While I has grinding my way up Alp du Zwift yesterday, I wondered what speed I would need to be doing at the bottom of the hill to enable me to coast all the way to the top. I have no idea how to begin or what to take into account but I thought of just the place to ask the question: Singletrack
So some assumptions to hopefully make this easier(ha):
Alp du Zwift lets imagine that it was perfectly smooth tarmac, straight (No corners) and an equal grade all the way to the top. 1036m of height gain, 12.6km, gradient 8.2%, no wind or rain forecast.
Rider - 100kg including the rider and bike. Timetrial position. Drag coefficient of 0.6??
If we figure that out would the speed needed be less or more if the rider weight was 200kg or 50kg??
322 mph I reckon
The gradient doesn't matter. As long as the transition is curved enough, obviously.
Also your drag coefficient is flawed too. The drag varies with the speed. If you started to take into account the air resistance then I reckon you'd quickly fry your cyclist.
The air resistance on a 322 mph cyclist would be huge. You'd need to massively increase that starting speed to factor in that drag.
But that new higher starting speed would have an insane drag, so you'd need to go even fasterer to overcome that.
Etc etc until he fries.
Would be interested if we have any serious cleverists who could actually work that out. It's way beyond me.
Assume no losses (massively unrealistic due to drag) and all your kinetic energy gets turned into potential energy at the top.
Potential energy = mgh
Kinetic energy = 1/2mv^2.
mgh = 1/2mv^2. Rearrange for v.
If you can find a formula for air-resistance drag at different speeds, this will be straightforward using conservation of energy (required KE = PE + losses).
Assuming 200kg rider has the same frontal area / CdA, and therefore the 'losses' do not include a function of rider mass: The (square root of) mass becomes a small factor on the 'bottom' of the equation - I.e. is inversely proportional to the speed required. So more mass = less speed required.
KE=PE + losses
0.5 MV^2 = mgh + losses
V^2 = (mgh + losses) / (0.5m)
V= sqrt (2*(gh + (losses /m)))
AINAPhysicist
I make it 142m/s assuming no losses.
Or 317mph, or 511 km/h.
But that's way too slow as losses will be massive at 317mph.
Also, you might need a fairing or you'll struggle to hold onto the bars 🙂
So if I was in a vacuum it's around 320mph..
Sorry yes the drag of 0.6 was copied from a website, I have no idea what it means so probably a red herring.
Stumpy with his g to 1 decimal place 😉
This confirms to me how boring Zwift must be.
Not at that speed it isn't!
What if the entire continental US was on a decreasing slope from West to East.
Christ on a bike. The author can't even express simple things like inclines properly.
A concave slope .... weird.
To be fair to him, that’s the question from a reader. Site author replies to it.
88mph
Nah, that would just let you go back in time to decide not to do a 320mph Zwift session.
What if the entire continental US was on a decreasing slope from West to East.
It actually is, well large parts of it. We left Wisconsin at 233m, crossed the Mississippi at 300m, and by the time we got to Boulder we were at 1600m without having gone up any obvious hills.
Well yes, agreed molgrips. Which illustrates why a decreasing slope wouldn't work. You'd just plummet out of the Rockies at 80mph and then grind to a halt a while later on the flat bit.
It needs to be a constant slope, ie decreasing height, not decreasing slope
thegeneralist
Stumpy with his g to 1 decimal place
2dp I'll have you know!
Simple energy conservation won't work here due to forces of drag, hence the standard sqrt(2gh) will be a huge underestimate. The correct method is to use forces, which are not constant.
For velocity, v, the drag on a cyclist is Fd(v) = 0.5*Cd*A*r*v^2, where A is area and r density. Power is Force*velocity hence scales as v^3.
So for AdH slope G, decceleration force will be g*Sin(Arctan(G)) + 0.5*Cd*A*r*v^2/m
that is dv/dt and you can integrated from initial velocity V0 to V(t) for time t - and of course v=0 to give time to stopping, T.
Once you have v(t), which is dx/dt for distance, x(t), you can integrate again from time t=0 to time t=T. Get the value of T the ascent time, and the total distance travelled up AdH and V0 will drop out. It's nasty and non-linear.
2dp I’ll have you know!
I knew you'd bite 🙂
I'm not convinced it is to 2 DP. If you use it to 2DP then you get 142.57 m/s which any fule no rounds to 143.
You gave the answer as 142, which you could only have got by using a value of 9.8 for g. (142.4977, which rounds to 142.)
