Physics

analyticcycling.com has some information on the physics of cycling. They have a few little calculators that give information about how you ride based on the info you give them. They also go into other scientific fields related to cycling. They also want you to own a power tap and a subscription to their site, neither of which I have. Maybe someone here who has one or the other (or both) could give us some more info.

Cycling makes physics class worth it.

Thanks for the link.

What I really want to determine as an incentive for weight loss is;

How much speed will I gain for each Kg I lose, providing I don't lose any power....

As a generic formula for weight vs. power, you might find something.

But you have thrown in the intangible of the human body and individual genetic disposition into the equation .... which makes it all a moot point. Every individual is different.

Basically, you need to get onto a generator to determine your max power output (or get one of the more expensive cyclometers that register kw/h) and then take your weight divided by the power to determine your individual power to weight ratio. Then to calculate what it would be if you were lighter, just change the weight number. The lower the number, the better power to weight (ie. - if you weigh 200 lbs, and can push 50 kw - just using some numbers as an example - then you are pushing 5 lbs for every kw. If you drop down to 150 lbs. with the same 50 kw/h power, then you are only pushing 3 lbs for every kw generated). Same power but lighter equals faster acceleration. That's why a '98 Integra Type R with 200 hp but only weighing 2500 lbs is as fast as a Scooby WRX with 240hp but weighing 3,100 lbs ......

Shedding 10 pounds off your body will do much more than shaving a pound off your bike components .......

Of course that is a very generic example and statement, but it should give you the basic idea.

I like to make the car analogies since I am a huge import car buff ....... and it is pretty relevant to the consversation usually.

What I really want to determine as an incentive for weight loss is; How much speed will I gain for each Kg I lose, providing I don't lose any power....

Using the defaults at analyticcycling.com, you gain 0.1mph dropping from 75kg to 74kg.

Well I've dropped about 2kgs since last week (sensibly) and I was in the lead bunch at training this morning... and was not puffing... I think I'm on to something!

Quoted post said:

you gain 0.1mph dropping from 75kg to 74kg.

Well that is a mile over 10 miles!!!! Enough to put me out in front a lot of the time!!!

Quoted post said:

Originally posted by JAPANic
Well that is a mile over 10 miles!!!! Enough to put me out in front a lot of the time!!!

Nope, it's not. If you were averaging say, 18mph before the weight loss and now you're averaging 18.1 mph, you were previously covering a mile every 3.33 minutes...therefore 10 miles was taking you 33.33 minutes.

Now you cover a mile every 3.31 minutes so you cover ten miles in 33.15 miles.

When the 18mph you completes 10 miles in 33.33 minutes, the 18.1mph you will have completed 10.054 miles. The faster you would be 285.12 feet ahead or just over 95 yards...not a mile!

It shouldn't have a big effect. Your weight will increase your inertia and rolling resistance and slightly increase the torque required to turn the bearings but these are small power losses at your top speed. The greatest use of you power is overcoming the drag force created by the air.

at any time:

F(provided by you) = d/dt (mv) + Frr + Faero = ma + Frr + Faero (for constant mass)

m - mass
a - acceleration
Frr - rolling resistance
Faero - aerodynamic parasitic drag force

Frr = aV where a is some constant

Faero = bV^2 - where b is almost constant

So from the formula above; (assuming you produce a constant force) Frr and Faero will be zero at rest so:

V = 0

F = ma

therefore

a = F/m to give your acceln

When you are at your top speed a = 0 (because you no longer accelerate)

F = Frr + Faero

As Faero is proportional to V^2 this will be the dominating term at high speed (so long as a is of similar magnitude to b). Therefore your mass is no longer a highly important factor in fact your Drag Co-efficient Cd is most important because reducing this will give the greatest increase in speed.

Faero = 0.5*A*density*Cd*V^2

So to go quicker your frontal area A can be reduced i.e. putting your head down. You could also ride through helium whic has a lower density and more significantly viscosity (viscosity affects Cd).

You can get an approximate indication of your power output from:

P - power
d - distance travelled

P = d/dt(W)

Integral of W dt is energy so

Average P = E/t E is energy due to work and t is time taken

With reference to earlier discussion about forces and noting that energy is F*distance - as long as speed is low then Frr and Faero is almost zero and thus no energy is used to overcome them. Therefore the force supplied by you is almost entirely going into your acceleration

i.e. F = ma

So below 10 mph all the energy you produce is approximately used for your acceleration

Therefore;

If starting from 0mph and ending up at Vmph in time t

P(av) = (1/2*m*V^2)*(1/t) n.b. V in metres/second = 4/9 * mph

I think I could get to 15mph from 5mph in a second without drag so (me and my bike weigh about 75kg);

P = 0.5*75*4/9*4/9*15*15*1/1 - 0.5*75*4/9*4/9*5*5*1/1
1.5kw = 2hp

Dont forget 1hp is the weight vs time a horse can lift whilst standing up, not including its own weight, so this seems reasonable.

