CargoDane

You added that parenthesis at the end.

Anyway, in still air, that holds true as speed over ground will match speed through air. That is the only time the two will meet. Which website did you copy/paste it from?

Of course the formula looks a little different when you have to take BOTH the speed of the vehicle and the wind into account (. It is, however, again, the same if only the wind is moving OR only the vehicle is moving. Both moving, you add/subtract the two (depending on headwind or tailwind) and then calculate the effective drag.

In any case, wind tunnels work for air drag because it doesn't matter if the air moves or the vehicle moves. The air hits and goes around the object at the same speed.
So that example does not in fact back up what you are saying. Especially not the parts where it matters (according to you) whether the vehicle is propelled by tyres or by, well, a propeller.

tomato coupe

No.
It's actually just from good ole Wikipedia, so you can confirm yourself I didn't add the parenthetical statement. I can provide other more rigorous references, but you should let me know your physics background so I can provide something appropriate to your level of expertise. A
Wind tunnels measure aerodynamic forces, not the power required to overcome aerodynamic drag.

CargoDane

Then in their example, they're calculating it in still air. The other example does not support your assertions at all.

Drag, lift, and so on ARE aerodynamic forces. And if you have the drag number, you can calculate the power needed to overcome it. So, yes, you can calculate the power needed. In fact, in your wikipedia example, they are calculating just that with how much power is needed to go through the air at a given speed. It's the same calculation in a frigging wind tunnel.

I can't be bothered anymore with you on this topic. I have given up on Cubewheels, and then you come in and make further multiple unsubstantiated claims, and completely misrepresenting what the various formulas actually says.

tomato coupe

I'm starting to wonder if you can read:



Yes it does. And it clearly states the power depends on the square of the airspeed (vw + vo) multiplied by the ground speed (vo).

The formulas are pretty clear.

Just out of curiosity, what is your level of physics training?

livedarklions

This whole discussion is kind of a throwback to conventional wisdom of the 1960s. Schwinns were heavy and relatively expensive, but generally built like a tank. Huffy and Murray were a little lighter and much cheaper, but also more prone to breakage. When families had more kids, a kid's Schwinn was considered a bit of an investment because it would be a hand me down.

Lemond1985

Very true. I was quite suspicious of the whole idea of light bikes back then, thinking it to be some Huffy-Murray collusion designed to get people to buy cheap bikes that broke faster. Besides, 4 out of 5 experts back then agreed, heavy Schwinns with balloon tires were the way to go.

CargoDane

Yeah, not being able to use math (as you two aren't capable of) somehow validates the notion that the two of you are able to cheat physics. Seriously, the both of you need to take a course - not only in math, but also in aerodynamics (i.e. the dynamics of air flow). Maths and physics are not "personal beliefs". They are facts. You cannot cheat physics even if you cheat with the maths. Reality will beat you down.

And, Cube, maybe you should take a hard look at yourself and your posts before spouting such things. You are once again projecting.. The irony of that last statement of yours is so thick you can cut it.

ofajen

One last try: yes, the force is the same. The power requirement is the dot product of force and rider speed. So the power requirements are not the same.

If I straddle the bike and stand in a 60 mph headwind, my power requirement is zero. If I ride in still air at 60 mph, the power requirement is enormous, but as you have noted, the force of air drag is the same. Don’t know if that will help explain things or not at this point.

Otto

CargoDane

Yes, one last time: Wind resistance (= air drag) is the result of the object going through the air. The reason it changes with rider speed is that your speed-through-the-air changes. It has naught to do with your groundspeed.


Yes, that's about torque. But the drag won't change. The wind resistant - the force pushing backwards are the same. Of course at 0 speed over ground, the force required to stay in the same place is superceded by friction. Hence why you use windtunnels. You can tie down the object and get the drag numbers and calculate the power needed to move.

Your explainations of why your misunderstanding are correct doesn't actually make them right.
Because for you to actually move in those winds you need a certain amount of power. For you to stand still, you just need friction. The drag, the wind resistance doesn't change, hence we can use wind tunnels. And we can use wind tunnels because when the air is going over an object at 60 mph, it doesn't matter if the object is still and the air moves at 60 mph or the object moves at 60 mph in still air. The power needed to MOVE in those circumstances are the same.

