The Numerator is for rapidly progressing through values of one of x,y or z,
selected via a drop down list.
It consists of the digits 0 to 9 in a row. As the mouse cursor is run along these,
the final digit of x,y or z changes to that digit, and all
the displayed formulas are recalculated.
Clicking on a digit moves on the the next lower power of ten, and the process continues.
Moving the cursor away from the Numerator sets the last digit to zero, so clicking a digit
has the same net effect as typing a digits directly.
Clicking in the area just above the row of digits will revert to the next higher power of ten,
and the current and any less significant digits will be cleared.
To the left of the row of digits is a - - box, and to the right is a + + box.
Clicking on either of these will subtract or add 1 to the next higher power of ten,
so that moving over the digits will work on the range to one side of the current.
This will continue across zero into negative or positive values.
it is compared with the previous maximum, minimum, and 'nearest zero' values,
which are updated if necessary.
The 'previous values' can be reset by the user at any time.
They will normally be used in conjunction with the Numerator
to find, say, the roots of an equation.
There are also Auto buttons which will home in on a root or max/min
once the x,y,z values have been set close enough by the Numerator.
The precision of the comparisons is the same as for the main textarea result.
More and addMore are additional functions
that can be inserted in the A B C D and main textareas along with other formulas.
They display short messages in a small textarea to the right of the x y z values.
They can be used to display intermediate results or, say, warning messages.
To clear the display, select the text (click on it, and press Ctrl A) and press Delete.
Below are 4 small textareas where values or formulas can be entered |
to define up to 4 more variables A,B,C and D.
Double clicking on a formula from the main list at the bottom of the screen
will copy that formula to the next empty A,B,C or D slot, if there is one.
An A,B,C or D formula can be appended to the main textarea by double-clicking it.
Any of the variable names x,y,z, or A,B,C D, or the result of T the main textarea,
can be used in the main textarea or in the A,B,C or D formulas.
Note that it is the value of the variable that is used in any calculation, not the formula,
so iterative methods can be used (by repeatedly clicking the Execute button).
The precision for the A,B,C D results extends to the preset formulas.
|STATISTICS: Normal, Binomial, Hypergeometric|
|Factorial||fact( x )||=|
x itself will normally be an integer, but if it is not any fractional part is ignored.
In mathematics, it is designated by x! .
The factorial function is useful in many calculations involving permutations and combinations of items, because it is the number of ways of arranging x items (- there are x ways of selecting the first item, and for each of those there are x-1 ways of selecting the next from the remaining items, and so on).
e.g. fact( 5 ) = 5*4*3*2*1 = 120.
N.B. By convention, fact( 0 ) = 1. Sort of logical - the number of ways of arranging zero items could be said to be 1.
In practice things work out better this way, and the infinities that would otherwise occur in some equations are avoided.
|Permutations of x objects from y||perm( x, y )||=|
There are y ways of choosing the first item, and for each of those there are y-1 ways of choosing the second item, and so on till all x have been chosen.
The number of ways is therefore y * (y-1) * (y-2) ... * (y-x+1), and so:
[Though for efficiency, the calculation is done using another function muft( a, b ), which multiplies together the numbers from a to b, again ignoring fractional parts, and also returning 1 if a > b, so that fact(0)=1, etc.]
e.g. perm( 3, 5 ) = 5*4*3 = 60.
N.B. perm( 0, n ) = 1 if n >=1.
|Combinations of x objects from y||comb( x, y )||=|
The number of permutations must therefore be divided by the number of ways of arranging x items,
which is fact( x ). And so:
N.B. comb( x, y ) = comb( y-x, y )
- selecting x items by removing them is the same as selecting the y - x items left behind.
e.g. comb( 3, 5 ) = 5*4*3/(3*2*1) = 10 = comb( 2, 5 ) = 5*4/(2*1).
e.g. The number of possible poker hands is comb( 5, 52 ) = 52*51*50*49*48/(5*4*3*2*1) = 2,598,960.
