How do we measure wind speed over the ocean waters?
84192
This is a challenging question. There are weather ships that try to measure wind speed, but of course the presence of the ship itself can produce orographic effects and produce biases. Another estimate can be made from satellite and airplane observations of cloud movement. Another technique is to fly balloons (radiosondes) up into the air column and use anemometers on the balloons to measure wind speed, or track the balloons using radio methods and try to infer wind speed from that. Dropsondes can be dropped from airplanes as well to estimate windspeed. \n\nA technique that is used from satellites and airplanes is to measure the roughness of the sea surface by boucing microwaves off it (a "scatterometer"). The windier it is, the rougher the sea surface. \n\nThere are probably other techniques that are in use or have been tried or at least considered as well\n\nProbably there is room for improvement in all of these, if you are so inclined to think about it!
809239
Here are some sites/links that may interest you.  I assume you are a pilot and know what these maps represents. (Past, current, forecast) \n\nThese are the type of maps I used for my pilot weather briefing days.\n\nPast winds aloft mainly using satellite estimated wind speed and direction. (This first link - winds are streamlines)\n\nhttp://cimss.ssec.wisc.edu/tropic/real-time/eastpac/winds/winds-dlm.html\n\nNext two links (need to zoom in)\n\nNCEP (4-Panel included) 200 mb, 250 mb, 300 mb, (Not on the label, but also include) 500mb, 700mb, and 850 mb winds.\n\nRegular single map - \n\nhttp://www.nco.ncep.noaa.gov/pmb/nwprod/analysis/epac/nam/00/model_m.shtml\n\nhttp://www.nco.ncep.noaa.gov/pmb/nwprod/analysis/npac/gfs/00/model_lu.shtml\n\n850 maps\n\nhttp://wwwt.emc.ncep.noaa.gov/mmb/mmbpll/opsnam/nam.npac.850wind_animate.html#picture\n\nPlease use with caution as computer model forecast winds can have higher error rate over the ocean due to lack of measured sounding data and few ship and low level pilot weather report data available.  \n\nHope this is what you are looking for.
452913
You could place a vertical rod in the pond (on a stand or stuck in the mud?) and start the experiment when the rod is dry.  After you create the wave and the surface returns to rest, measure the height of the wave on the rod.  This will be the maximum height of the waves at the given point and will depend on the location of the disturbance (radiates outwards like a bullseye?).  Also, there should be nothing near the rod to magnify the wave (including shore effect or irregular bottom?).  Applying chalk to the rod (especially blue chalk?) may make the wetted height more visible and resist drying too fast.  Including a scale on the rod may allow you to read the wetted height above calm surface easier, especially using binoculars.  The trough should approximately equal the wave height.  Good luck.
890224
Velocities are vectors, so you´ll have to proceed according to vector rules for addition/subtraction.  Draw a horizontal line, with an arrowhead pointing to the left. This represents the (unknown) velocity due West. Now draw another line, vertical, this time.  Put an arrowhead in the lower end of this line.  This stands for the 50 kph wind speed to the South.  These two lines should make a right (90°) angle.  The lines haven' t to be drawn to scale.\n\nFinally, draw an oblique line connecting both remaining ends.  This is the aircraft speed; to indicate direction, put an arrowhead  at its leftmost end.\n\nNow, you have a triangle of which two sides are known.  This, and the fact it is a right triangle, suffices to solve the entire triangle. The angle you seek is that formed between the horizontal and oblique lines.  Let θ denote this angle.  From elementary trigonomety, sin θ = opposite side/hypotenuse = 50/205 = 0.2439... All you have to do now is to find the angle whose sine is 0.2439...  This is 14.1169...°.  From the figure you have drawn, this angle lies N of W.\n\nThere are several ways to figure out the speed to ground.  You may use Pythagoras' Theorem; or, having found the opposite angle, you may compute it as 205 × cos 14.117° = 198.809...\n\nJust follow the same procedure to solve the other problem.  You'll have to decide which direction is "river across" and which is "downstream"; I´ll suggest "right" and "down", respectively.  Just plug in the proper figures.
518861
"The approximate increase of speed with height for different surfaces can be calculated from the following equation:\nv2 = v1 x (h2/h1)n where v1 is the known (reference) wind speed at height h1 above ground, v2 is the speed at a second height h2, and n is the exponent determining the wind change.\n\nIn your calculation v is equal to 10 mph, h1 is 30 ft and h2 is 100 ft. Since your surroundings consist of low woods, the correct value for n is .20. Plugging these values into the formula, the average wind speed at 100 feet is:\n\n    v2 = 10 mph x (100 ft/30ft).20\n    v2 = 10 x (3.33).20\n    v2 = 10 x 1.27\n    v2 = 12.7 mph"
630748
The more wind the higher the waves.\n\nWhen wind blows over the ocean's surface, it creates waves. Their size depends on how far, how fast and how long the wind blows. A brief, gently breeze forms patches of tiny ripples on the surface called catspaws; strong, steady winds over long distances create large waves. But even when you feel no wind at all, you may encounter large swells created by distant storms.
568304
Wind speed is never zero.  It seems like it somedays, but higher up it is windy. Just look at the clouds.
1063787
"Wind is the roughly horizontal movement of air (as opposed to an air current) caused by uneven heating of the Earth's surface. It occurs at all scales, from local breezes generated by heating of land surfaces and lasting tens of minutes to global winds resulting from solar heating of the Earth. The two major influences on the atmospheric circulation are the differential heating between the equator and the poles, and the rotation of the planet (Coriolis effect)."
237929
You take lots of measurements of wind velocity, temperature and barometric pressure at many grid points, each of which defines a "cell" (a small volume of air -- small compared to the size of the hurricane, that is). From pressure and temperature you can estimate density of each cell, and from the cell volume, the cell's mass. Kinetic energy = 0.5*M*V^2.  Compute for each cell and add 'em up.
60733
simply use a hydrometer.
422042
NASA WorldWind:\nhttp://worldwind.arc.nasa.gov/
531695
Well an easy one is the thermometer of corse for the temperature \n\nAnother will be your barometer this instament measures pressure of the atmosphere (lower pressure a good signal for bad weather) \n\nThe anamometer we use to measure wind speed and direction \n\nNow one of the ccolest instaments we have is a laser cieliometer...aka(LBC) this device measures cloud heights by firing an invisible laser into the air, however it is only good up to 12,000ft AGL (above ground level)\n\nTransmisometer-runway visual range\nWSR-88D(Radar)- storm location, movement, intesity, direction...ect\n\nThere are several instaments we use....one of the best resources other than this might be to go to the local news station.
733758
A small test tube type of rain guage.\nTape a small calibrated thermometer to it ans us that for the flag pole for a small wind sock.
165378
Standard sea level barometric pressure is 1013 Millibars. When the needle goes above this, it indicates a High pressure. When the needle drops below, it indicates Low pressure.\nGo to a weather page like http://www.wunderground.com/global/CI_Index.html\nIt will tell you the pressure.
948487
air naturally comes from high pressure fronts to low pressure fronts\n\nalso if you look at weather maps those wierd looking squiggly lines \nthe closer together they are the higher wind speed you'll have
234213
http://winds.jpl.nasa.gov/missions/quikscat/index.cfm
1140889
Its maximum sustained winds (measured or best estimate due to a general lack of data over oceans) would reach 39 mph. It will almost certainly affect some residential areas either way - but not threatening at this time.