No, gravity has no different effect. But the effect of a headwind is like climbing a hill, a 10 mph (16 kph) headwind is like a 2% grade.

I think if it is windy enough, it will still hold you up a bit when you go downhill, overriding some of the effects of gravity.

Wind and gravity act in different directions, so while it may feel that way, it doesn’t. A stiff wind on level ground will bog you down like a steep hill without wind and make it feel like you are pedaling through a mud pit, but it won’t actually alter gravity.

Thanks, guys!
I am an airhead when it comes to physics, but you cleared things up for me.

I think as long as we reside on planet earth, gravity always enters the equation.

Gravity is the same regardless of the wind (barring minute changes like the possuble effect @captainsmooth described) so it need not be considered.

E=MC2
this is now the third time I have got to use this as an answer

@judochop Actually, E=mc2 is irrelevant for this particular answer because mass is not being destroyed or created.

@rebbel It enters into the equation if you are going uphill or downhill, but otherwise it’s a constant. The major effect is that the force is proportional to the square of the speed.
http://sheldonbrown.com/brandt/wind.html

@Rarebear is the mass not wind and the energy created the force of the bike pushing through it?

I would think that since most winds have a downward vector (almost no winds are purely horizontal, and none that I’m aware of emanate from the earth to have a vertical component), then there will be some added air pressure in a downward direction, which will seem to make you heavier.

So my short answer is, “Yes, usually.”

You might find the opposite effect if you cycle along the top of a ridge, where the wind could actually seem to come from beneath you and give you a lighter feel.

Given the choice between riding uphill into a headwind (and downhill with a tail wind) or downhill into a headwind (and uphill with tailwind) I would chose to ride uphill into a headwind. The force of wind resistance (as illustrated in the brandt article above) is not linear to speed – it’s worse than that. So, assuming we travel slower uphill, the force of the headwind is less than the force of the headwind riding (faster) downhill. With the tailwind downhill there is also a lot less drag.

Also, if I am huffing it up a hill, the headwind can help cool me off. It sucks going 8MPH up a hill with an 8MPH tail wind. You broil.

@judochop yes but that equation still does not apply.

@CWOTUS Even if the wind is pushing you straight down, gravity remains constant.

@Rarebear how does it not? If the answer is yes then also how no? I am not arguing, I am trying to learn something here. Teach me please.

Gravity force will be acting always no matter if wind is present or not.

We used to solve mechanics problems by resolving forces (which are actually vectors) into its components, like it is showed in this picture.

While cycling, there need to be considered three forces, namely gravity force, air resistance (or wind), friction (between tires and road) in addition to peddling. Now, direction of gravity force components can be known. Direction of wind is deciding factor if the wind is working against you or with you. Friction will act in opposite direction to the motion.

Calculation of air resistance is a complicated phenomenon. Aerodynamics is the subject if anybody wants to refer to it.

To minimize wind acting against you, you can lean forward so that you will make approximately bullet shape like this to reduce the air drag. And, this diagram illustrates it on airplane wing.

@judochop To put it in simple words, the equation (e equal to mc square) is used to bodies that have very high velocity; to be precise, it is used for bodies that have velocity equal or greater than that of light (3 x (10^30) m/s). For lesser speeds, Newtonian physics is used.

@jerv of course you’re right. I was apparently answering a different interpretation of the question (in my own mind) such as “Will I feel heavier when…?”

Clearly, gravity is a local constant (and I say “local constant” because it does vary from place to place on the planet, but it’s constant at those places – at least “constant in human time scales”), but one’s apparent weight can be changed by atmospheric conditions.

I answered the wrong question.

Good catch.

@judochop ok. When I’m not typing on an iPad and I have time I will explain. If it makes you feel better @prasad isn’t correct either.

Doesn’t e=mc^2 involve the conversion of mass to energy or vice versa? And wind is not mass converting to energy. The mass is still there. It is a just a basic Newtonian force (momentum of the air particles being transferred to the bicyclist).

I get that. I understand most physics and I understand e=mc2. This is why I’m confused by your answer. I don’t feel bad though. If you want, im me the answer.

klOkay, I’m at a computer.

E=mc2 is a mass/energy equivalence equation. When mass is destroyed, it turns into energy, and vice versa. Mass being destroyed happens in very specific situations, such as a nuclear reactor, nuclear bomb, particle accelerator, or the sun. (Mass can be created also, theoretically, but we don’t yet know how to do that—that’s Star Trek technology.) For bicycle headwinds, mass is neither being created or destroyed, so E=mc2 is not relevant.

Oh, wait. Mariah said the same thing I just did, just more elegantly. Sorry.

Hhmm…this really made me to go back and look into it. So, I decided I would hear it directly from Einstein here.

If you find hard to hear that, then following is the paragraph for reference.

“It followed from the Special Theory of Relativity that mass and energy are both but different manifestations of the same thing – a somewhat unfamiliar conception for the average mind. Furthermore, the equation E is equal to mc², in which energy is put equal to mass, multiplied with the [by the] square of the velocity of light, showed that very small amounts of mass may be converted into a very large amount of energy and vice versa. The mass and energy were in fact equivalent, according to the formula mentioned before [E = mc²]. This was demonstrated by Cockcroft and Walton in 1932, experimentally.”

jello. thanks :)

That’s what Mariah and I said. It’s still not relevant to the question, though.