N.B. comb( 0, n ) = 1 if n >=1.
|height Normal curve at value y if mean=x, SD=z||exp(0-(y-x)*(y-x)/(2*z*z))/(z*sqrt(2*pi))||=|
The spread of values of many things (peoples height's, the IQ of politicians) follow the shape of the Normal Distribution curve.
- The 'Napolean's Hat' shape, with its peak over the mean (average) value.|
What is of most interest is the area under the curve. The area to the left of a given value is the probability that a randomly selected measurement will be less than that value, and the area to the right the probability that it will be more.
The total area under the curve is therefore 1.
The Normal Distribution is closely related to the Binomial Distribution below - see the discussion at the end of that section.
The Standard Deviation (SD) of the Normal Distribution is a number such that about 68% of the measurements are within one SD of the mean, about 95% are within 2 SD's, and over 99% are within 3 SD's.
It is the square root of the Variance,
and the Variance is the average of the square of the difference between each measuement and the mean.
|Significant Skewness||2 *sqrt( 6 /z )||=|
With positive skewness, the peak is displaced to the left; with negative, to the right.
A rough estimate of the significance of Skewness is that it should not be further from zero (in the positive or negative direction) than 2 "standard errors of Skewness" (ses), where the ses is roughly √(6 / Sample size ) (Tabachnick & Fidell, 1996).
|Significant Kurtosis||2 *sqrt( 24 /z )||=|
A high kurtosis distribution has a sharper peak and longer, fatter tails, while a low (negative) kurtosis distribution has a more rounded peak and shorter thinner tails.
The formula for Kurtosis is adjusted so that the Normal Distribution has Kurtosis equal to zero.
The lowest value for Kurtosis is then -2 (which occurs, for example, in a single flip of a coin).
The highest value is infinite.
A rough estimate of the significance of Kurtosis is that it should not be further from zero (in the positive or negative direction) than 2 "standard errors of Kurtosis" (sek), where the sek is roughly √(24 / Sample size ) (Tabachnick & Fidell, 1996).
If the Kurtosis is not close to zero, then reliance on the Normal Curve may lead to risk being underestimated,
and some other method of data collection or processing should be considered.
|Probability of being x or more SD's above mean||pNsd( x )||=|
(The polynomial used here is provided by sitmo).
pNsd( x ) is an additional function that calculates the area under the Normal Distribution curve which is more than x Standard Deviations to the right of the mean.
The first step in determining probabilities is to find the mean, and then how many SDs a value of interest is away from the mean.
The value of interest = mean +(x *SD).
If x is negative, the value is to the left of (i.e. less than) the mean, and the probability will be greater than 0.5.
|Probability of being between x and y SD's above mean||pNsd( x ) -pNsd( y )||=|
|Probability a random value is >y, if Mean=x SD=z||pNsd( (y -x)/z )||=|
The same as above, but the number of Standard Deviations away from the mean is calculated.|
Finds the area under the Normal Curve to the right of (y-mean)/SD, the probability that a random measurement will be greater than y.
A small value means that y itself is an unlikely measurement.
|Value with probability y of being more than a random measurement, if Mean=x SD=z||normInv( y )*z +x||=|
normInv is a polynomial approximation to the inverse of the Normal distribution with mean 0 and standard deviation 1, so:|
normInv( 1 -pNsd( (y -x) /z ) )*z +x = y.
Same answer as Excel NORMINV( y, x, z ).
|Sample z, zx have it. Prob( at least y of population have it)||pNsd( (y -x)/Math.sqrt( x*(1 -x) /z ) )||=|
|Binomial: z trials, each trial is just a hit (fixed probability x) or a miss (probability (1-x).|
|- Probability of exactly y hits||comb(y,z)*Math.pow(x,y)*Math.pow(1-x,z-y)||=|
The mean is z * x.
The probability of getting y hits (and z -y misses) in a particular order (e.g. hit, miss, miss, hit,...) is
The probability of exactly y hits is therefore
It obviously makes no difference if the trails spread over time or all done at once, providing no trial depends on the result of any other.
