Watts up with the wind?
 

If I am spinning 110rpm and going 20 kmh and my HR is 170 or so and the wind is blowing 40-50 kmh into my face am I exerting the same effort it would take to ride at 60 kmh on flat ground?
Just asking!

Short answer: no.

http://www.kreuzotter.de/english/espeed.htm

If you're going 20km/h with a 40km/h headwind, you're outputting 464W. 464W with no headwind gives you a speed of 40.8km/h. Of course, all of this is an ideal world, so this isn't strictly true.

I'm gonna have to disagree. Short answer: yes.

Remove from the equation the distracting variables - heartrate and cadence - and just focus on the variables at play. The primary forces that work against you are gravity (in the event you're riding at an incline; assuming you're flat, this is zero), various forms of friction within components and rolling resistance (these are usually taken as pretty nominal, especially if you're comparing the difference in forces between these at 20 and 60 kmh), and relative wind resistance, relative being the key word.

If you're pedaling at 20kmh into a 40kmh headwind, relatively speaking, this exerts the same force as a 60kmh wind if you were stationary. The flaw is that, as I said, you would be stationary. Considering wind resistance is the primary force operating against a cyclist, especially at relative speeds of 60kmh (since force from wind resistance increases exponentially with speed), I'm inclined to say yes, they are effectively the same thing.

Interesting. The reason I included the HR and cadence as that is my lowest gearing. If I was spinning 110 rpm on the big ring and the 12, I would probably have the gearing to ride at 60kmh. That is, if I even could do that.
I just wanted to know if the perceived effort would be the same.

What a great calculator. And the difference between riding in the drops and up top is bigger than I thought.

If you were spinning at 110rpm and your relative speed against the wind is 60kmh, then yes, perceived effort should be the same, regardless of if it's 10kmh/50kmh (ground/headwind speed), 20/40, 30/30, etc. But it sounds to me like there are two questions here. One is a physics question, the other about biomechanics.

The physics question has been answered. Now, assuming you have the same forces applied against you, the next question it sounds like you're asking is about cadence. The biomechanics of your body are more efficient at different cadences, and not everybody is the same. There have been a couple of studies done on this if you're interested in reading up.
http://www2.bsn.de/Cycling/articles/cadence.html

In attempting to eliminate variables, you forgot to include the ones he didn't mention. Rolling resistance accounts for a significant percentage of power consumption. Rolling faster and experiencing the same airspeed, you have increased rolling resistance in your tires, bearings, etc. So in short, JoeMetal was correct.

It also doesn't take into account the effects of different riders, bikes, and positions on aerodynamics. If it were necessarily the same thing to go 60 into no wind as it were to go 20 into 40kph wind no one would bother with aero gear or positions.

Then you have wind direction. Fat chance you're headed directly into the wind constantly (though I think we've all had rides that feel that way. This is all compounded by the fact that real wind changes speed and direction frequently and does not present a steady opposing force.

Most importantly, force and speed do not have a linear relationship (though force and acceleration do).

I made mention of rolling resistance in my post. It is considered close to nominal when compared with wind resistance. I do not recall the exact numbers (that calculator link probably distills them), but there have been plenty of posts in this forum supporting this.

Why not do the calculation and be done with it. http://www.analyticcycling.com/ForcesSource_Page.html

Rolling resistance using defaults, 304.5 gmf (2.98e5 dyne). So at 20 km/hr it takes 16.6 W to overcome rolling resistance; at 60 that grows to 49.8. Whether a difference of 33 W matters is up to the individual.

Gravity has nothing to do with this, or rolling resistance, it is drag that is the factor.

Why would you say rolling resistance (which is caused by gravity) is not a factor? It is a ******ing force which has to be overcome by the rider's power and is speed dependent.

Using the defaults except in the drops with 110 rpm cadence.

20 kph and zero wind-> 52 watts
20 kph and 40 kph wind-> 334 watts
5.3 kph and 40 kph wind-> 52 watts
60 kph and zero wind-> 1022 watts

The math clearly shows that 20kph with a 40kph headwind is not the same effort as 60kph.

Simple intuition gets you to the same result. Ride 2kph into a 40kph wind. It's going to be harder than riding at 2kph with no headwind, but I guarantee you it will be easier than riding at 42kph with no wind.

Obviously there's more at work than just adding the headwind to the speed.

The difference is that when you are riding at a certain speed in still air, everything, the pavement, the air is going by you at that speed. With a headwind, there is a velocity profile to the wind with lower wind closer to the ground. Measured at the surface of the road, the wind speed is always 0. You are not necessarily experiencing the same effective wind profile.

So put a number to it. What do you think the boundary layer thickness is at 40 km/h?

40 km/h where? Boundary layer is not a well defined term relative to the surface of the Earth because there is no universal "free stream velocity". Everything is a complex system of currents depending on which layers of the atmosphere you look at. Also we tend to have trees and buildings that block out a lot of the wind close to the ground.

As it is commonly used in environmental and atmospheric science, the boundary layer of the Earth is approximately 1/4 mile.

at 60km/h you generate 1000 watts
with 40-50km/h winds in your face and going at 20km/h you generate 360-480 watts

I guess we have a terminology issue here. Rolling resistance, what exactly is that defined as? Drag + friction? Factors affecting such? I don't know.

Drag is your biggest problem when on flat land. When going uphill, not gravity is the issue.

I don't know what "rolling resistance" means.

I take it to mean, basically where the rubber meets the road. Friction.

But, the topic is talking about "wind". Which I would take this to mean, we are talking about drag.

going up over an incline of 6% at 12mph you use 11% of the power to fight friction, 9 % to fight aerodynamic and 80% to fight gravity

Rolling resistance is basically the energy lost towards deforming the wheels/tire frame, and overcoming bearing friction, for a closed cycle on a distance basis. It is independent of the wind speed. Mostly it comes from the tires. So when your tires get squished, it takes some energy. Some of that gets returned as the wheel goes around and the tire pops back again, but not all of it. Same for frame flex. Ultimately all of that energy is lost to heat.

Does that calculation of 1000 watts at 60km/h seem right? I thought pro-level sustained power output was circa 400-450 watts and they rode at speeds close to that fairly often. That's 37 mph.

Perfect.

Common usage holds that rolling resistance refers to the force due to energy lost through tire deformation and hysteresis. Energy lost through the drivetrain and bearings is referred to as frictional losses, e.g., Martin et al. http://www.recumbents.com/WISIL/Mart...%20cycling.pdf

I'm glad we could clear that up for you. Next time, roll some dice.