"I'll be take it easy from then on, coming off at that speed just doesn't bear thinking about."
Why? Its more likely to mean less broken bones than if you fall of at 10mph. Of course there would be no skin left on your body...
Bike Forum
Nearly cracked 50mph today...
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Posted 6 months ago #
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Have hit around 50 on the bit of road between Abinger and Westcott on the road bike. Didn't feel too bad as the road is smooth like a snooker table. 40 off the top of Ranmore and I was not happy at all; very nervous in fact.
You mean Hollow Lane? From Abinger Common down to the A25 near the Wootton Hatch? I pinch flatted coming down there in the pouring rain on the road, by the houses at the bottom. A pinch flat at 40 is a lot scarier than not pinch flatting at 60!
I've regularly done 50-something coming down from Coldharbour to Dorking. Was noticeably quicker with some 50mm deeps. That they had a penchant for rolling fully inflated tyres was something I tried to forget!
IMO it's the gradient as much as the length that's key for good top speed. Did c53 in a 40 (setting off a speed camera) on the A283 on a fairly innocuous looking hill. PB was in Lanzarote, perfect road really, absolutely dead straight, wide, perfect surface and a good gradient. Forget the exact figure, 60+. Felt less scary than some of our roads at 50 though! That was in a very tight pace line with some nutter triathletes. Annoyingly we got to the bottom, round the roundabout and came back up again, 2 miles of utter torture!
Not really sure about folk thinking it's harder/more scary on an MTB though. When your hands are right by the stem, and your nose is on the bars it's fairly irrelvant what shape your frame is!
Posted 6 months ago # -
41mph down Mastiles Lane descent earlier this year. Yes, that's off road for those that don't know.
Posted 6 months ago # -
Okay 13thfloormonk I'll try and explain.
We will use freefall as an example.
So you have an 80kg sky diver gravity acts on him with a force of 800N (with some rounding) He jumps out of the plane and initially and accelerates at 800 / 80 = 10ms2 (accelaration being equal to force divided by mass). However as his speed builds so does air resistance. It builds until the point where it equals the force of gravity on him, so he stops accelerating, he has reached terminal velocity.
No imagine we stick 20kg of lead on his back and he is falling in the same position, Gravity is now exerting a force of 1000N on him therefore the air resistance requred to stop him accelerating is also 1000N, in order for the air resistance to equal 1000N he needs to be falling a bit faster so his terminal velocity will be higher.
Same basic principle holds true for blokes on bikes freewheeling down a hill
Posted 6 months ago # -
Posted 6 months ago #
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Man, road bikes are exciting.
Posted 6 months ago # -
Eh... really? Gravity is not a force, its an acceleration. Acceleration due to gravity is 9.81 m/s2.
In your example the 80kg sky diver will experience a force (due to the accleration of gravity) equal to 80*9.81=785N. The 100kg skydiver will experience a force of 981N. The net result however is that the two skydivers will both accelerate at 9.81 m/s2.
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).
Posted 6 months ago # -
13th - add in the effects of wind resistance than given the same frontal area a heavier object will reach a higher terminal velocity
Posted 6 months ago # -
TJ, I agree, when descending on a road. However, this would not happen in freefall, as has been proven repeatedly (experiments on the moon with a hammer and a feather, air resistance non-existent, both objects land at the same time).
I wasn't actually disagreeing with the fact that heavier riders could achieve higher speeds on descents, I was just trying to establish how, and, for fear of turning this into a thread that never ends, still believe that my intial argument is correct.
Posted 6 months ago # -
Try the cycling specific paper above ^^^
Posted 6 months ago # -
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!!
Posted 6 months ago # -
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.
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!
Posted 6 months ago # -
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
Posted 6 months ago # -
Eh... really? Gravity is
See almost rightnota force,its an acceleration. Acceleration due to the force of gravity on Earth is 9.81 m/s2
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 oftenDo i need to explain big G and small g as well?
Posted 6 months ago # -
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
ridersobjects 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?
Posted 6 months ago # -
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 foamPosted 6 months ago # -
ignore
Posted 6 months ago # -
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^^ Posted 6 months ago # -
...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.
Posted 6 months ago # -
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!
Posted 6 months ago # -
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?
Posted 6 months ago # -
Not one argument with my logic?
What's wrong with this forum these days?
Posted 6 months ago # -
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
Posted 2 weeks ago #
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