The probability of y or more hits is obtained by adding the probabilities of exactly 0, 1, ... y-1 hits, and subtracting the answer from 1.
e.g. The probability of throwing exactly two sixes with 3 dice is
The probability of throwing two or more sixes =
|- Standard deviation||Math.sqrt( z*x*(1-x) )||=|
Since the trials are independent, the variance for z trials = z*x*(1 -x), and the SD is the square root of this.
|- Skewness||(1 -(2 *x))/Math.sqrt( z*x*(1-x) )||=|
|- Kurtosis||(1 -(6 *x *(1 -x)))/( z*x*(1-x) )||=|
See STATISTICS Normal, Binomial, Hypergeometric - Significant Kurtosis.
|- Probability of y or more hits||=|
e.g. The probability of throwing exactly two sixes with 3 dice is
The probability of throwing two or more sixes =
probability of exactly 2 + probability of exactly 3 = 0.06944 +(comb(3,3) *pow(1/6,3) *pow(5/6,0)) = 0.06944 +1*(1/216)*(1) = .07407
|- same, using Normal approximation||pNsd( (y -(z*x) -0.5)/Math.sqrt(z*x*(1-x)) )||=|
Imagine the Binomial Distribution drawn as a histogram, the height of each rectangle being the probability of getting a particular exact number of hits.
The mean is at the left hand edge of the rectangle for z*x hits.
The Normal curve would make a better fit if the the peak was positioned over the middle of the rectangle.
This can be achieved if by adding 0.5 to the mean in the Normal approximation formula:
Each trail is, say, a single toss of a coin, one roll of the dice, or the inclusion of a particular gene in someone's DNA.
For each trial, a particular event (a hit) may occur (the coin lands with heads up, the dice rolls a six, the gene is included).
Each test consists of a (fixed) number of trials (z), and the outcome is a number of hits (y), from 0 to z.
Consider a histogram of the probabilities, with z+1 columns (for 0 and 1 to z).
If z is a large number, there is only a small chance of getting a particular number of hits exactly,
and a large number of tests would be needed to get reliable relative heights for the columns.
But each column would be very narrow, and the tops of them would form a smooth curve, peaking at the mean and tailing down to zero on each side.
If the probability of an individual event is 0.5, the peak will be central, and the curve will be that for the Normal Distribution.
If, say, a person's height depended on a single gene, and people with the gene were 2m tall and those without were 1.5m tall, then a histogram of peoples' heights would have two peaks with nothing between or to the sides, and be not at all like the Normal Curve.
Since height histograms are like the normal curve, this means that many copies of the same gene or many different genes are involved in creating a person's height.
|Hypergeometric: Sample of z items taken (without return) from a batch of A items which contain x 'defectives'|
|- Probability of exactly y defectives in sample||comb(y,x)*comb(z-y,A-x)/comb(z,A)||=|
If a sample has y hits (and z-y misses), for each of the
[See Wikipedia: Hypergeometric Distribution.]
The Hypergeometric has diverse applications.
e.g. A batch of 1000 items is received in which it is claimed no more than 30 are defective.
A sample of 20 is found to contain 2 defectives. What is the chance of finding 2 or more if the manufacturer's claim is true?
A=1000, x = 30, y=2, z=20. Probability y or more defectives = 12%
If the sample size were doubled to 40, and the same proportion of defectives (4) were found, the probability would drop to 3%.
e.g. What is the probability of being dealt a 4 cards to a flush in a poker hand?
For a particular suite, A=52, x=13, y=4, z=5, Probability = 1.07%
Probability for any suite = 4*1.07% = 4.28%.
The probability of getting a card of the required suite in the exchange is 9/47 = 19%
In Texas Hold'em, in the same situation (4 to a flush after the the 'River' has been dealt),
for a card of the required suite in the remaining two: A=47, x=9, y=1, z=2, Probability = 35%
|- Standard deviation||Math.sqrt(z*x*(A-z)*(A-x)/(A*A*(A-1)))||=|
Only to give some estimate of the quality of the probability results.|
|- Probability of y or more defectives||=|
The probability of y or more defectives is obtained by adding the probabilities of exactly 0, 1, ... y-1 defectives,
and subtracting the answer from 1.|
|STATISTICS: Poisson, Exponential, Queues|
|Poisson: random events, average x events per time interval|
|- Probability of exactly y events||Math.pow(x,y)/(fact( y )*Math.exp( x ))||=|
There is no limit on the number of events that can occur in a particular period.
