My last resort Please Help me!

[DougF]

CSlagle I'm not very good at Aeronautics, But I'll Post it for you maybe the forum can help you out.

PS. try to give us more time next time OK, Thanks.

 

CSlagle

I have some extra credit due Monday and the equation is too hard for me. I don't even know how to work a scientific calculator I bought one and tried' date=' but no luck. I am attaching a file with my velocity equation and the figures. If you have time and are able to help, please let me know what to do. This meteorology class has more math than I anticipated and I really could use all the extra credit that I can get.

Well I hope that you can help. My last resort will be going to our school resource center and asking for help. That was really embarrassing last time. I had to ask for help on a volume question from some kids studying Calculus.

Thanks cslagle.

 

 

MET 1010

September 28, 2007

Web Project 6

 

Due October 1st

 

Any object can become airborne if the wind velocity is sufficient. The following is a formula that provides a reasonable estimate of liftoff in mph:

 

V = (195 × (weight /area))^½ (weight (lbs) area (square feet facing the wind)

 

Use the internet or any other source to find the data for the calculations. Calculate V for the following objects: single brick, single grain of sand, standing human being, round garbage can lid. Document your data sources and show your calculations.

 

Brick Area and Weight

V = (195×(3.8lbs × 19.0625)^ ½

(195× 72.4375)^ ½

141253.3125^½

376

 

Brick Masonry Specs, Kings Material Inc., Serving the Midwest

 

Sand Area and Weight

V= (195×(13mg × .2mm)^ ½

 

 

.2mm diameter

13mg.

 

Diameter of a Grain of Sand diameter

Mass of a Grain of Sand weight

 

Human Body Area and Weight

 

1.8m²

150lbs.

 

Surface Area of Human Skin

 

Round Garbage Can Lid Area and Weight

 

.2 lbs

359.49 inches

 

Measured and weighed this one in Wal-Mart

[modest]

CSlagle I'm not very good at Aeronautics, But I'll Post it for you maybe the forum can help you out.

PS. try to give us more time next time OK, Thanks.

 

Sorry I didn't see this earlier to help, but here it is -

You present 2 different equations:

V = (195 × (weight /area))^½ --- with weight divided by area, but when you work the problem:

V = (195×(3.8lbs × 19.0625)^ ½ --- you multiply the weight and area.

 

Intuitively, I would think division is correct because this would give you less wind velocity for a larger surface are, and I found a source on the internet that agrees: A World of Weather: Fundamentals of ... - Google Book Search

 

Also, I noticed you had trouble with the surface areas. Only include the amount of surface area facing the wind. If a brick is 3.5 x 7.625 on one side (facing the wind) then it is 26.69 sq. in. There are 144 sq. in. in a sq. ft. So we divide 26.69 / 144 = 0.19 sq. ft.

 

Besides the conversion trouble with surface area and the mixed up division and multiplication - you were definitely on the right track. Good sources on the internet too.

 

Here the problems are worked out:

 

[math] \text{V(mph) = } \sqrt{195\frac{weight}{area} }[/math]

 

Brick:

Weight: 3.8lbs

Surface Area: 0.19sq. ft.

 

[math] \text{V(mph) = } \sqrt{195\frac{3.8}{0.19} } = \sqrt{195 * 20} = \sqrt{3900} = 62.45mph [/math]

 

Sand:

Weight: 0.000 028 66 pound [from 13mg]

Surface Area: 0.0095 sq. in. [r = 1mm = 0.039in (using 4πr^2) = .019 sq in / 2 (facing wind) = 0.0095 sq. in]

 

[math] \text{V(mph) = } \sqrt{195\frac{0.00002866}{0.0095} } = \sqrt{195 * 0.0030} = \sqrt{0.585} = 0.76mph [/math]

 

Human Body

Weight: 150 lbs

Surface Area: 9.69 sq ft [1.8 sq meters = 19.37 sq feet / 2 (facing wind) = 9.69 sq ft]

 

[math] \text{V(mph) = } \sqrt{195\frac{150}{9.69} } = \sqrt{195 * 15.48} = \sqrt{3018.6} = 54.94mph [/math]

 

Round Garbage Can

Weight: .2 lbs

Surface Area: 1.25 sq. ft. [359.49 sq. in. = 2.50 sq. ft / 2 = 1.25 sq. ft.]

 

[math] \text{V(mph) = } \sqrt{195\frac{0.2}{1.25} } = \sqrt{195 * 0.16} = \sqrt{31.2} = 5.59mph [/math]

 

 

Don't know what good this will do now if it's late - drop a line next time :)

[cslagle]

Modest,

 

Thank you for the reply. I turned in what I could do, and got partial credit. I was proud of that much. I am overlooking your equations for future reference, and I have written down, A World of Weather. Thanks so much for taking the time. It is appreciated greatly.

 

Sincerely,

Carla Slagle

[modest]

No prob -

You got at least half way there yourself. Your post shows a definite effort to understand.

 

-modest