So, which was it. Did you use 1DP or screw up the rounding?
We need to know.😛
B
thegeneralist
I’m not convinced it is to 2 DP. If you use it to 2DP then you get 142.57 m/s which any fule no rounds to 143.
You gave the answer as 142, which you could only have got by using a value of 9.8 for g. (142.4977, which rounds to 142.)
"I'm not convinced......" Ha ha. Be unconvinced all you want.
It was obviously such an incorrect number to start with due to the 'ignore losses' simplification, that I didn't really see rounding up would add anything to it.
For an incorrect, illustrative thumb in the air number, it works for me.
TiRed seems to be on it with some proper maths.
Although, we (ahem, the royal "we") also need to take into account the reduction in air density as we ascend, if we're worrying about decimal places.
Still having fun?! 🙂
TiRed seems to be on it with some proper maths.
Hmmm. I'm not convinced. I did reply earlier to his post, but deleted it when I realised how clouded it was by my bitterness at how shit my physics degree turned out to be 🙂
But I think he's generally wrong in worrying about the slope angle. Using exceedingly rough numbers suggests a deceleration force due to air resistance of around 5g on the cyclist at our notional 322mph. That would slow him down to a fraction of that speed within a few seconds. The deceleration due to gravity/ slope angle is around 1/50 of that, ie inconsequential.
Since he's decelerating so fast we need to update his starting speed. But as stated in my initial post, that just increases the drag.
My view is that there is no speed that you could send a cyclist off at ( even theoretical speed) that would get him up AdH.
Not unless he was reeeeeeally fat n heavy , and at the same time realllllly small and on an even smaller bike.
Simple energy conservation won’t work here due to forces of drag
Being pedantic (I think this is the right thread for it), energy conservation always works. I suppose the point is that you'd have to measure frictional losses in terms of energy rather than as a force, but that requires only multiplying the force by distance (which we know and I think as you say would cancel out, anyway).
Given that we know V0=0, is there a way of generating an average force over the whole distance? We know this is proportional to V^2, so you could perhaps use the geometric mean of velocity to calculate this? Thinking out loud, I suppose this approach would assume a linear deceleration which probably isn't correct. I dunno. I'm interested in these sorts of questions but it's not my day job! I'd have probably had a better answer 20 years ago.
Part III: Imagine the road is a conveyor belt, and the cyclist has wings. At the top of the hill is a VW Transporter dealership. How fast must the rider pedal on the conveyor belt to heat a Kelly kettle to 98°C for optimal (artisanal) coffee brewing? Please give your answer to the nearest integer number of chocolate hobnobs.
I've got it.
Take your hypothetical perfect slope, put it in a bell jar and vacuum all the air out. You will, of course, need an assistant for this.
Start at the top of the slope, swap to frictionless tyres and freewheel down. Look at your speedo at the bottom. That's your answer.
For proof, at the bottom of the slope place a 100% energy-efficient trampoline, vertically. Repeat as above, see how far you boing back up.
Actually, you could avoid the need for frictionless tyres by constructing a 1036m tower and then riding off the top of it. You'll be able to measure your speed at the bottom, once.
Much as it pains me to say it, I think Cougar's answers are the best.😃
You’ll be able to measure your speed at the bottom, once.
The two forces are in opposite directions, hence you will meet steady-state terminal velocity. Going up is not the same as going down.
Being pedantic
Ouch, but true. Need the energy loss from drag. And tyre resistance, which is proportional to v. And v will be large. Hence my "Simple".
Integral over velocity needs u = Tan(v) substitution. Then invert for v(t) over incline. Anyone closer yet?
you will meet steady-state terminal velocity.
I don't really know what 'steady-state' means here, but is there a terminal velocity in a vacuum?
not in a vacuum, but then it's just the simple sqrt(2gh) answer. If you haven't suffocated on the way down...
... which is what I said. 😜
So the important questions now are how long do you need to hold your breath to be still able to read the speedo at the bottom and which helmet and knees pads to assist with the impact with the ground / magic trampoline?
You presumably could estimate the answer with a set of iterations. Start at the top and consider the last meter. Guess a speed that you want to end at so you don’t wobble off. By Estimating the work done over that meter work out the energy kinetic energy needed to enter that meter. From that estimate the speed needed to enter the last meter. Repeat for all previous meters.
But the answer will be impossible in real terms. No part of the system could move through the air that fast without damage.