Imagine a large circle divided int n segments.
In the centre is a cannon gun that fires y*n shots in random directions.
The probability of one particular segment being hit is 1/n for each shot.
As with the Binomial Distribution, the probability of a particular segment getting x hits (and y*n -x misses) in a particular order (e.g. hit, miss, miss, hit,...) is pow( 1/n, x ) *pow( 1-1/n, y*n -x ).
There are comb( x, y*n ) ways of arranging that many hits and misses, so the probability of getting exactly x hits is:
Now if n is very large, but the other numbers are not, then y*n -1 is nearly the same as y*n, and so are y*n -2 ... down to y*n -x +1,
so multiplying all these together is nearly equal to pow( y*n, x ). And so:
Now the pow( n, x ) cancels with the pow( 1/n, x ) in the first formula, to leave:
Again we can discard the -x following y*n in the first term, which becomes:
This is an interesting expression. It relates to compound interest, and how often that is calculated.
Suppose that in times of immense prosperity or rampant inflation, your bank was to pay you 100% interest per annum.
If this were calculated annually, each £1 would be worth £2 after a year.
But if it were calculated twice a year (at half the annual rate), after 6 months it would become £1.50, and after 12 months £2.25 because of the interest on the interest (50% of 50p).
Calculating the interest each month, say, would give pow( 1 +(1/12), 12 ) = £2.61
Calculating n times a year would give pow( 1 +1/n, n ),
and continuous recalculation (i.e. very large n) would give £2.72.
[- Try it using MaxiCalc.]
This number 2.72, or more accurately 2.718281828 crops up in many places
(it isn't a recurring decimal, the repeated digits are just by chance, and the decimal expansion continues indefinitely without pattern).
In mathematics it is designated by the letter e.
[For convenience, MaxiCalc also defines e as Math.E, and exp( x ) as Math.exp( x ).]
Now (1 +a)*(1 +a)*(1 +a)... for n times is:
Similarly (1 + a) multiplied by itself n*y times would be very close to 1 +n*y*a.
But (1 + a*y) multiplied by itself n times is also very close to 1 +n*y*a.
So it must be the case that for very large n,
This also works for negative values of y, so for large n:
And since pow( z, p*q ) = pow( pow( z, p ), q )
(e.g. 3 squared, and the answer cubed, = 3 to the power 6.)
|- Standard deviation||Math.sqrt(x)||=|
The Poisson Distribution is the same as the Binomial above if the number of trials (z) is very large,
but the probability of a hit (X) is very small, such that the average number of hits (zX) is a more reasonable number,
in fact the x for the Poisson Distribution.|
The Variance is therefore zX(1 -X) = zX -zX² = zX if X is very small, = x
and the Standard Deviation is the square root of this.
|- Skewness||1/Math.sqrt( x )||=|
|- Probability of y or more events .||=|
The probability of y or more hits is obtained by adding the probabilities of exactly 0, 1, ... y-1 hits,
and subtracting the answer from 1.|
|- same, using corrected Normal approximation||pNsd( (y -x -0.5)/Math.sqrt(x) )||=|
Because of the high values of factorials and powers, the formulas above fail if x or y are greater than about 150,
and to prevent lockup in the calculation, they are blocked if x or y are greater than 200.|
Fortunately the Normal approximation is reasonable for x,y above 100.
