The French call it formule des trois niveaux.  In the language of calculus :

For any polynomial P  of degree  3  or less :

b    P (x)  dx      =       ba   [  P (a)  +  4 P ( a+b )  +  P (b)  ]     a 6 2

Becauseintegration is linear, it's enough tocheck this for  P = 1, x, x² or x³.

In elementary terms,  the volume  V  of a solid of height  H, bottom section  B,  middle section  M  and top section  T  is approximately:

V   =   ( H / 6 )  ( B + 4M + T )

The formula is exact  when the surface area of an horizontal cross-section varies with its altitudeas a polynomial function of degree  3  or less,  which includes the following important special cases;

• Sphere:  H = 2R,  B = T = 0,  M = R²
• Conical frustum:   M = ¼ (B + T)²
• Barrel of Oughtred:  

Here is our version of what's called the [second] principle of Cavalieri.
(Thefirst principle of Cavalierideals with planar areas instead of volumes.)

If an horizontal plane always intersects two given solidsin
sections of equal areas, then the solids have equal volumes.

Two such solids are said to beCavalieri congruent. This old-fashioned "principle" can be made to sound trivial nowadays (the integrals of equal functions are equal)  but it helped define  the very concept of integration.

Some direct  applications are still interesting. For example, we can deduce the volume of a sphere from the formulas giving the volumeof a cylinder and the volume of a cone: The section of a sphere of radius R on a plane at a distance z (<R)above its equatorial plane is a disk of area:

( R ² z ²)

This is also clearly the area of a ring  with outside radius R andinside radius z,which corresponds to a solid that consists of acylinder of radius R and height R fromwhich acone of the same base and height is removed (the apex of the cone being on theequatorial plane).  The volume of this solid  (2/3 the volume of the cylinder)is thus the volume of the hemisphere. For the whole sphere, this does gives the familiar formula ofwhichArchimedes of Syracuse was most proud of:

V_sphere   =   (4/3) R³

This was obtained two millenia beforeCavalieri's principle got its name.

  A tetrahedron ABCD can beconstrued as a cone of apex  D  and base  ABC. As such, its volume is a third  of the product of the base area bythe height, which is also the dot product  of  AD  by thevectorial area of  ABC (itself equal to half the cross product of  AB  by  AC).  All told, the volume is:

V   =  ¹/₆ | ( AB AC ). AD |  =  ¹/₆  | det ( AB, AC, AD ) |     

A vertical cone of height  H,  radius  R  and apex  (0,0,0) has cartesian equation:

( x² + y²) / R²   =   z² / H²

On the vertical wall where  x  is a positive constant not exceeding R, the above can be interpreted as the cartesian equation  (in y and z)  of thehyperbola where the plane and the cone intersect.

Between the altitude -H  [of the cone's base]  and the altitude  -xH/R [of the hyperbola's apex]  the surface area of the hyperbolic segment is:

H/R xH/R

2  (zR/H)2 - x2    dz

This solid may be seen as thedifference of two cones with the same apex:

• The larger cone is of height H.  Its base is a half-circle of area R²/ 2.
• The height of the smaller cone is  h = H cos . Its base is a segment of aconic section of area A (aparabola when the plane's slope is H/R, an ellipse for a less inclined plane, anhyperbola for a steeper one).

The volume (V) of the wedge thus reduces to the computation of the area A:

V   =   (H/3)  [ R²/ 2 A cos ]

With suitable choices of coordinate systems, a point ofcoordinates  (x,y)  in the inclined plane has spatial coordinates(x, y cos , y sin )... Plug these into the cone's spatial equation to obtain the equation ofits intersection with the plane:

H²( x ² + y ²cos ² )  =   R²( H y sin ) ²

The curve's apex  (x=0, y>0)  is at  y = r = [ (sin ) / H +(cos ) / R ] ^-1
The segment's area A is thus given by either of the following expressions, wherex and y are positive quantities related to each other by the above equation:

For example, in the special case of a parabolic segment(tan  = H/R) we have:

y   =   (R²-x²)/ 2R cos         and        A   =   2R² / 3 cos

Therefore, in thisparabolic case, the volume boils down to:

V   =  HR² ( 3 4 )/ 18

[ Thanks toSteve Battison for catching anembarrassing typo in the last step. ]

In theelliptic case (tan  < H/R)...

In thehyperbolic case (tan  > H/R)...

Generally,any conical wedge bounded by planar sectionsof respective areas A₀ and A₁ [not necessarily joiningwithin the cone] and whose respective planes are at distances h₀  and  h₁  from the cone's apex,has the following volume V:

V   =    ¹/₃   | h₀A₀ h₁A₁|

In this, the cone is understood to besingle-sided  [its surface is generatedby half-lines originating at the apex]  but it need not be a circular one...

That book has been said to be to Chinese mathematics what Euclid's Elements is to the mathematics of Europe and the Middle East. However, it's more similar to Fibonacci'sLiber Abacci (1202) in style and scope.

The 26^th question is a fairly easy one:

A cistern is filled through five canals. Open the first canal and the cistern fills in 1/3 day. With the second, it fills in 1 day. With the third, in 2 1/2 days. With the fourth, in 3 days, and with the fifth in 5 days.
If all the canals are open, how long will it take to fill the cistern?

