Why Is the Calculated Velocity for Lifting a Person in a Wind Tunnel So Low?

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Discussion Overview

The discussion revolves around the calculated velocity required to lift a person in a wind tunnel, with participants exploring the physics behind air mass, velocity, and forces involved in lifting. The scope includes theoretical calculations, conceptual understanding of air dynamics, and practical applications related to wind tunnels and natural phenomena like thunderstorms.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation using the formula mass of air times velocity equals mass of man times gravity, questioning why the resulting velocity appears low.
  • Another participant suggests that the size of the wing could influence the required velocity for lift.
  • Some participants inquire whether the discussion pertains to indoor skydiving or natural wind conditions, specifically referencing an approaching thunderstorm.
  • Concerns are raised about the choice of 10,000 kg as the mass of air and the validity of the calculations presented, with a request for clarification on the source of the formula used.
  • Momentum is discussed as a relevant concept, with one participant suggesting that the momentum of air should equal the force exerted on a person by gravity.
  • Another participant challenges the initial calculations, indicating that the equation may be incorrect and emphasizing the importance of air resistance in balancing forces.
  • A participant shares a personal experience with vertical wind tunnels, noting that a velocity of about 100 mph was required to suspend a person, providing a practical reference point.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and concepts presented, with no consensus reached on the validity of the initial formula or the implications of the mass of air used. The discussion remains unresolved regarding the correct approach to understanding the forces at play.

Contextual Notes

Participants have not fully addressed the assumptions behind the mass of air and its interaction with a person in the wind. There are unresolved questions about the applicability of momentum in this context and the specific conditions under which the calculations are made.

Suyash Singh
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mass of air X velocity = mass of man x gravity
(10000)x(v)=60x10
velocity=600/10000
how come velocity is so less?
I am still in school so forgive me if my calculations look stupid.
 
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Suyash Singh said:
mass of air X velocity = mass of man x gravity
(10000)x(v)=60x10
velocity=600/10000
how come velocity is so less?
I am still in school so forgive me if my calculations look stupid.
It depends on the size of the wing... :smile:
 
Or do you mean like indoor skydiving?

https://www.iflyworld.com/

https://www.tmtmtour.com/wp-content/uploads/2018/01/ifly2.jpg
ifly2.jpg
 

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berkeman said:
Or do you mean like indoor skydiving?

https://www.iflyworld.com/

https://www.tmtmtour.com/wp-content/uploads/2018/01/ifly2.jpg
View attachment 225341
i was talking about thunderstorm
there is a storm coming to Delhi with speed 60 kmph.
I was thinking if people would start flying. Or maybe the small dogs on street.
 
Suyash Singh said:
i was talking about thunderstorm
there is a storm coming to Delhi with speed 60 kmph.
I was thinking if people would start flying. Or maybe the small dogs on street.
Looks like 60-80mph is pretty close to the limit...

https://i.ytimg.com/vi/ITBq82c6RzY/maxresdefault.jpg

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It seems that nobody has really addressed your attempt at calculation. Where did you obtain that formula from, and also what made you chose 10,000kg as your mass of air?
 
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cjl said:
It seems that nobody has really addressed your attempt at calculation. Where did you obtain that formula from, and also what made you chose 10,000kg as your mass of air?
I thought that
momentum of air= force on man by earth
also mass of air per square metre=10000kg
 
cjl said:
It seems that nobody has really addressed your attempt at calculation
Hey, who you calling "nobody"?! :biggrin:
 
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berkeman said:
Hey, who you calling "nobody"?! :biggrin:
I meant directly addressing the calculation rather than the end result. You did address a number of the factors though
 
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  • #10
Suyash Singh said:
I thought that
momentum of air= force on man by earth
also mass of air per square metre=10000kg
Momentum isn't the worst place to start here, but I'll address that more after we look at the mass some more. 10,000 kg is about right if you're looking at the entire column of air from the ground up to space, but what bearing would that have on its ability to lift a person? If a person is standing in the wind, what mass of air actually interacts with them?
 
  • #11
Suyash Singh said:
I thought that
momentum of air= force on man by earth
On that track you will find out what happens when you hit somebody with 10 tons of air in one go. (But the equation is still wrong.)

What you really need is indeed the "force on man by earth", but what balances it is the air resistance (try look it up).
 
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  • #12
Rive said:
On that track you will find out what happens when you hit somebody with 10 tons of air in one go. (But the equation is still wrong.)

What you really need is indeed the "force on man by earth", but what balances it is the air resistance (try look it up).
conservation of momentum
2 x mass of air x velocity (since air velocity won't change)=m(0)+ m(v)
2x 10000 x velocity=60v
 
  • #13
Suyash Singh said:
conservation of momentum
That's an excellent tool for collisions, for example: but for this, you better to swim instead.
 
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  • #14
My grandson "flew" in one of those iFLY vertical wind tunnels and he and his adult "instructor" where suspended with a air velocity of about 100 mph. The force of wind applied perpendicularly on a one sq foot surface is about 49 lbs.
 
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