(- See Binomial Distribution for why 0.5 is added to the mean.)
|Exponential: time between random events, average x events per time interval (λ) =1/average time between events (1/μ)|
|- Mean time between events||1/x||=|
The Exponential Distribution is for the time between events, rather than the number of events.
x events on average per time period, so the mean time between events is 1/x.
|- Mean time till next event||1/(2*x)||=|
|- Probability that time to next event is > y||1/Math.exp( (x*y) )||=|
|- Probability that time to next event is < y||1-(1/Math.exp( (x*y) ))||=|
From the Poisson Distribution section above, the probability of no events in a particular period is
|- Standard deviation||1/x||=|
|Queues x = average occupancy of the server (0 to 1). Poisson (random) distribution of arrivals.|
|(a) Service times exponential. i.e. Probability service time > t = exp( -u*t ), where u is the average service time.|
|Average queue, including item being serviced||x/(1-x)||=|
This is the easiest to work out, because, if an item is being serviced, the probability of completing the service in the next time interval δt does not depend on the time already spent servicing this item.
The simplest formulas are those for all the items in the system, in the queue or being served.
Let p(n) = probability that there are n items in the system, in the queue or being served,
At equilibrium, all p(n) must be constant.
If a is the arrival rate, and d the server rate (assuming the queue is never empty), then in a short interval t,
N.B. If n >0, then there is at least one item in the system, so it is appropriate to use d, the rate of a fully occupied server.
The case n==0 is dealt with below.
These are the only things that can happen to p(n), and for p(n) to remain constant, the probability getting out of this state (of n items in the system) must equal the probability of getting into it. i.e.
except that p(0) cannot go to p(-1), so
N.B. Again, as there is one item in the system going from p(1) to p(0), it is appropriate to use d.
If x is the fraction of time the server is busy, then
But all the p(y) probabilities must add up to 1, so
This is called the Geometric Distribution (see Wikipedia, but note that the p in that article = 1-x).
The mean is
[- The proof of the mean and variance formulas involve summing y*pow(x,y) for y=0 to infinity, which is done by a calculus twiddle which is short to do but tedious to explain. See Wikipedia.
So simple a result must surely allow a straightforward explanation.]
(See Queueing Theory, Ivo Adan and Jacques Resing.
Also Amazon: Queues by D.R.Cox and Walter L.Smith (- not all pages displayed).)
|- Standard deviation||Math.sqrt(z*x*(A-z)*(A-x)/(A*A*(A-1)))||=|
|- Probability of queue size being y or more||Math.pow( x, y )||=|
and of there being y or more = pow( x, y ).
|- same, using SD and Normal Distribution||pNsd( (y - (x/(1-x)))/Math.sqrt( x/((1-x)*(1-x)) ) )||=|
|- Standard Deviation of queue size||Math.sqrt( x/((1-x)*(1-x)) )||=|
|Average queue, excluding item being serviced||(x*x)/(1-x)||=|
|- Probability of queue size being y or more||pNsd( (y - ((x*x)/(1-x)))/Math.sqrt( (x*x*(1+x-(x*x)))/((1-x)*(1-x)) ) )||=|
The average queue size is then
|(b) Service times constant.|
|Average queue, including item being serviced||x*(1 -x/2)/(1-x)||=|
For Constant Service Times the reduction in variability reduces the average queue size by a factor of (1 -x/2).|
|- Standard Deviation of queue size||Math.sqrt( (x*x*(1+x-(x*x)))/((1-x)*(1-x)) )||=|
|GEOMETRY and TRIGONOMETRY|
|sin( x° )||Math.sin( Math.PI*x/180 )||=|
|cos( x° )||Math.cos( Math.PI*x/180 )||=|
|tan( x° )||Math.tan( Math.PI*x/180 )||=|
|Hypotenuse of right angled triangle||Math.sqrt(x*x+y*y)||=|
|- other side if hypotenuse = x and side = y||Math.sqrt(x*x-y*y)||=|
|Other side of triangle sides x,y at angle z°||Math.sqrt(x*x+y*y+2*x*y*Math.cos(Math.PI*x/180))||=|
|Area of a triangle with base x and height y||x*y/2||=|
Cut the triangle into thin strips parallel to the base.|
The strips can be moved, without changing the area, to make, say, a triangle with a right angle at the base.