Well, the number of cisterns per day  filled by all canals togetheris the sum of the rates supplied by every canal separately,  namely:

3 + 1 + 2/5 + 1/3 + 1/5   =   74/15

The number of days to fill the cistern is the reciprocal of that:  15/74. In modern terms, that's  4 hours, 51 minutes and 53.513... seconds.

This illustrates one type of problems which still puzzles students and adults. Two other examples are given below. Thefirst one is straightforward,thesecond one is a little bit more complicated.

2 hours. In that amount of time,you'll have completed 1/3 of the job and your friend 2/3 of it.Of course, the assumption is that the "job" is of such a naturethat it can be "distributed" efficiently.

This result is obtained by adding the rates [the number ofjobs per hour]of both workers to obtain the rate for the team.In this case, your rate is 1/6 and that of your friend is 1/3,so the combined rate of the team is 1/6+1/3, or 1/2; the team does half a job in an hour,so it takes 2 hours for the whole job.

We assume that the "job" is of such a nature that it can be distributed between thetwo clerks. (A modeling assignement would not be such a "job"...)The first clerk completes 1/8 of a job per hour, the second 1/10 of a job per hour(if you want to be silly, you could say their respective "speeds" are 1/8 jph and 1/10 jph).

After t hours (with t>2), the first clerk will have completed a fraction t/8 of the job,and the second clerk a fraction (t-2)/10. The whole job will be completed when t is such that:1 = t/8 + (t-2)/10 Solve for t and you get t=16/3 hours, that's 5 hours and 20 minutes.Check your answer:The first clerk has worked 16/3 hours and has completed 16/24=2/3 of the job.The second clerk has worked 16/3-2=10/3 hours, completing 1/3 of the job.

To get ana% solution from a strong solution at p% and a weak one at q%,you should mix (p-a) parts of theweak one and(a-q) parts of thestrong one.

This will indeed give you(p-q) parts of a solution rated at:

a   =	(p-a)q + (a-q)p
	(p-a) + (a-q)

For example, if p=30% and q=3%, you obtain a solution ata = 12%by mixing30-12 = 18 parts of the stronger solution with12-3 = 9 parts of the weaker one(that's 2 parts of the strong solution and 1 part of the weak one,if you must have proportions expressed inlowest terms).

Metallic alloys and almost all other chemical mixtures arenormally rated by mass. The percentage given is the ratio of the mass of the ingredientof interest to the mass of the whole mixture.

Alcohol andvinegar solutions, on the other hand, are rated by volume.

The volumes involved are understood to be volumes of pure alcoholor pure water before mixing.  Therefore, the basic  relevant computations  don't depend on the well-known fact thatmixing one liter of pure alcohol with one liter of water yields only 1.92 liters of liquid!

Let  d  be the relative density of alcohol with respect to water. The value of  d  is practically equal to  0.79  (nearly 4/5). More precisely, the relative density  d  is the ratio of the absolute densities  of alcoholandwater:

At 20°C,   d   =   (789.45 g/L) / (998.2071 g/L)   =   0.79087
At 25°C,   d   =   (785.22 g/L) / (997.0479 g/L)   =   0.78754

The relative density is exactly  0.79 at a temperature of about  21.3°C.

A rating of  x  by volume  (ABV = alcohol by volume)  would be obtainedby mixing  x  volumes of pure ethanol with  (1-x)  volumes of water. This yields a liquid whose rating by weight is  y (ABW = alcohol by weight)  where:

y   =   xd / (xd + 1 - x)      or      1/y   =   1 + 1/xd - 1/d

For a dilute alcohol  (x and/or y are small) each bracket is well approximatedby its first term and, therefore,  y  is approximately equal to  xd. For example, 5% by volume is approximately 4% by weight  (using d = 4/5).

Of course, 100% by volume is 100% by weight. The above formula does hold for   x = y = 1  (the approximation   y = xd   isn't applicable;it would be off by  20%  or  21%).

US proof is twice  the percent of alcohol by volume :

Vodka is normally  40% by volume  (sold as "80 proof", in the US) which correspondsto  34.5%  by weight  (using  d = 0.79). Everclear  is 95% by volume (190 proof)  which is 93.75% by weight.  For an N-proof spirit:

x   =   N / 200         y   =   0.79 / ( 200 / N - 0.21 )

Standard answer:  30 cc. Better answer: 31 cc.

The computation would be easy if the nursehad  6 cc  of pure alcohol and 4 cc  of water instead of the  10 cc  mix. She would simply need to add  z cc  of pure alcohol tomake the volume rating  (definedabove)  equal to 0.9:

0.9   =   (z + 6) / (z + 6 + 4)

Solving for z, we obtain  z = 30 cc. That's what they teach in nursing school (possibly with the niceshortcut that's valid either forratings by mass or for mixing liquids whose volumes don't change significantly upon mixing).

What a stellar  nurse would do :

Because water and ethanol shrink significantly  upon mixing,there's more stuff  (both water and ethanol)  in her 10 cc mixture than wouldhave been obtained from mixing 6 cc of ethanol with 4 cc of water...

Let's assume that the original solution had beenprepared at 20°C and that it's being adjusted at the same temperature. We use asingledata source, according to which, at 20°C :

• The specific gravity of pure water is 0.99823
• The specific gravity of pure ethanol is 0.78934
• The relative density is thus  d = 0.78934 / 0.99823 = 0.79074
• By weight, a 60% ABV solution is y = 1 / (1+2/3d) = 54.2566...%
• The specific gravity of a 54% solution by weight is 0.90485
• The specific gravity of a 55% solution by weight is 0.90258
• The interpolation 0.90258(100y-54)+0.90485(55-100y) is 0.90427
• The unmixed average 0.99823(0.4)+0.78934(0.6) is 0.872896
• The mix is 0.90427/0.872896 = 1.035942 denser than its components.

