## Flying man!

Ok, propably everyone here knows that IN THEORY it is possible for a human to fly. This question has been bugging me a lot and I know it can be solved using some kind of aerial physics (Dont know the term :D).

Therefore my question is: How fast must 75 kilograms weighing human wave his hands up and down while standing like a cross to fly above the ground.

Here is a picture of the position:
http://images.jupiterimages.com/comm...5/22833596.jpg

Let's assume that an average sized hand is about 20cm*10cm = 200cm^2 and the arm itself is about 60cm*10 = 600cm^2 which means that the total waveable parts make 800cm^2 and for both hands its 1600cm^2. Therefore the surface that is waved up and down is 1600cm^2 and I need figured out how fast must that sized object go up and down to lift a 75kg weighing human off the ground. The hands will be waved at the range of 1 meter (from hips to a bit above head).

This will be done in normal air, in water it doesn't take alot, like 1 move per second.

I doubt this will be solved for me but if anyone would atleast try, this is a physics forum isn't it :) I would like the answer's units to be: 1 unit = 1 times hands from down to up and then back down.

For now let's assume that the person turns his hands when lifting his hands so he will not create the same force to push him back to ground he just created to get up from the ground. Let's say when he puts his hands back up from down it shall be arms = 600cm^2 and hands = 40cm^2 (hand on it's side for good aerodynamic ability).

I wish someone could answer this :)

THANKS

-Joppe

 PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
 Mentor Welcome to PF. You can use Bernoulli's equation to calculate the required speed of your hand to generate a force great enough to lift you off the ground by equating the velocity pressure to static pressure. I can tell you without calculating it that it would be several hundred m/s.
 Joppe, It is not at all clear that any primate could ever fly. Our bones are solid, our wings are small, and our metabolism low. A couple books you might want to look at are "Life in Moving Fluids" S. Vogel "Animal Flight"C. Pennycuick.

## Flying man!

 Quote by Joppe Ok, propably everyone here knows that IN THEORY it is possible for a human to fly. This question has been bugging me a lot and I know it can be solved using some kind of aerial physics (Dont know the term :D). Therefore my question is: How fast must 75 kilograms weighing human wave his hands up and down while standing like a cross to fly above the ground. Here is a picture of the position: http://images.jupiterimages.com/comm...5/22833596.jpg Let's assume that an average sized hand is about 20cm*10cm = 200cm^2 and the arm itself is about 60cm*10 = 600cm^2 which means that the total waveable parts make 800cm^2 and for both hands its 1600cm^2. Therefore the surface that is waved up and down is 1600cm^2 and I need figured out how fast must that sized object go up and down to lift a 75kg weighing human off the ground. The hands will be waved at the range of 1 meter (from hips to a bit above head). This will be done in normal air, in water it doesn't take alot, like 1 move per second. I doubt this will be solved for me but if anyone would atleast try, this is a physics forum isn't it :) I would like the answer's units to be: 1 unit = 1 times hands from down to up and then back down. For now let's assume that the person turns his hands when lifting his hands so he will not create the same force to push him back to ground he just created to get up from the ground. Let's say when he puts his hands back up from down it shall be arms = 600cm^2 and hands = 40cm^2 (hand on it's side for good aerodynamic ability). I wish someone could answer this :) THANKS -Joppe

Airline industry is in danger of facing extinction just based on your question :)

 Hehe good one, anyways this question came to my mind when i was working out doing like-flying moves with dumbells :)
 Mentor Ok, for more, the equation is at the top of the page here: http://www.princeton.edu/~asmits/Bic...Bernoulli.html The first term is static pressure (pressure of non-moving air), the second is velocity or dynamic pressure (pressure of moving air), and the third is pressure due to elevation, which in this case is zero. Setting the static pressure equal to the dynamic yields: p=1/2 rho* v^2 The static pressure is the weight of the person divided by the area of his hands (75*9.8/(400/10,000)= 18,000 n/m^2. And you only generate it half the time, so lets double it. Air has a density of about 1.2 kg/m^3, so now you have: 36,000 = 1/2 * 1.2 * V^2 V= 245 m/s Assuming your arms move 1m up and 1m down and accelerate quicly, that's a rate of 490 flaps per second. I'm making a simplifying assumption here, though: that the back of your hand doesn't have a negative pressure on it, which it does. The actual lift generated will be somewhat less than I allowed.

 Quote by russ_watters Ok, for more, the equation is at the top of the page here: http://www.princeton.edu/~asmits/Bic...Bernoulli.html The first term is static pressure (pressure of non-moving air), the second is velocity or dynamic pressure (pressure of moving air), and the third is pressure due to elevation, which in this case is zero. Setting the static pressure equal to the dynamic yields: p=1/2 rho* v^2 The static pressure is the weight of the person divided by the area of his hands (75*9.8/(400/10,000)= 18,000 n/m^2. And you only generate it half the time, so lets double it. Air has a density of about 1.2 kg/m^3, so now you have: 36,000 = 1/2 * 1.2 * V^2 V= 245 m/s Assuming your arms move 1m up and 1m down and accelerate quicly, that's a rate of 490 flaps per second. I'm making a simplifying assumption here, though: that the back of your hand doesn't have a negative pressure on it, which it does. The actual lift generated will be somewhat less than I allowed.
Wow.. 490 flaps per second :/ Now everyone just imagine that :D Russ you're bout the smartest guy i've ever met ;).

What does it mean when u say 75*9.8/(400/10000)
I understand that 75*9.8 is the force you put on ground, but what is the 400/10000?

EDIT In my post i say its 1600cm^2 for both hands (the area) so why 400? I noticed the /10000 just makes it m^2 but shouldn't it be 1600/10000?

 Mentor You said 200 cm^2 for each hand - the rest was in the arms and since you get the same up force as down force with the arms, they don't help you fly.
 Recognitions: Gold Member Science Advisor Don't know if this helps any, but it is pretty cool... http://epic.weather.com/epicPlayer.h...=141&from=epic

 Quote by russ_watters You said 200 cm^2 for each hand - the rest was in the arms and since you get the same up force as down force with the arms, they don't help you fly.
Oh yeah then I understand, pretty cool tho (Stops trying to fly on a daily basis..) :)