Two such triangle would make a rectangle, so the area of the triangle is base * height /2.
|Area of a triangle with sides x,y,z||s=(x+y+z)/2; Math.sqrt(s*(s-x)*(s-y)*(s-z))||=|
|Area of triangle with sides x,y at angle z°||x*y*Math.sin(Math.PI*z/180)/2||=|
|Area of a rectangle with sides x, y||x*y||=|
|Area of a trapezium parallel sides x,y height z||(x+y)*z/2||=|
Average width is (x+y)/2, so area is this times z.|
|Surface area of a box||2*(x*y +x*z +y*z)||=|
|Volume of a box||x*y*z||=|
|Volume of a pyramid base x,y height z||x*y*z/3||=|
Consider a pyramid with a sqare base. Cut it into many thin horizontal slices.|
These can be slid around to make a slanting or a symmetric pyramid, without altering the volume.
Six symmetric pyramids with height half the length of a side of the base, will fit together with their tips at the centre to form a cube.
The volume of each is therefore base area*height/3.
The volume of a pyramid with a given base is proportional to the height (e.g. doubling each of the thin slices will double the height and the volume).
The volume of any pyramid with a square base is therefore base area*height/3.
But a pyramid with any shape of base can be built up from a large number of pyramids with small square bases.
The volume of any pyramid is therefore base area*height/3.
For instance, the volume of a cone is πr²h/3.
|Area of a circle radius x||x*x*Math.PI||=|
Divide the circle into many small thin triangles with bases on the circumference and tip at the centre.|
The area of each is ½ *base *radius (r), and the sum of the base lengths is 2πr.
So the area of the circle is πr².
|Area of a circle diameter x||x*x*Math.PI/4||=|
|Area of an ellipse of width x and height y||x*y*Math.PI/4||=|
An ellipse with width 2a and breadth 2b is just a circle with unit radius on a plane that has been stretched
a times in one direction and b times in the other. The area of anything on the plane is therefore increased by a*b,|
and the stretched 'unit circle' then has area πab.
|Surface area of a sphere radius x||x*x*Math.PI*4||=|
(Alternatively, forget the cylinder: the surface area of each strip is 2πr times the width of the strip.
The sum of all such widths is 2r, so the area is 4πr².)
|Volume of a sphere radius x||x*x*x*Math.PI*4/3||=|
Divide the sphere into many small pyramids, with their tips at the centre.|
The volume of each pyramid is 1/3 * r * the area of the base (see above).
The sum of all the bases is the surface area of the sphere, 4πr² (see above),
so the volume is 4πr³/3.
|Volume of an ellipsoid width x depth y height z||x*y*z*Math.PI/6||=|
An ellipsoid with widths 2a, 2b, 2c is just a sphere with unit radius in a space that has been stretched
a times in one direction and b times in the another and c times in the third.|
The volume of anything on the space is therefore increased by a*b*c,
and the stretched 'unit sphere' has volume 4/3 πabc/2³ = πabc/6.
|Volume of a cone radius x height y||x*x*y*Math.PI/3||=|
See discussion above of the volume of a pyramid.|
|Surface of a cone radius x height y (inc base)||x*x*Math.PI +(Math.sqrt(x*x+y*y)*Math.PI*x)||=|
Divide the curved surface into many samll thin triangles with base at the bottom and top at the apex.|
The area of each is ½ * length of base * slanting height of the cone.
The total curved area is therefore ½ * total base length * slanting height = ½ *2πr *√(r² + h²).