Increasing her previous estimate by that last factor, the nurse will add:

(30 cc) 1.035942   =   31.08 cc

We leave it to the reader to work out what the volume of the resulting 90% ABV solution willbe at 20°C  (no, it's not 41.08 cc).

(2015-07-20)   
For historical reasons, vinegar is rated by volume  (just likealcohol).

In chemistry labs, acetic acid is best rated in mol/L.  Elsewhere, we have to deal with the traditional volumetric ratingoriginally designed to measure the acidity of vinegar.

Only a food-grade solution of acetic acid obtained by natural fermentation,with or without distillation, can legally be called vinegar. Vinegar must contain at least 4% of acetic acid, by volume  (5% is typical). In the US, distilled white vinegar is sold watered down to  6%. In France, cheap  8%  white vinegar is widely available.

Industrially, acetic acid is produced by carbonylation of methanol, using a catalyst containing iodine and either rhodium (theMonsanto process, 1966) or iridium (the more economicalCativa process,promoted by ruthenium):

CH3OH	+	CO		CH3COOH
Methanol		Carbon monoxide	cis- Ir(CO)2I2-	Acetic acid

Pure 100% acetic acid is a liquid commonly called "glacial" acetic acid (CAS 64-19-7)  with a density of 1.0497 kg/L. It freezes at  16.6°C  and boils at  117.9°C. It can be mixed with water in any proportion. At retail, diluted acetic acid and vinegarare rated by volume,  as explained below.

By definition, a solution rated  x  by volume is achieved by mixing  x volumes of pure acetic acid with  (1-x)  volumes of water at 20°C (as is the case with alcohol, this definition doesn'tdepend on whether the total volume varies upon mixing). The correspondence between the rating by volume (x) and the rating by weight (y)is the same as withalcohol, except that the calibration density involved is nowthe relative  density  d = 1.0516  of pure acetic acid at 20°C. (recall that water weighs 0.99823 kg/L at 20°C).

Rating acetic acid by volume (x) or by weight (y)

(1/y - 1) d   =   (1/x - 1)      where   d = 1.0516

The traditionalgrain of a vinegaris equal to  1000 x,  which is to say that it's ten timesthe volumetric percentage of pure acetic acid in it.  For example,ordinary 5% vinegar can be labeled "50 grains". Chemists will be delighted to know that a molar solution of acetic acid is precisely 56.881 grains.

The density of acetic acid is maximum at a concentration of about 80%.

The number of moles of acetic acid contained in a liter of solution (c)is equal to the number of moles in a kg of solution multiplied into the densitygiven by the density table at the relevant temperature [2 ].

(2015-07-24)     
Soluble crystals often incorporate a fixed proportion of water molecules.

Citric acid is one of the most common organic acids. It's a triprotic acid of formula (CH₂ COOH)₂ COH COOH or  C₆ H₈ O₇  which would implya molar mass of  192.12 g/mol.

However, the crystal structure of this substance is a hydrated form which featuresas many water molecules as acid molecules (C₆ H₈ O₇ ,  H₂ O ). Therefore, you need  210.14 g  of the crystal to get a mole of the acid.

All published studies refer to this monohydrated form of citric acid when ratingthe citric acid solutions "by weight".

For example, Alexander Apelblat (Citric Acid, 2014, p.42) states that the density at 20°C  (in kg/L)  of a dilute solution of citric acid is wellapproximated by the following quadratic function of the weight fraction  w:

0.99823  +  0.4023 w  +  0.1590 w ²      for   w < 0.63

In this, the weight fraction  w  is so defined that the number of moles of citric acidcontained in  M  grams of the solution is equal to:

w  M /  210.14

dva1270 (2002-03-29)    
A vehicle goes from A to B at 60 mph, and returns at 40 mph.Why is the average rate of speedequal to 48 mph? [not 50 mph]

That's because average speeds areharmonic, notarithmetic...  Read on.

Theaverage of several things is whatever single value you could replace all of thesethings with, and still obtain the sameeffect.Obviously, the way you would actually compute such an average may very welldepend on exactly which type ofeffect you are interested in.In various cases, there may be hidden assumptions,which must be made explicit before a reliable computation can be made:

Arithmetic Mean:  For example, if you are receiving two checks, the only thing you normally care aboutis the total amount of money you are receiving.In this case, it does make sense to consider that anaverage check is equal tohalf the sum of both, because two suchaverage checks would have exactly thesameeffect on your bank account as the two checks you actually got. This is called anarithmetic average and it's, by far, the most common formof averaging. It's clearlynot the only one, though...

