Retardation of a Balloon's Vertical velocity

AI Thread Summary
Balloon ascent and descent speeds have been historically reported at extremes, with some 19th-century flights reaching 15 m/s and -40 m/s, while typical modern speeds are around 10 m/s for ascent and 6 m/s for descent. A recent email suggests that balloons can exceed these speeds, prompting a discussion about the role of air resistance in limiting velocity. The forces acting on a balloon include gravitational force and lift, with friction opposing motion. The law of Stokes is proposed for calculating the effects of air resistance on a spherical balloon, though there is uncertainty about whether Newton's law should be applied instead. Weather balloon launches are noted for potentially achieving initial velocities greater than 10 m/s, indicating that higher speeds may be feasible under certain conditions.
Lunar_Lander
Messages
33
Reaction score
0
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, \textbf{F}=6*\pi*\eta*v, according to the law of Stokes. Is that correct or should Newton's law be used?
 
Physics news on Phys.org
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, \textbf{F}=6*\pi*\eta*v, 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 baloon launches would provide some of the fastest initial velocities. I googled vidoes on weather baloon 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+baloon+

.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?

Similar threads

Replies
5
Views
181
Replies
4
Views
7K
Replies
9
Views
2K
Replies
22
Views
3K
Replies
4
Views
2K
Replies
53
Views
4K
Replies
10
Views
6K
Back
Top