To put it another way: When you are standing still in 60 mph wind, the force, the power is in the wind (literally). That you can use your brakes and contact patch to stay still doesn't mean there is no force applied. That is physics 101. Hell, you leaning into the 60mph wind while straddling the bike is "force applied" just to stand still.

tomato coupe

Have you taken Physics 101?

ofajen

Perhaps Jobst Brandt does a better job at explaining this:

”Wind drag is a nonlinear function of the relative wind. That is, drag does not increase in proportion (linearly) to wind but rather with the square of its speed relative to the rider. Relative wind-speed is the sum of wind and rider speed, which in still air is equal to rider speed. Direct head- and tailwinds can be combined by addition with rider speed to give relative wind-speed.Although inline winds add to or subtract directly from drag, they do not similarly affect the power required to overcome that drag. For instance, standing still in a 15mph wind requires no power because power is the product of inline drag and rider speed.

Drag is proportional to the square of relative wind-speed while power is the product of the inline portion of that drag and rider speed.”

For the full discussion that also addresses crosswinds and tailwinds, see:

https://www.sheldonbrown.com/brandt/wind.html

Update: I went back and emphasized the main point on this topic.

Otto

wphamilton

I think you know what "Physics 101" almost always means when someone throws it out as they "argue physics".

BillyD

This is all we ask, that our posters look at unorthodox posts and posters with an open mind before labeling them a troll. Everyone isn't the same, so don't be surprised to find someone who thinks differently than you.

Now just look at the relatively wonderful discussion that has materialized.

CargoDane

That it is basic physics that shouldn't need to be explained to anyone arguing aerodynamic drag?

ofajen

Yeah, I experienced this wind thing first hand today and having run some numbers during this discussion clarifies why the power requirement varies for the same relative wind speed.

I did a nice long ride with steady 15 mph wind from the south. Towards the back half of the ride I’m riding out directly into the wind and had to work really hard to keep up a speed of 15 mph.

This handy wattage calculator

https://www.omnicalculator.com/sports/cycling-wattage

estimates that I (175 lbs plus 25 lb bike on gravel surface) had to pedal at about 350 watts to keep my speed at 15 mph. I believe it. Hard enough to turn around and be glad to go home.

On the same bike and surface with no wind, this calculator indicates I’d need to do twice that (700 watts) to ride that same stretch at 30 mph (the same relative wind speed).

It will probably not come as a surprise to know that I don’t ride that stretch anywhere near that speed.

And of course, if I drop the speed to zero in the face of a 30 mph wind, the calculator properly indicates zero watts.

Update: out of curiosity, in the face of a 30 mph headwind, if I pedal at 350 watts, I can go 9.2 mph (for a little while...)

Otto

livedarklions

I'm an admitted physics illiterate, but I've now been convinced that I was wrong to disagree that the power requirements for riding 15 kph into a 15 kph headwind are considerably less than riding 30 kph with no wind. I also think I've figured out how to explain why (not how much) to physics dummies like myself.
Perhaps a physics literate can be so kind as to check my logic.

​​​​​​It's actually very simple--you expend no energy to create the drag from the 15 kph headwind when you ride 15 kph into it. When you increase your speed from 15 kph to 30 kph with no wind, you are both generating the energy to increase the drag AND the energy needed to overcome the additional drag. That make sense?

Another way to put this is I've got to spin the rear wheel the same rate wind or no wind to maintain 15 kph regardless of air speed so I'm doing that against a combined 30 kph air flow with a 15 kph headwind, while increasing from 15 to 30 kph with no wind requires me to spin my rear wheel twice as fast into the same 30 kph air flow.

Note- edited to fix typo in first paragraph pointed out below.

ofajen

The second part does make sense, but the earlier statement isn’t correct. The power requirement for riding 15 kph into a 15 kph wind IS less than riding 30 kph in still air, though the drag force is the same in both cases. Perhaps you said the opposite of what you intended? Or maybe I’m reading it wrong?