Harmonic Mean:  In the case ofspeeds, the important thing ishow long [how much time]it takes to accomplish a journey whose legs are traveled at different speeds. In other words, theaverage speed has to be whateveruniform speedwould allow the journey to be completed in the same amount of time as the actual one[where the speeds on different legs may have been widely different]. This is clearly equal to the total distance divided by the total duration of the journey.That number is obtained as the so-calledharmonic average of the speeds,namely the speed whosereciprocal is thearithmetic averageof the reciprocals of the speeds involved.Using the numerical example in the question, 48 is theharmonic average of60 and 40, because:

1/48   =   ( 1/60 + 1/40 ) / 2

Geometric Mean:  The geometric mean oftwo nonnegative numbers is the square rootof their product. More generally, geometric averaging is appropriate for successiverelative increases: Consider, for example, a first-year increase of10% (a factor of 1.1),a second-year increase of 21% (a factor of 1.21),and a third-year increase of 700% (a factor of 8). Over three years, the increase corresponds to a factor of = 10.648.The yearly average is the cube root of that, namelya factor of 2.2, corresponding to anaverage yearly increase of 120%. The logarithm of a geometric average is thearithmetic average of the logarithms:

Log(2.2)   =   [ Log(1.1) + Log(1.21) + Log(8) ] / 3

The fact that people often express a relative increase in terms of apercentage difference is best discarded, except  for input and output.

Quadratic Mean:  This is the [positive] quantity whose square is the arithmetic mean ofthe squares of the quantities under consideration. It's commonly called theroot mean square (RMS),as it's the square root of the mean of the squares.

This averaging is common in physics. For example, the RMS speed of a [large] set of molecules may be used todefinetheir temperature (which is classically proportionalto their average kinetic energy). Also, the RMS value [over time] of the current through a pure resistorequals the continuous current that would dissipate the same heat as thevarying current observed.

Hölder Mean:  The German mathematicianOttoL. Hölder (1859-1937) investigated, for any given exponent p,the quantity H whose p-th power is the arithmetic mean of the p-th powersof the quantities under consideration. (For arbitrary exponents,this must be restricted tononnegative quantities.)

H   =  H_p(a₁,a₂, ...a_n)  =  [ (a₁^p +a₂^p + ... +a_n^p)/ n ] ^1/p

• Exponents 1, -1, 0 and 2 correspond, respectively, to the aforementionedarithmetic, harmonic, geometric and quadratic means. 
• The case p = 0 for thegeometric meanis definedby continuity as the limitof thep-exponent Hölder mean  H_p,when p tends to zero.
• If  p < 0  andany a_i  is zero,then H_p = 0   (defined by continuity).
• H_p(a,b)H_-p(a,b)   =  ab

For aninfinite exponent,the Hölder mean of a finite set of nonnegative quantities is defined as eithertheirmaximum (p = )or theirminimum (p = ).

Generalized Mean:  A bijective functionf can be introduced so that thefollowing relation defines the "mean" m of n quantitiesa₁,a₂ ...a_n:

f (m)   =  (1/n)  [ f (a₁) + f (a₂)+ ... + f (a_n) ]

f (x)  is  x  for the ordinaryarithmetic mean,x²  for thequadratic mean,1/x  for theharmonic mean,Log(x)  for thegeometric mean [only when x>0], etc. 

Averaging over a spheroid:  Nautical mile as an "average" minute of latitude.

Batting Averages:  

...Farey Series, etc.

(2003-01-08)  
What's theaverage number of days in aGregorian month?

This is a good way to expand the discussion in theprevious article. Theaverage ormean value of anything is often ultimately definedas the "expected value" of that thing according tosome probability distribution. For most pratical purposes, the distribution to use is clearly understood andmay be left unspecified. For example, when askedout of context about the average of28, 29, 30 and 31, we would normallyassume that these 4 numbersare equally likely and would readily equate theirmean (their expected value)with theirarithmetical average (namely 29½). On the other hand, this approach is clearly unacceptable if we have to work outthemean number of days in a month, since durations of 28 or 29 daysshould clearly weigh less than months of 30 or 31 days,on account of their lesser likelihood.

Our modern secular calendar (the Gregorian calendar) repeats with a periodof 400 years which includes 97 leaps years of 366 days and 303 regular yearsof 365 days. Therefore, this period consists of 146097 days, or 4800 months. The average number of days in a month would thusseem to be146097/4800, namely 30.436875. But is is really so?

Well, yes and no. The question is not precise enough to have a definite answer. It depends on exactly what type of event you expect the "mean" to predict. If eachmonth is equally likely, the above is indeed the correct answer. However, it would beat least as reasonable to assume that what you really want is theaverage duration of thecurrent month for a random Gregorian date. In this case, eachday is equally likely in the Gregorian cycle of 146097 days and theresult must be somewhat higher than the above, because longer months are more likely.

More precisely, in a Gregorian cycle there are 97 months of 29 days (1 perleap year), 303 months of 28 days (1 per regular year), 1600 months of 30days (4 in any year) and 2800 months of 31 days (7 in any year). Therefore, there are 2813 days (97 times 29) which fall in a month of 29 days,8484 days (303 times 28) falling in a month of 28 days, 48000 days fallingin a month of 30 days, and 86800 (2800 times 31) falling in a month of 31 days. The "mean" is obtained by summing up all possible values "weighted" with theirrespective probabilities. For example, the value 29 is weighted by theprobability of 29, which is 2813/146097 (roughly 1.925433%). Work it out and what you obtain for the mean length of a month, in that sense,is 4449929/146097, or about 30.4587294742534...This is slightly more than our previous result, as predicted.

Neither answer is better that the other; they are just answers to different questions. Which answer you pick depends on which question you feel is"intended" when pople ask about the mean length of a month. Given enough time to think it over,most mathematicians would probably find the latter interpretation of the questionmore "natural", but they would still acknowledge that other interpretationsare better adapted to specific contexts. For example, our earlier value of 30.436875 days is the onlyacceptableconversion factor between durations expressed in days anddurations expressed inaverage calendar months;any other number leads to a systematicbias for long durations...