The bottom circle adds a further πr².
|power. [N.B. x^y is "x XOR y"]||Math.pow(x,y)||=|
|First root of quadratic x*X*X +y*X +z = 0||(-y +Math.sqrt(y*y -4*x*z))/(2*x)||=|
|- Second root||(-y -Math.sqrt(y*y -4*x*z))/(2*x)||=|
|x to the power ? = y to the power z||z*Math.log(y)/Math.log(x)||=|
|Term z of of A, (A+y)x, (A+2y)x², ...||(A+((z-1)*y))*Math.pow(x,z-1)||=|
|- Sum of first z terms||(y*x*(1-Math.pow(x,z-1))/((1-x)*(1-x)))+(A-((A+(z-1)*y)*Math.pow(x,z)))/(1-x)||=|
|- Sum of first z terms if x=1||z*(A+(y*(z-1)/2))||=|
|- Sum to infinity||(A/(1-x))+((y*x)/((1-x)*(1-x)))||=|
|feet to metres||0.3048*x||=|
|metres to feet||3.2808399*x||=|
|inches to centimetres||2.54*x||=|
|centimetres to inches||0.393700787*x||=|
|miles to kilometres||1.609344*x||=|
|kilometres to miles||0.621371192*x||=|
|pounds to kilograms||0.45359237*x||=|
|kilograms to pounds||2.20462262*x||=|
|degrees to radians||Math.PI*x/180||=|
|radians to degrees||180*x/Math.PI||=|
|Centigrade to Fahrenheit||(x*9/5) +32||=|
|Fahrenheit to Centigrade||(x-32) *5/9||=|
|PHYSICS (A=Amp, F=Farad, J=Joule, m=metre, N=Newton, s=second )|
|Speed of light c m/s||(c =299792458)||=|
|Plank constant h J s||(h =6.62606896e-34)||=|
|reduced Plank constant h/2π J s||h /(2*Math.PI)||=|
|Magnetic constant (vacuum permeability) N/A/A||Math.PI*(4e-7)||=|
|Electric constant (vacuum permittivity) F/m||1/(Math.PI*(4e-7)*c*c)||=|
|Absolute zero (° Centigrade)||-273.15||=|
|Velocity if initial vely x, accn y, after time z||x +y*z||=|
|Distance if initial vely x, accn y, after time z||x*z +y*z*z/2||=|
|Velocity if initial vely x, accn y, distance z||Math.sqrt(x*x+2*y*z)||=|
v = u + at
v² = u² +2uat +a²t² = u² +2a(ut +½at²) = u² +2as
|Charge on the electron e Coulombs||1.602176487e-19||=|
|ASTRONOMY (m=metre, s=second )|
|Gravitational constant G cubic m/Kg/s/s||(G =6.67428e-11)||=|
|Acceleration due to Earth gravity g m/s/s||(g =9.80665)||=|
|mean Radius of the Earth m||6371000||=|
|equatorial Radius of the Earth m||6378137||=|
|polar Radius of the Earth m||6356752||=|
|Mass of the Earth Kg||5.9742e24||=|
|Radius of the Sun m||695500000||=|
|Mass of the Sun Kg||1.98892e30||=|
|average Distance from Earth to Sun m||1.496e11||=|
|Days per year||365.242199||=|
|Seconds per day||3600*24||=|
|Seconds per year||365.242199 *3600*24||=|
|Orbit radius if mass=x Kg orbital period=y s||Math.pow( (y*y*G*x)/(4*Math.PI*Math.PI), 1/3)||=|
In a small time dt it makes a small angle θ (in radians) at the centre.
The velocity changes in direction by the same angle (the velocity direction must change by 2π each revolution).
The change in velocity (towards the centre) is therefore vθ (if θ is small), and so the acceleration is vθ /dt.
But rθ = vdt, so the acceleration towards the centre is vθ/ (rθ /v) = v²/r.
If T is the time for one orbit, then T = 2πr /v, and so the acceleration is 4π²r /T.
By Newton's law of gravitation, the acceleration of the satellite is mG/r², where m is the mass of the planet or star.
So 4π²r /T² = mG/r², or 4π²r³ = mGT²
and T = √( 4π²r³ /mG ) and r = ³ √( r²Gm /4π² )
and v = √(Gm/r).
|Orbit period if mass=x Kg, orbit radius=y metres||Math.pow( (4*Math.PI*Math.PI*y*y*y)/(G*x), 1/2)||=|
|Velocity if mass=x Kg, orbit radius=y metres||Math.pow( (G*x)/y, 1/2)||=|
|Population of the Earth (aug 2009)||6780000000||=|
and World Clock.|
|Births per year (2009)||134200000||=|
|Deaths per year (2009)||57200000||=|
|Life expectancy (2009)||67||=|
and Wikipedia: Mortality Rate.|
|Population of the UK (2009)||61383000||=|
Population statistics are usually for mid-year to mid-year.|
See Office fo National Statistics: Population;,
and BBC News 24aug09: UK population now more than 61m.
