Is there a chart showing altitude versus height in millibars?
360638
I'm thinking that you meant to type pressure instead of height. because millibars do not measure height. they measure pressure. and altitude and height are pretty much the same thing.\n\nthe relationship between altitude and pressure can be modeled by this equation\n\nlog(.01*P)~5-(h/50853.0185)\n\nwhen P=pressure in millibars and h=altitude in feet
1154907
since P1.V1 / T1 = P2.V2 / T2 ... u don't actually care about the altitude. so V2 = 8.88 cm^3 (btw those degrees are °C, °F ?? cuz u gotta take them to Kevin before using them)\n\n1*10^5 Pa * 2cm^3 / 270 °K = 2*10^4 Pa * V2 / 240 °K\nV2 = 8.88 cm^3
93783
I think one has to recalibrate your barometer for every different location (including altitude), according to what I read on the internet and my intuition. \n\nI am working on a more quantitative analysis which I will try to complete to post here later.
768918
prssure measured in athmosphere
874775
ignor the atmospheric changes, it would be 4.8s if I'm not mistaken
906431
Just type "2140 meters in feet" in google:\n\n7021 feet
1146161
The barometric formula, sometimes called the exponential atmosphere, is a formula used to model how the pressure (or density) of the air changes with altitude. It is based on the simplifying (not very realistic) assumption that the temperature does not depend on altitude. However, this formula agrees reasonably well with the actual pressure and density variations above the earth's surface up to a height of about 450,000 ft (140 km).\n\n \np=p0exp(-Mg0z/RT)\n\n\n\n \nwhere h is the scale height, ρ (rho) is density, P is pressure, P0 is pressure at ground level (mean sea level pressure is 1013.25 hPa), M = 0.029 kg mol-1 (the mass of 1 mole of air), R = 8.314 J K-1 mol-1 is the gas constant, T is temperature, g0 is the acceleration due to gravity (about 9.8 m s-2 depending on your location) and z is the vertical height above the earth's surface.\n\nUsing the same principles, the above equation can be solved for altitude as a function of pressure. This formulation is known as the hypsometric equation.\n\nAs a rule of thumb, the pressure decreases by about 1% for every 80 metres increase in altitude.
309749
Altitude -0.8° 58.2° -0.8°
251868
Depends on the level of elevation.  The higher elevations have less air pressure.  At sea level it's about fifteen pounds for every square inch.\n\nConversion to mph is about 31.07 mph or 546.83 inches per second.\n\nHere are some drag formulas\nhttp://carini.physics.indiana.edu/E105/drag-force.html
91941
If you want to see your home from the sky try google earth.\nIt will show you any place from 100 meteres up to several miles up.
66785
Latitude, altitude, prevailing winds.
909410
Strange formula, since it has no consideration of the changes when the engine runs out of fuel.\n\nSuggestion, if you have Excel or something like that...\n\nPlug in a series of times in one column - perhaps seconds, or for more precision, try tenths of seconds.\n\nHave the next column calculate the height based on your formula.\n\nThen you could calculate vertical speed in the next column.\nYou can also have Excel plot graphs of this....\n\nSounds like your homework is the usual kind.  Maybe this will help.  \n\nClearly, if you differentiate the expression, you can get a formula for vertical velocity (dH/dT)  \n\nIf you solve for t when vertical velocity = 0, that will be the time of maximum altitude...\n\nAnd so on.  Good luck.
533658
Altitude is the height above the horizon, usually measured in degrees. The horizon is at 0 degrees and the point directly overhead, called the Zenith, is 90 degrees.\n\nAzimuth is the direction along the horizon, with North usually taken as 0 degrees. So East is 90 degrees, South is 180 degrees and West is 270 degrees. Sometimes Azimuths over 180 degrees are expressed as negative numbers, so that West is -90 degrees, meaning 90 degrees from North but counting to the left instead of to the right.
976314
OS maps with contours will do the trick to a reasonable accuracy. If you want real accuracy you will need to level to the nearest OS bench mark or get a GPS reading from an instrument which gives height as well as grid location.\n\nFurther info after reading another answer.\nI am talking about UK here where the ground levels can change quite dramatically from street to street unlike the previous US respondant who must live in a very flat area of the States.
194460
Atmospheric pressure is the pressure above any area in the Earth's atmosphere caused by the weight of air. Standard atmospheric pressure (atm) is discussed in the next section.\n\nAs elevation increases, fewer air molecules are above. Therefore, atmospheric pressure decreases with increasing altitude. The following relationship is a first-order approximation:\n\n    &#92;log_{10} P &#92;approx {5-{h &#92;over 15500}},\n\nwhere P is the pressure in pascals and h the height in metres. This shows that the pressure at an altitude of 31 km is about 10(5-2) Pa = 1000 Pa, or 1% of that at sea level1.\n\nA column of air, 1 square inch in cross section, measured from sea level to the top of the atmosphere would weigh approximately 14.7 lb. A 1 m2 column of air would weigh about 100 kilonewtons\n\nHere is a link to an online Java calculator that converts pressure to altitude\nhttp://www.csgnetwork.com/pressurealtcalc.html
875826
From Wikipedia:\n\nThe bar and millibar are defined as:\n\n1 bar = 100 000 pascals (Pa) = 1 000 000 dynes per square centimetre \n1 mbar = 0.001 bar = 100 Pa = 1 000 dyn/cm² \n(A pascal is one newton per square meter.)\n\n\n[edit] Origin\nThe word bar has its origin in the Greek word βάρος (baros), meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.\n\nThe bar and millibar were introduced by Sir Napier Shaw in 1909 and internationally adopted in 1929.\n\nAtmospheric air pressure is often given in millibars where "standard" sea level pressure is defined as 1013.25 mbar (hPa), equal to (1.01325 bar). Despite millibars not being an SI unit, they are still used locally in meteorology to describe atmospheric pressure.\n\nThe article goes further to explain that 1 mbar = 1 hPa, or 1 millibar = 1 hectoPascal.\n\nIn terms of inches Hg, 30.00 has long been accepted as the standard, or long-term average pressure at sea level.  30.04 in. is slightly above standard pressure, as 1017hPa is slightly above 1013.25.  In weather terms this would indicate clear or clearing skies, with winds shifting from counter-clockwise to clockwise.\n\nThe popularity of the millibar (now hPa) unit seems to stem from the coincidence that 1 bar is almost equal to one atmosphere.
894864
http://en.wikipedia.org/wiki/Darts#Height_and_distance