For example, riding 15 mph into 15 mph wind today needed 350 watts, while riding the same stretch at 30 mph in still air would require 700 watts. With no wind, at 350 watts, I could make about 23.2 mph (for a little while...)

Otto

rsbob 

Fizik. Nice bike seats, purchased without resistance. My last physics class was in high school, so I will stick to bike seats. But not literally.

tomato coupe

I think you pretty much have it. I'd restate it as: The force required to overcome the aerodynamic drag is the same in both cases, so the torque at the back wheel is the same. Traveling at 30 kph requires twice the angular velocity at the back wheel, however, so the power (torque x angular velocity) is double.

CargoDane

Not really. The only additional drag when you yourself are moving are the rolling resistance. The drag through the air is the same whether you are going 15 + 15 headwinds as you going 30 in no wind. You are moving through the air at the same speed.
You do create drag when riding 15 kph into a 15 kph headwind. That is a combined airspeed of 30 kmh. When you go 15 kph in still air, you still create the drag of going through the air at 15 kph. It is simply not true that if you go 15kph (even in still air) that you do not create drag.


[No, that's not true. It doesn't matter one bit as the bottom of the wheel will then move relatively slower (because the ground then moves faster under your bike) into the same speed of air going over it.
If you're talking about acceleration, then that is of course a matter of weight vs. power too and have absolutely nothing to do with aerodynamics.

Again, if any of the nonsense the other were saying were true, we couldn't calculate power needed based on drag, we would have no fluid dynamics (air is a fluid too), no testing tanks would work, nor would wind tunnels, and planes, boats, bicycles, and so on would suddenly experience less resistance the moment they stopped their propulsion. Planes would fall out of the sky because according to those three, the resistance (lift vs. drag) suddenly increase when power is applied, and hence is decimated when the propulsion is cut off.

It is beyond ridiculous. They have no idea about the implications of their claims.

Edited to ADD:

It may be beyond the three others, but you can test all of this out yourself very easily:

RIde 15kph in 15 kph TAILwind. You will feel no wind because of the very simple reason that you move at the exact same speed as the wind. The only thing you have to overcome is the rolling resistance.

Also, there is a reason that planes always try to land and take off into a headwind: Because when they do, they don't have to get as much over-ground speed in order to create the lift needed to take off and land (= shorter landings and take-offs). Because what matters when it comes to drag and lift is the speed through the air.

wphamilton

No, it's because people who want to argue and then say "it's basic physics" tend to be missing something. Nothing personal - I myself will re-examine my own conclusions if I'm ever tempted to say "basic physics".

I'm not going to argue. Physics is not an argument. Look at the well-known power equation already supplied by tomatocoupe, understand it, and that's the physics.

CargoDane

Another way to test it is to go out in a boat in waters with a well defined current going on. You will find that it is not your speed over the bottom of the sea that matters to your steering and speed, but the speed you're going through the water.

tomato coupe

That's absolutely, 100% correct. But, it doesn't apply to bicycles ...

CargoDane

You do that. Fluid dynamics is a settled science, and when people can't even get their head around the basics of physics, maybe they shouldn't attempt to tackle fluid dynamics.

I do understand it. But that equation can only be equated to speed through the air if the air is still. It's from there he applies all sorts of hoops, revealing he has no grasp of what it is he copy/pasting.

I already agreed that at zero winds, speed over ground is the same as speed through the air. Just like speed over ground will be the same as speed through the water if there are no currents. But that is the ONLY point that is true.

How fast can a hot air baloon fly through the air? 0 mphs (apart from inertia in gusts etc.). Now, how fast ca a baloon fly in a jet stream? The answer is the same, only if it was in the fastest jet stream ever, it would have gone 231 mph over the ground.
Now, if ground speed mattered to how much drag it has and even has no propulsion, how would that even be possible?

If you guys want to argue that it matters that your propulsion is connected to the ground, you have to show why that would matter one iota.

But fine. The other three doesn't want to argue either. They are happy with copy/pasting things and making up additional stuff to support their claims.

tomato coupe

I don't think it's possible for it to be stated any more explicitly. Maybe bold, red ink will help ...

There are many, many references available that state the same results.