Also The population of England and Wales, 1911-2001.
|UK population increase (2008 to 2009).||408000||=|
Growth rate 0.7%. An increase over previous years.|
See BBC News 24aug09: UK population now more than 61m.
Also see WikiAnswers.com and CIA World Factbook.
|UK net immigration (2008 to 2009)||118000||=|
2007 net immigration = 210000.|
|UK Births per year (2008)||791000||=|
Births in 2007 were 758000.|
See BBC News 24aug09: UK population now more than 61m,
|UK Deaths per year (2008)||580000||=|
See Office for National Statistics: Births and Deaths for England and Wales (509090)|
and General Register Office for Scotland: 2008 Births, Marriages and Deaths - Preliminary Annual Figures : Table P2 (55699)
and Mortality Statistics for Northern Ireland (2008) (14900).
But estimates vary. The CIA World Factbook puts the death rate at 10.02 per 1000, which would be about 615000.
|UK Life expectancy (Male) (2009)||76.5||=|
|UK Life expectancy (Female) (2009)||81.6||=|
|Euler's constant (e)||Math.E||=|
|Natural logarithm of 10||Math.LN10||=|
|Natural logarithm of 2||Math.LN2||=|
|Base 10 logarithm of e||Math.LOG10E||=|
|Base 2 logarithm of e||Math.LOG2E||=|
|1 / square root of 2||Math.SQRT1_2||=|
|square root of 2||Math.SQRT2||=|
|Arccosine (x from -1 to 1, answer in radians)||Math.acos( x )||=|
|Arcsine (x from -1 to 1, answer in radians)||Math.asin( x )||=|
|Arctangent (answer in radians)||Math.atan( x )||=|
|Arctangent of (x * y) (answer in radians)||Math.atan2( x, y )||=|
|Integer greater than or equal to x||Math.ceil( x )||=|
|cos( x ) x in radians||Math.cos( x )||=|
|e to the power x||Math.exp( x )||=|
|Integer less than or equal to x||Math.floor( x )||=|
|Natural logarithm of x||Math.log( x )||=|
|Greater of x and y||Math.max( x, y )||=|
|Lessor of x and y||Math.min( x, y )||=|
|x to the power of y||Math.pow( x, y )||=|
|Random number between 0 and 1||Math.random()||=|
|x to the nearest integer||Math.round( x )||=|
|sin( x ) x in radians||Math.sin( x )||=|
|Square root||Math.sqrt( x )||=|
|tan( x ) x in radians||Math.tan( x )||=|
|Bitwise AND (if x&y fails)||~((~x)|(~y))||=|
|Left Shift x by y bits||x<< y||=|
|Left Shift x by y bits (if x<<y fails)||x*Math.pow(2,y)||=|
|Right Shift, sign bit propagated||x>>y||=|
|Right Shift, sign bit propagated (x>>y fails)||x/Math.pow(2,y)||=|
|Right Shift, 0s from left||x>>>y||=|
|Right Shift, 0s from left (x>>>y fails)||Math.abs(x)/Math.pow(2,y)||=|
|Cubic with roots 1,2,3||x*x*x -6*x*x +11*x -6||=|
|Cubic with roots near 1,2,3||x*x*x -6*x*x +11*x -5.8||=|
|First root of differential of cubic above||q=3;r=-12;s=11;(-r+Math.sqrt(r*r-4*q*s))/(2*q)||=|
|Second root of differential of cubic above||(-r-Math.sqrt(r*r-4*q*s))/(2*q)||=|
|< and > and & problem||(( x < y )&&( x > 0 ))?0:(2<<1)||=|
|Special characters used in help text||0||=|