Retardation of a Balloon's Vertical velocity

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SUMMARY

The discussion centers on the vertical velocity of balloons, specifically addressing ascent speeds of up to 15 m/s and descent speeds of -40 m/s reported in the 19th century. It is established that typical ascent speeds do not exceed 10 m/s and descent speeds are around -6 m/s due to air resistance. The relevant formula for calculating the forces acting on a balloon is F = 6πηv, derived from Stokes' law. The conversation also references weather balloon launches, which may achieve initial velocities exceeding 10 m/s.

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  • Understanding of Stokes' law and its application to fluid dynamics
  • Basic knowledge of gravitational forces and lift
  • Familiarity with the principles of air resistance
  • Experience with balloon physics and dynamics
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Lunar_Lander
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Because I am working quite a lot with balloons (still theoretically though), I have come across reports of some balloon flights in the 19th century, which gave balloon ascent speeds of up to 15 m/s and descent speeds up to -40 m/s. I agreed to the comment to that text, that normally balloons would never exceed ascent speeds of 10 m/s and descent speeds of about 6 m/s.

However, I now have received an E-Mail saying that there should be no problem at all that a balloon could ascent or descent faster than 10 m/s. I think that is not possible due to air resistance, and I would like to calculate for an example. If we consider the forces acting, we got the gravitational force acting straight downward, and the lift straight upward. If the balloon ascents, friction will point downward too (in case of descent upward). The relevant formula should be, given that the balloon is a sphere, [tex]\textbf{F}=6*\pi*\eta*v[/tex], according to the law of Stokes. Is that correct or should Newton's law be used?
 
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Lunar_Lander said:
Because I am working quite a lot with balloons (still theoretically though), I have come across reports of some balloon flights in the 19th century, which gave balloon ascent speeds of up to 15 m/s and descent speeds up to -40 m/s. I agreed to the comment to that text, that normally balloons would never exceed ascent speeds of 10 m/s and descent speeds of about 6 m/s.

However, I now have received an E-Mail saying that there should be no problem at all that a balloon could ascent or descent faster than 10 m/s. I think that is not possible due to air resistance, and I would like to calculate for an example. If we consider the forces acting, we got the gravitational force acting straight downward, and the lift straight upward. If the balloon ascents, friction will point downward too (in case of descent upward). The relevant formula should be, given that the balloon is a sphere, [tex]\textbf{F}=6*\pi*\eta*v[/tex], according to the law of Stokes. Is that correct or should Newton's law be used?

I'm no help on the calculations, but it would seem that weather balloon launches would provide some of the fastest initial velocities. I googled vidoes on weather balloon launch, and got plenty of hits. Take a look at the 2nd video on the hit list -- initial takeoff velocity looks to be more than 10m/s:

http://www.google.com/search?q=weat...1&source=vgc&hl=en&aq=1sx&oq=weather+balloon+

.
 

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