(2009-02-08)  
The AGM of a  andb is the AGM of (ab)^½  and  (a+b) / 2

The AGM of two positive numbers can be obtained by replacing them iteratively by theirgeometric mean and their arithmetic mean until thosetwo bounds coincide at a given precision. For example, starting with 4 and 9:

WHILE A+B <> A+A   X=SQRT(A*B) : B=(A+B)/2 : A=XWEND

The above is an efficient algorithm with a quadratic  convergence (i.e., the number of correct digits roughly doubles  with each iteration).

The arithmetic-geometric mean is tightly related to the complete elliptic integralof the first kind  (K)  as follows. (See also: Gauss's constant.)

K(k)  =		/2    0	d	=	/2
			1 k2 sin2		agm ( 1k , 1+k )

.shows it to be equivalent to the Legendre normal form that appears elsewhere on this site.

Conversely,    agm (a,b )   =   (a +b ) / 4 K( |ab|/|ab| )

wankman (2002-04-10) 
How far away is the ocean horizon line?
Is there a formula for figuring that out?

Suppose the Earth is a perfect sphere of radius R (that's not quite true, but close enough). If your eyes  are at a height H above the surface of the ocean,then the horizon is at a distance D which is such that yourline of sightis tangent to the sphere at that distance.See figure at right.

So, you have aright triangle whose vertices consist of your own eye,a point on the horizon and the center of the Earth. The hypotenuse is  R+H  and the sides are  R  and  D. Therefore, (R+H)² = R² + D², which boils down to:

D²   =   (2R+H) H

In practice,  H  is much smaller than the diameter of the Earth  (2R), so that  (2R+H)  and  2R  are virtually the same and we have 2RH D². The distance  D  to the horizon seen from an altitude H is thus :

D	2RH

In this, R is the (conventional) radius of the Earth:R = 6371000 m orR = 20902231 ft(choose whichever unit of length you use to express Hin order to have the result D expressed in the same unit).For example, at an altitude of 2 m [standing on a small boat],the horizon is at 5048 m(about 5 km, 3.2 statute miles, or 2.7 nautical miles).If you climb on top of a tall mast at 20 m [ten times higher],the horizon is at 15964 m [3.16 times farther].From a hilltop on the seashore at 200 m,the horizon is 10 times farther than if you are standing on your small boat.From a high mountain at 2000 m,you could see the ocean 31.6 times farther than when standing on a small rock on the beach:The horizon line is then almost 100 miles away! [86 nautical miles]

For a quick estimate, compute the square root of your altitude in feet.Add 20% or 25% to that result  [more precisely 22.455%]  toobtain the distance to the horizon in statute miles (for a result innautical miles, add 5% instead, or6.41% to be more precise).

For example, if your eyes are 9 feet above the ocean[the square root is 3], the horizon is about 3.7 miles away(roughly 3.2 nautical miles).

The distance to the horizon in kilometers is roughly 3.5 timesthe square root of the altitude of your eyes in meters (a more precise coefficient is 3.5696).

(2016-07-14)  
Vertical temperature gradients bend light.

In a vacuum or in an homogeneous medium, light travels instraight lines. Otherwise, we may invoke either the Euler-Lagrange equations or the more elementary Fermat principle of least time (c. 1655)  toshow that light rays curve toward the gradient of the index of refraction.

We may use the latter directly, with no need for higher mathematics, in thespecial case of an horizontal ray in the presence of a vertical gradient:

The curvature  1/R  of such a ray is the same as that of a circular ray around a spherewhere the gradient of the index is such that the circular ray of radius R is the ray which istraveled in the least amount of time compared to all nearbycircular rays in the presence of said gradient in the radial direction.

The travel time around a circular ray at altitude  z  above our basic circle is 2[R+z]n(z)/c.  For this to be minimal at  z=0  the corresponding derivative  must vanish. Dividing that zero quantity by the constant 2/c,  we obtain:

0   =   n  +  R  dn/dz  

R is positive when the center of curvature is downward,which happens when  dn/dz  is negative  (superior mirage). Otherwise,  we have an inferior mirage  (R negative, dn/dz positive). Let's consider only the former case.  Since the index of refraction is a decreasing  function of temperature, this corresponds to a temperature whichincreases with altitude locally.  For example, that's what happens near the surfaceof the ocean when the water is colder than the ambient air.

In particular, an horizontal ray will stay indefinitely at the same altitudeif the above curvature constantly matches the curvature of the Earth (R = 6371 km).  In that case,an observer just above sea-level will see the images of distant objectsas if the Ocean was rigorously flat (erroneously assuming that the observed light travelled in straight lines).

