The thing that stops me, is my confidence in the brakes (excuse the pun).

DH bike – hydraulic brakes, 9″ front rotor, 8″ rear rotor, 6-pot calipers (plus nice rigid dual crown forks.. + 20mm axle..)

Road bike – mechanical calipers + rim brakes!!!!

I’ve had great fun bombing down some hills up in the Lakes with the DH bike – knowing I can stop pretty quickly, even with just one finger… but on a road bike?? No chance!!

Sorry KCR, I did, meant to refer to it in my last post.

It confirms that yes, heavier riders will go faster downhill. But it just says ‘because of their greater mass to area ratio’. Suggesting that ultimately, being heavier helps. Doesn’t explain HOW.

I’m aware I’m totally killing this thread though so I’m going to drop the subject. 8)

For what its worth though, my top speed is a mere 47mph. Crosswinds on the Lechtd, traffic on Ben Lawyers and heavy rain and mist on the Bealach Na Ba have all conspired to prevent me breaking that record! 😀

teh how is they have a greater force acting on them to push them downhill but similar wind resistance to the skinny guys – so have a higher terminal velocity

Eh… really? Gravity is not a force, its an acceleration. Acceleration due to the force of gravity on Earth is 9.81 m/s2

See almost right 😉

What you are suggesting (heavier falls faster) is in direct disagreement with established principles (or at least, what I have assumed are established principles from memory).

But those estabilished principles only hold true for a vacuum. This seems stange because we don’t experience the effects of terminal velocity very often

Do i need to explain big G and small g as well?

Do i need to explain big G and small g as well?

I doubt that will help somehow 😆

My clumsy/wrong definitions aside, will everyone PLEASE drop the subject of wind resistance, I keep stating, or trying to state, I am considering the hypothetical situation of two otherwise identical riders of different mass. Therefore wind resistance should not be a factor, they experience it equally.

The established principles I refer to do not only hold true for a vacuum, they also hold true in the instance of two otherwise identical riders objects of different mass.

The point I have been trying to make is this:

If we agree that in freefall, two aerodynamically identical objects off different masses will still fall at the same speed, then why should it be that two aerodynamically identical riders of different masses will achieve different speeds on a downhill?

If we agree that in freefall, two aerodynamically identical objects off different masses will still fall at the same speed,

they won’t unless its a vacuum because of the wind resistance. think cannonball and similarly sized ball of polystyrene foam

ignore

Shit, its finally twigged. Apologies all!

edit: I was going to say ‘the ball had finally dropped’ but sensed that wouldn’t help.

To be fair you’ve got the basics right its just one aspect that you got completeley backwards

If we agree that in freefall, two aerodynamically identical objects off different masses will still fall at the same speed,

^^This bit^^ 😀

…yeah… thats why I edited my last post, I’m still having trouble with that aspect.. 😳 🙄

I’m going to drop it now, without some heavy duty research (or maybe just a children’s encyclopedia) I doubt I’ll get to the bottom of it.

13thfloormonk

Without going too far into it, it’s fair to say the biggest factor affecting an individual’s top speed down a road is wind resistance, or aerodynamic drag. No matter the person’s mass, the coefficient of friction of the road is the same, and gravity is the same. The variable is a person’s mass, and more importantly, their mass/frontal area ratio.

Now effectively, the aerodynamic drag is an acceleration force acting to slow you down, where gravity is the acceleration force speeding you up. Both are independent of a person’s mass, but whilst gravity is a constant, aerodynamic drag is affected heavily by frontal area.

So given that F=Ma, we can all agree that a fat person will have more force acting upon them from above than a skinny person. But fat people generally don’t have significantly higher frontal areas (bigger yes, but not usually proportionately so), so whilst there is more force acting to slow them down, it is not acting to slow them down as much as a skinny person… If that makes sense?

Terminal velocity occurs when forces due to gravity equal forces acting up due to aerodynamic drag. Increase the frontal area, the object will slow down. Increase the mass for the same frontal area, the object will speed up.

In the case of a guy riding a bike down a steep hill, take a 10st man versus a 15st man. The heavier guy has 50% more mass acting as a force downwards (you then need to work out its effect as a result of the gradient of the slope using trigonometry) than the skinny guy. But the fat guy will probably only have 20% more frontal area (hence experience only 20% more drag)… The result…?

Fatties go faster down hills!

To expand on what I just said a moment ago…

Adding some maths in…

If we say the 2 riders are on the same bike, wearing the same clothes, using the same tyres, and that neither are pedalling (its too steep to have any effect any more), then the only variable is their weight and frontal area.

We’ll say the 2 riders are 65kg and 100kg (roughly 10st and 15st)

We’ll say the hill is a constant 15 degree slope.

We’ll also say that the heavier guy only has 20% more frontal area (hence only 20% more drag) when they’re both in an aerodynamic tuck.

If we say that the max velocity reached for the thinner guy is 80km/h (50mph), we should be able to work out the predicted max velocity of the heavier guy… (I’m going to assume drag goes up in a linear fashion, I know it doesn’t, but my maths doesn’t extend that far).

Using F=Ma… Force acting upon the smaller guy vertically is… 9.81m/s2 x 65kg = 637.65 Newtons

But we’re not interested in the force vertically, we’re interested in the force in line with the slope…

Which is… Sin15degrees x 588.6 = 165.04 Newtons propelling him forward in line with the slope.

So what we’re saying is that when he reaches his terminal velocity of 80km/h, he has 165.04 Newtons still acting to propel him forward, but he is also now experiencing 165.04 Newtons of deccelerative force due to aerodynamic drag.

For the heavier guy, we’re saying he is experiencing the following accelerative force…

sin15 x 9.81 x 100kg = 253.90 Newtons

Which is directly proportional to the amount he weighs.

If he has a 20% larger front area though, at 80km/h he is only going to be experiencing the following amount of deccelerative force…

165.04 x 1.2 = 198.05 Newtons

Clearly, he is still accelerating!

So assuming drag is a constant (as I’ve stated before I know it isn’t, but for the purposes of making it easy to explain I’m saying it is), we can then apply the following…

253.9 Newtons / 198.05 Newtons = 1.282

1.282 x 80km/h = 102.56km/h (or 64.1mph)

So in conclusion, once again, Fatties go faster!!! 😉

If we’re to actually do it properly, as aerodynamic drag goes up as a square of the speed increase, the fat guy is likely to only hit 90.58km/h on the same stretch of road (by my crude calculations), which is still some 10 and a bit km/h (or 6.6mph) quicker than the skinnier guy… All just cos he weighs more!

That help?

Not one argument with my logic?

What’s wrong with this forum these days? 😕

😉

in my younger thinner days i hit 56mph on a rigid 1996 rockhopper with slicks blown up 110psi on a london to brighton charity bikeride