A "cloudy" puzzle........for those with too much time on their hands

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carson

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Need some exercise for your neurons? Read on....

    As I was sitting in my stationary Motorhome, contemplating life on a beautiful, bright but partially cloudy day, my eyes wandered towards the windshield. In front of the RV I had parked my Minivan at a right angle to the RV.

  I noticed the reflection of beautiful whisper clouds racing across the essentially flat side window of the van. I asked myself: wonder at what speed the clouds are moving from West to East. Many thoughts went through my mind for calculating the speed. I gave up after I felt my neurons getting tangled up. Good question for all the sharp folks here on the forum, I thought.

  Anyone wish to tackle this problems?  Here are the parameters that may play a role.

    1. The window is 4 feet (1.2 m) in length.
    2. The angle from vertical about 20 ?
    3. Est. altitude of the clouds about 2000 feet (670 m)
    4. Time for the edge of a cloud to move across the window in straight line, left to right, 60 seconds (1 minute)
            (Cloud formation virtually unchanged during this interval)

What is the speed of the clouds?  Would that be the wind speed as well?

Call this a trivia question; I don't think this will solve anything except maybe give you a headache.

(see image attached)

carson FL
 

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  • Cloud speed test.jpg
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Missing an important piece of information -- how far was it from your eyeball to the van's window?  Hard to tell in the photo since I don't know the focal length of the camera.
 
OK, I would say 9 feet (2.7 m)

This is what it says on the camera lens housing: f=6.3 - 12.6mm  1:3.8
  I used 0 telescopic action.

Camera is SONY Cybershot DSC-P52    -  3.2 MP

Phase of the MOON - I don't know :)

carson
 
Carson, On a beautiful day like that, you need to get out of the motorhome, go for a walk or drive, and quit worrying about cloud speed.  ;)
 
I know, Wendy, but somebody has to do it.

I am in A-1 shape now for my age, I'm afraid that walking will only wear out my bearings.:)

carson
 
Lessee...
4 ft/min x 60 = 240 ft/hr
240 ft/hr / 5280 ft/mi = 0.454545 mi/hr
Scale factor 2000/9 = 222.22222
.0454545 x 222.222= 10.101010mi/hr
Which is the wind speed at the cloud altitude
Art
 
Okay, here's a rough solution...
With the van window tilted back 20 degrees from vertical, you are looking up at the sky at about a 40-degree angle, neglecting the
small down angle you are viewing the van at.

If you are looking up at 40-degrees and the clouds are 2000 feet above you, then you are looking at clouds in the center of the window that are 3,111 feet away from you.
[ 2000/sine(40-degrees) = 3111 ]

Neglecting the complication of stereo vision, if the van window is 4-feet wide and 9-feet away from you, then your field-of-view of the clouds
is 1,383 feet wide.  [ 4 * 3111/9 = 1383 ]

Therefore the cloud speed is 1,383 feet per minute or about 15.7 mph.  There are some seconday issues that would change the number a bit, but that's a ballpark answer.

Sorry to take so long, but my granddaughter's birthday party took priority.

 
King said:
Lessee...
4 ft/min x 60 = 240 ft/hr
240 ft/hr / 5280 ft/mi = 0.454545 mi/hr
Scale factor 2000/9 = 222.22222
.0454545 x 222.222= 10.101010mi/hr
Which is the wind speed at the cloud altitude
Art

You're off by the factor of 3111/2000.  The clouds are 2000 feet high, not 2000 feet away.

I agree that cloud speed matches wind speed -- I forgot to put that in my answer.
 
I am impressed!  Thanks Frank and Art, sure makes sense to me.

Now I can concentrate on measuring the speed of an airliner or birds in the sky. Might have to wait a while before one shows in the window in just the right spot.  ;D  Think I'll leave that for a while.

  And all this while we on Earth are traveling at 887 mph ( Earth rotation) at this latitude 28? N.

(1070 (speed at equator) X cosine 28 (.88295) = 887.565 mph). Whowudathunk  :)

Back to work, guys,

carson FL  59.2 F




 

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