At pressure  p = 101325 Pa,  the refractive index of airis a function of the absolute temperature  T. To a good approximation, we have:

n   =   1 + 0.079678°C / T   =   1 + 0.079678 / (t +273.15)

Since change in hydrostatic pressure is relevant,  let's use the full expression:

n   =   1  +  ( p / 101325 Pa ) ( 0.079678°C / T )

Differentiating that at the point  p = 101325 Pa,  we obtain:

dn   =   0.079678°C  [ dp / (T 101325 Pa)    dT / T² ]

In still air, the change in pressure is proportional to the change in altitude dz,in such a way that the forces applied to a small box of air add up to zero. Using a box formed by a vertical wall and two congruent horizontal surfaces we see thatthe difference in pressure between the top and the bottom is equal tothe weight of the enclosed air per unit of horizontal surface. That's equal to the height of the box multiplied into the gravitational fieldand the mass density of air  (obtained as its molar massdivided by the molar volume at pressure  p = 101325 Pa, derived from theideal gas law).  All told:

-dp   =   dz  (9.80665 N/kg)(0.028966 kg/mol)  /  (8.31446 T / p )

Equivalently:    -dp / p   =   ( 0.034164 / T ) dz

Plugging this into the previous equation, we have:

dn/dz   =   ( -0.079678°C / T ²) [ 0.034164  +  dT/dz ]

Let's solve for  dT/dz when  dn/dz  =  -n / R  (with  R = 6371 km) :

n / 6371000   =   ( 0.079678°C / T ²) [ 0.034164  +  dT/dz ]
n T ² / 507630   =   0.034164 +  dT/dz
dT/dz   =   n ( T / 712.48 ) ²    0.034164

Let's make the physical dimensions clearer:

dT / dz   =  [ n  (T / 131.69) 2   1 ]  0.034164 °C / m

Vertical temperature gradient  dT/dz  (in °C per meter)  for which
horizontal light rays remain parallel to the surface of the Earth  (in dry air)

t		n	n/T	dT/dz (R = 6371 km)
T - 273.15°C		1 + 0.079678°C / T	- 0.079678°C / T2
0°C	32°F	1.00029170	-1.06791  ppm / °C	0.11286 °C/m
5°C	41°F	1.00028646	-1.02986  ppm / °C	0.11829 °C/m
10°C	50°F	1.00028140	-0.99381  ppm / °C	0.12382 °C/m
15°C	59°F	1.00027652	-0.95962  ppm / °C	0.12945 °C/m
20°C	68°F	1.00027180	-0.92717  ppm / °C	0.13518 °C/m
25°C	77°F	1.00026724	-0.89633  ppm / °C	0.14100 °C/m
30°C	86°F	1.00026283	-0.86701  ppm / °C	0.14693 °C/m
35°C	95°F	1.00025857	-0.83910  ppm / °C	0.15295 °C/m
40°C	104°F	1.00025444	-0.81252  ppm / °C	0.15907 °C/m

If the temperature gradient is stronger than the values tabulated above (e.g.,  0.135 °C/m = 0.243 °F/m) then water located well beyond thegeometric horizon can be seenand some distant terrestrial objects could become visible,  as if the surface of the Ocean had an upturned rim.

Superior Mirages without a Temperature Gradient :

For a cold enough uniform temperature, the pressure gradient due to gravitywould be enough to warp light around the Earth!  Theoretically at least.

The above formulas are certainly not very accurate atextremely low temperatures because they use the ideal gas law (for one thing, carbon dioxide freezes out of the gaseous phase below -57°C). However, we may still take them at face value to obtain a rough  ideaof the uniform temperature we seek, by solving a quadratic  equation in T:

( 1 + 0.079678 / T ) ( T / 131.69 ) ²   =   1

The positive solution is  T = 131.65 K = -141.5 °C, which is well below the lowest ever recorded on the surface of the Globe  (-93.2°C at this writing) although it's not nearly low enough to liquefy nitrogen, oxygen or argon.

Moist Air andHumidity Gradient :

By far, the most important variable component of clear  atmospheric airis water vapor. In still air over a body of water, we must consider the possibility of a significantvertical humidity gradient capable of bending horizontal light rays.

To a good enough approximation for our present purpose,moist air is simply a mixture of water vapor and dry air (consisting itself of fixed proportion of all other atmospheric gases,including carbon dioxide). The molar ratio of those two components is very nearlyequal to the ratio of their respective partial pressures (the sum of the partial pressures is equal to the total pressuremeasured by an ordinary barometer).

At some given ambient temperature  T,  the partial pressure ofwater vapor  p_w cannot exceed the saturation pressure  p_w (T) at which vapor would condensate into liquid droplets  (as dew, fog or clouds). The composition of clear moist air is thus most commonly specified in termsof its relative humidity  (RH)  defined as the followingratio, which can vary between  0  and  1 and is usually expressed as a percentage  (0 to 100%).

RH   =   p_w/ p_s(T)

When  RH  is low enough,  so is  p_w  and vaporbehaves within dry air as if it, too, was an ideal gas. The index of moist air is then obtained as a weighted average of the index ofdry air and the index of that low-pressure vapor. This yields an expression similar to what we had for dry air,except that the total pressure  p  is now replaced by the followingexpression involving the ratio  k  of the index of vaporrelative to the index of dry air:

p'   =   ( p - p_w)  +  k p_w  =   p  +  (k-1) RH p_s(T)

The measured value of  k-1  is    .  Therefore:

n   =   1  +  ( p' / 101325 Pa ) ( 0.079678°C / T )

(R. S. of Austin, TX.2000-10-25)  
How can I find the distance (in feet) between two exact locations byentering latitude/longitude coordinates to 6 decimal places?
lancelizabeth (2001-07-11)
Using latitude and longitude, what is the formula that calculatesthe distance [in meters] between two points on Earth?

Assuming the Earth to be a perfect sphere whose conventional  radius is R = 6371000 m 20902231 ft, a point of latitude   and longitude    has the following cartesian coordinates (in a well-chosen coordinate system) :

[x, y, z]   =  [ R cos   cos   ,  R cos   sin   ,  R sin ]

In this, a northern latitude is positive, a southern latitude is negative. An eastern longitude is positive and a western one is negative. (Youcould use any other convention, provided you do so consistently.)

