving light bulbs will never climb as well as your more traditional incandescent bulb - the latter have both a power and weight advantage.

OfficerDigby

TO the OP there are loads of useful physical parameters that either have apparently meaningless SI units, or can be manipulated into apparently meaningless ones.

Some figures of merit that have real physical significance are even dimensionless.

In reality you've dumped the physical concepts and are thinking the units are key <- the mistake!.

The physical concepts are power and mass.

SI units are generally to check what you have done for error. In this case the fact that kg has gone means we are talking about a property which is independent of weight. Which is a what we want - analyzing climbing independent of weight!

Well done on your A-level

For example: did you know what this one can be.... m2/s2 ?

Absorbed dose rate.

OK not meaning to play actually there is a way out for you.

Specific energy is m2/s2 - which is actually energy of body per kilo. So (m2/s2)/s specific power of a body.....
..... Or power of a body per mass.

Right back where you started.

Sweet Dreams..ZZzzz..

Alex_Simmons/RST

OfficerDigby wrote:

TO the OP there are loads of useful physical parameters that either have apparently meaningless SI units, or can be manipulated into apparently meaningless ones.

Some figures of merit that have real physical significance are even dimensionless.

In reality you've dumped the physical concepts and are thinking the units are key <- the mistake!.

+1

For example, one of the physical parameters that helps us determine how much energy is required to overcome air resistance is called the Coefficient of Drag (Cd). It's a dimensionless quantity and helps us assess things that are "bricks" versus "arrows".

Another is effective frontal area (A) - unit m^2. The more area an object (of same shape) presents to the wind, the more energy is required to push through at same velocity.

The two combined tell us an important information about the cyclist Cd x A = CdA, unit m^2.

So A and CdA have the same unit (m^2) but tell us different things about the physical properties of an object moving through air.