It shouldn't have a big effect. Your weight will increase your inertia and rolling resistance and slightly increase the torque required to turn the bearings but these are small power losses at your top speed. The greatest use of you power is overcoming the drag force created by the air.

at any time:

F(provided by you) = d/dt (mv) + Frr + Faero = ma + Frr + Faero for constant mass

m - mass
a - acceleration
Frr - rolling resistance
Faero - aerodynamic parasitic drag force

Frr = aV where a is some constant

Faero = bV^2 - where b is almost constant

So from the formula above; (assuming you produce a constant force) Frr and Faero will be zero at rest so:

V = 0

F = ma

therefore

a = F/m to give your acceln

When you are at your top speed a = 0 (because yo uno longer accelerate)

F = Frr + Faero

As Faero is proportional to V^2 this will be the dominating term at high speed (so long as a is of similar magnitude to b). Therefore your mass is no longer a highly important factor in fact your Drag Co-efficient Cd is most important because reducing this will give the greatest increase in speed.

Faero = 0.5*A*density*Cd*V^2

So to go quicker your frontal area A can be reduced i.e. putting your head down. You could also ride through helium which has a lower density and more significantly lower viscosity (viscosity affects Cd). Out of interest all drag is due to viscosity.

You can get an approximate indication of your power output from:

P - power
d - distance travelled

P = d/dt(W)

Integral of W dt is energy so

Average P = E/t E is energy due to work and t is time taken

With reference to earlier discussion about forces and noting that energy is F*distance - as long as speed is low then Frr and Faero is almost zero and thus no energy is used to overcome them. Therefore the force supplied by you is almost entirely going into your acceleration

i.e. F = ma

So below 10 mph all the energy you produce is approximately used for your acceleration

Therefore;

If starting from 0mph and ending up at Vmph in time t

P(av) = (1/2*m*V^2)*(1/t) n.b. V in metres/second = 4/9 * mph

I think I could get to 15mph from 5mph in a second without drag so (me and my bike weigh about 75kg);

P = 0.5*75*4/9*4/9*15*15*1/1 - 0.5*75*4/9*4/9*5*5*1/1
1.5kw = 2hp

Dont forget 1hp is the weight vs time a horse can lift whilst standing up, not including its own weight, so this seems reasonable.

Sorry for posting that twice. It was an accident on my behalf 🙁

The higher the hill the better the benefit...

Thanks for the posts.

Thanks for the posts.

Quoted post said:

Originally posted by JAPANic
The higher the hill the better the benefit...

Thanks for the posts.

Thanks for the posts.

You have a point there. Weight loss won't make a whole lot of difference on the flats, especially if they are pancake flat, but it will make a huge difference in the mountains. And remember, it's not just Newtonian Physics we are talking about in the hills; we're also talking about keeping cool. In the hills, especially the very steep ones, overheating is a big problem if you are overweight. This is why the biggest moose are where it's coldest; it's easier for them to retain their body heat. You, on the other hand, don't want to overheat and will find it much easier to stay cool on a hot day if your body mass is lower. This advantage doesn't show up in those force and mass equations, but it is also a huge factor.

If you want motivation for losing the weight, and speed - - - forget the flats and head for the hills. Getting dropped going UP - - lose the weight. Unable to shift DOWN during the hill, lose the weight. You will love it.......

Quoted post said:

Originally posted by JAPANic
As a high school student many years ago, I remember studying formulas for Force, Momentum, Gravity, Speed, Mass, Acceleration, Friction etc etc...

Now I failed physics, and can only remember a few... F=MA & S=D*T
(is that right?)

What I want to know, is there anywhere on this site or on the net or do any of you know some interesting formulas that pertain to cycling?

Vectors!!!!



😕

My trainer told me to forget the hills and concentrate on sprints and to get some heavier wheels....

The weight is coming off too, so it will all balance out as long as I keep riding harder & harder.... I'm getting there....

Went hill climbing about 2 weeks ago and although I struggled, my legs were fine with no pain, but the down hills were scary as I am still carrying more weight than the rest of my team and I can't take the bends with the same arcs, as I'm just packing to much potential energy.....

Like I said, I'm getting there....a race on the 27th...to work for....