If you have two such points, the scalar product ("dot product") of two such vectorsgives you R² times the cosine of the angle betweenthem  (along a great circle, which is the shortest route on a sphere's surface).

In other words, two points whose spherical coordinates (latitude,longitude)  are respectively ()  and (',' ) are separated by an angle given by:

cos   =  cos cos cos' cos'+cos sin cos' sin'+sin sin'

From that cosine of  , you may obtain the angle    by usingthe inverse trigonometric function on a scientific calculator. [However, you may prefer to use the better formulabelow,which remainsaccurate for nearby points!]

If    is in radians,the distance between the two points along a great circleis simply equal to  R. If    is in degrees,the distance is  R/180.

A Better Formula (especially for small distances):

The above formula is not  adequate for computing distances betweennearby points, because   is small and  cos   is so close to unitythat the use of the reverse trigonometric function arccos would cause an unacceptable loss of accuracy in the final result. To skirt the difficulty, we may use the following equivalent formula,which turns an accurate knowledge of('-)and('-)into a similarly accurate floating-point value for :

Angular separation  () of distant  or nearby points on a sphere :

sin2      =     sin2  '-   cos2  '-    +   cos2   '+   sin2  '- 2 2 2 2 2

 Theidentity  cos x  =  1 - 2 sin²(x/2)  allows the following transformation of our original formula:

sin2(/2)	=   sin2(['-]/2)+ sin2(['-]/2)cos cos'
	=   sin2(['-]/2)+ sin2(['-]/2){ ½ cos('+)+ ½ cos('-) }
	=   sin2(['-]/2)+ sin2(['-]/2){ cos2(['+]/2)- sin2(['-]/2) }
	=   sin2(['-]/2)cos2(['-]/2) +sin2(['-]/2)cos2(['+]/2)

(Ukulele8421. 2010-02-09) 
What's the length of a straight line from Grand Rapids to Melbourne?

In theabove spherical approximation, the two citiesare adequately represented by two points (at zero elevation) on a perfectsphere of radius  R = 6371 km.

	Latitude ()	Longitude ()	Elevation (h)
GrandRapids, MI	42.961250°	-85.655719°	195 m
Melbourne,Australia	-37.813611°	144.963056°	31 m

Each point is located in space by the cartesian coordinates of a vector:

U   =  (R+h)  [ cos cos   ,  cos sin   ,  sin ]

The distance  d between the two points through  the Earth is simplythe length of the difference  between two such vectors. The square  of that distance is:

||UU' || ²  =   2 R² - 2U . U'    (assuming h=0)

That expression would be easy enough to spell out but it suffers from thesame flaw as the first formula of the previous section (i.e., numerical loss of precision at small distances). Instead, it's much better to compute the angular separation   between the two cities  (using the second formula given above)  with the followingrelation, obtained from elementary geometry  (in the plane of the great circle):

d   =   2 R sin (/2)

Happily, the formula we are recommending gives  sin (/2)  directly,via its square! In the case of Grand Rapids and Melbourne, we obtain numerically:

sin (/2)   =   0.94462778

Multiply this into the diameter of the Earth (2R) to obtain:

d =  12036.448 km.

At this level of precision  (1 m)  the difference of altitudes comeback into play: If Grand Rapids was at the same altitude  (31 m)  asMelbourne, we would obtain an exact result  (for a spherical Earth) by using  R = 6317.031 km.  This yields d = 12036.506 km.  The difference in elevationsof  164 m  yields  (in the main)  an additionalcorrection equal tothat difference multiplied into the negative cosine of the angular separation ( = 141.6877°). All told, the distance between the two cities would be  12036.635 km on a spherical  Earth.

The true oblateness correction  (-7.372 km) can be worked out as follows:

Full precision, using the Reference Ellipsoid (IUGG, 1980) :

Ingeography, latitude is always  defined as the angle  from the plane of the equator to the local vertical (northward angles being positive). If the surface of the Earth is approximated by a spheroid  of equatorial radius a = 6378137 m  and polar radius b = 6356752.3141 m,  then the local verticalis perpendicular to the surface of that ellipsoid.  It's a simple exercise in calculus to express the cartesian coordinatesof the  [elliptic]  meridian as functions of the parameter :

Those formulas can be used to compute the components of the vector whichgoes from Grand Rapids to Melbourne. The straight-line distance is its length, namely :

		=   12029.263 km
	(x'-x)2 +(y'-y)2 +(z'-z)2

Incidentally, the above formulas also give the relation (at zero elevation)  between the geodeticlatitude   and the (useless) geocentric  latitude ₀ :

tg₀  =  (b ²/ a ²) tg

To see that, notice that we have  tg₀ = z/x   when  = 0  and  h = 0.

(Heather of Canada.2000-11-06)   
How do I find the radius of the Earth at 51° North?

The term is ambiguous.  I'll compute two related quantities: 

1. The distance from the center of the Earth to a point at alatitude L of 51°.
2. The radius of that latitude's parallel.

Before we do a precise computation, let's do the usual rough one by consideringthat the Earth is a perfect sphere whose radius is the "conventional" radiuswhich isdefined to be exactly R=6371000m. With this approximation,we have: 

1. A (constant) distance to the center equal to R, namely 6371 km.
2. A parallel of radius R cos(L), or about 4009.4 km when L is 51°.

A much better approximation is to consider that the Earth matches exactlythe regular shape against which its irregularities are charted.That shape is called the"reference ellipsoid" and its dimension have beenpreciselydefined once and for all by theIUGG in 1980:

The meridian is a perfect ellipse whose equatorial radius isexactlya = 6378137m and whose polar radius is "approximately"b = 6356752.3141m. So far so good.

Now, what does a latitude L correspond to on that ellipse?
It isnot the so-called "geocentric" latitude(the angle between the line to the center of the Earth and the plane of the equator).Instead the latitude L is thegeodetic latitude:the angle between your localvertical (which is the line perpendicular to theellipsoid's surface) and theequatorial plane.L is the latitude you would get from local astronomical observations.With the spheric approximation, we did not have to worry about this fine point,because both definitions amount to the same thing.In the case of the ellipse they are slightly different.How different? Well, if we call C the geocentric latitude,the difference L-C is given by:

L - C = 692.74" sin(2L) - 1.16" sin(4L).

For L = 51°, L-C is thus about 678.07378" or about 0.1888358°.So the geocentric latitude C is about 50.811646°.

Let x be the radius of your parallel and y its distance to the equatorial plane.We have y = x tan(C) = (1.22663011194) x,while the equation of the ellipse gives you (with the above valuesfora andb) x²/a²+ y²/b² = 1.

Replace y by x tan(C) and solve for x to obtain the value of x,whereas the distance to the center of the Earth is simply x/cos(C)(its square is x²+y²): 

1. The distance to the center of the Earth is 6365264.58 m (that's about 5.7 kmless than the conventional radius of the Earth R).
2. The radius of the parallel is x = 4022.031 km(that's about 12.6 kmmore than what we got from the spherical approximation).
    

yonda1234 (2001-04-28) 
How can I find the area of a spherical polygon formed by a series of pointson the Globe, given by their latitudes and longitudes?

A basic result of spherical geometry  (calledGirard's Theorem) states that the surface area of aspherical triangle on a sphere of radius Ris equal toR²(A+B+C-) ,where A, B and C are theinside angles of the triangle.You could dissect your polygon into triangles and add up the results obtained fromGirard's formula, but there's a more practical way to proceed(whose validity may be established using such avirtual dissection):

Consider the sum of all the inside angles at each of the n vertices of your polygon(see below for a recipe to compute these angles from your list of spherical coordinates).The surface area of your polygon is simply:

R² [ (n-2) ]

The polygon need not be convex,it's only assumed that the polygonal line does not intersect itself(or else, the formula would only hold for a critical matching definition ofinside angles andinside area; see footnotebelow).

Therefore, the only real problem is to compute the relevant angles from yourlatitude/longitude list.Well, on a sphere of unit radius, a point of latitude u and longitude vhas cartesian coordinates(cos(u)cos(v), cos(u)sin(v), sin(u)).To compute the angle at point B in your polygon,you need to consider this vector for B as well as for the previous point (A, say)and the next one (C, say). The cross products BxA and BxC give you the directionsperpendicular to the planes of the dihedral angle you're after.Divide these two vectors by their length to obtain two unit vectors U and V.The scalar product of U and V is the cosine of the angle x you want(if your polygonal line is an approximation to a smooth curve,there won't be too muchbending at each corner and x will be close to).Note, however, that the positive angles x and(2-x)have the same cosine; you must choose whichever corresponds to theinside of your polygonal line. That's it. I hope this helps...

 and the longitude  as (periodic) functions of an arbitrary parameter t.

Signed Area Encircled by a Closed CurveDrawn on a Spheroid :

A   =   [ f (/2)f () ] d  =   [ f (/2)f ( (t) )  ] ' (t)  dt

Weighing function  f () for an oblate spheroid of eccentricity  e

f ()   =  (b 2/ 2 ) atanh (e sin ) / e  +  sin / (1 - e2 sin2 )

(2002-12-01)  
The relation between distances and periods of orbiting bodies.

Johannes Kepler (1571-1630)was the first to observe that thesquare of the period ofa planet around the Sun is proportional to thecube of the size of its orbit. More precisely, the size of an elliptical orbit should be defined as itsmajor radius (a),also called itssemimajor axis (which happens to be equalto the average distance of the planet to the Sun).

This is now known asKepler's Third Law of planetary motion. It occurred to Kepler on March 8, 1618; more than 12 yearsafter he had formulated his first two laws. [He didn't accept his own idea until better computations finally allowed him to formulatethe third law, on May 15, 1618.] Kevin S. Brownremarks that the introduction of logarithmsbyJohn Napier (in 1614)is likely to have been instrumental in Kepler'sdiscovery of the third law...

Kepler's laws are now seen as a straight consequence of Newtonian mechanics,involving the masses of two orbiting bodies  (Kepler's original laws being recast as the specialcase where one mass is vastly larger than the other). In its most general form, the third law may be expressed in terms of the following relation,involving Newton's constant of gravity (G),the masses of the two bodies that orbit each other (M and m),their average distance(R = themajor radius of the orbit of one body,if the other is considered fixed)and the period of revolution (T):

4² R ³  =   G (M + m) T ²

This relation holds for SI units,or within any other consistent system of physical units. However, it's worth noting that, for the motion of a small mass around the Sun,we may choose to express distances inastronomical units and times insidereal years, which makes the relation boil down to:

R ³   =   T ²