Practical Formulas
RN_Renee2003 (Missouri.
2002-07-08)
How is a Celsius temperature converted to the
Fahrenheit scale?
If C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit,
then the following exact relation holds:
F + 40 = 1.8 ( C + 40 )
This allows an immediate answer to the trivia question:
"What temperature is the same in both scales, Fahrenheit and Celsius?"
(Answer: "40 below").
More importantly, it's useful to remember the conversion in the form given above,
because the numbers involved are easy to manipulate mentally:
- To multiply by 1.8 (from C to F), you multiply by 2 (easy)
and subtract 10% from the result (almost as easy and perfectly accurate).
- To divide by 1.8 (from F to C), you divide by 2 and add 10% to the result.
(Well, 11% is better, since the exact percentage is 11.11111...%)
For example, to convert 20°C you double 20+40 which gives you 120,
subtract 12 (that's 10%) to obtain 108,
and finally subtract 40 to obtain 68°F.
The result is exact and not much more difficult to obtain
than the dubious approximations given by overly "simplified" formulas...
With a simple calculator, exact conversions are performed either way
in only 5 keystrokes, since it's easy to account for the simple 40° translation
(the same in both scales) as you enter or read the data.
| °C |
-40 | -35 | -30 | -25 | -20 | -15 | -10 |
-5 | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
35 | 37 | 40 | 100 |
|---|
| °F |
-40 | -31 | -22 | -13 | -4 | 5 | 14 |
23 | 32 | 41 | 50 | 59 | 68 | 77 | 86 |
95 | 98.6 | 104 | 212 |
|---|
Alternately, you may memorize [part of] the above table and notice that
an increase of 5°C corresponds exactly to an increase of 9°F...
(S. P. of Piscataway, NJ.
2000-07-15)
Can you determine the speed of a car by knowing the rpm of the motor,
the gear [ratio] and tire [diameter]? If so, what is the formula? [...]
If D is the diameter of the tires, the car moves forward a distance
pD with each revolution of the tires.
If there are F such revolutions per unit of time, it moves forward at a speed of
pD´F per unit of time.
If D is in inches and F is in revolutions per minute (rpm),
the above result will thus be in inches per minute!
Now, there are 63360 inches in a mile and 60 minutes in an hour,
so one inch per minute equals 60/63360 miles per hour or 1/1056 mph.
Therefore, we have:
V(mph) = D(in) ´ F(rpm) ´
(p/1056)
In the automotive industry, the coefficient 1056/p
is usually taken to be equal to 336
(it's actually 336.1352398...) and the formula corresponding to the above is memorized as:
(mph)(gear ratio)(336) = (tire diameter in inches)(rpm)
With the above numerical approximation, both formulas are identical if we consider that
(rpm)/(gear ratio) is the rate of rotation of the tires.
This means that the industry defines a "gear ratio" as the number of rotations
of the driveshaft for each turn of the tire.
This ratio is normally more than 1, except in overdrive gear.
So, calling R this "gear ratio", your final answer is:
V(mph) = D(in) ´ F(rpm)
´ (p/1056) / R
or, approximately:
V = (D ´ F) / (336 ´ R)
where V is the velocity (in mph), D the tire diameter (in inches), F the engine rpm,
and R the gear ratio (the number of engine revolutions for each turn of the tires).
(Carlos of Jersey City, NJ.
2000-11-15)
[...] If my car can do a 1/4 mile in 15.9 seconds at 94 mph,
how fast is it going at 0.1 or 1/10 of a mile?
It accelerates to 60 mph in 7.5 seconds.
What you are asking is to give a point on a curve (speed as a function of time)
about which we have preciously few properties.
(A total of 3 points --including the trivial 0s,0mph--
and the overall integral between 0 and 15.9 s, which is 1/4 mile.)
This is not enough to do a totally accurate job,
but we can make a fairly precise educated guess:
First, let's dispose of the issue of units.
If we measure times in seconds and speeds in mph,
the consistent unit of distance is then the mph-second (mph-s),
which is simply 1/3600 of a mile.
A quarter-mile is 900 of these units, a tenth of a mile is 360, and the distance from the
1/10 mile point to the 1/4 mile point is 540.
In particular, since a quarter-mile is 900 mph-s,
the entire integral of our speed vs. time curve is 900.
For example, let's try a curve of the type V = C(1-exp(-t / T)),
where V is the velocity and t is the ellapsed time (whereas C and T are some constants).
With such a speed curve, the distance traveled in time t to reach speed V is
Ct-VT = 900 for the quarter mile.
The 1/4 mile in 15.9 s at 94 mph imposes [trust me!]
the following parameters on such a curve (forgive the ludicrous precision):
T=12.639196110577s and c=131.32606505623 mph.
With this formula, you do a distance of 360
(that's 1/10 of a mile with our "weird" units)
in 9.3475 seconds at 68.64 mph.
That's our first approximative answer to your question.
Note, however, that this approximation would mean a speed of only 58.775 mph
is reached after 7.5 s or, alternately,
that 0-60 mph is achieved in 7.715 seconds.
Our crude approximation, is thus not too far from what your car is actually doing,
but you may not like the 0.2s and/or 1.2 mph discrepancy.
I don't think I am far off in guessing that your car would do the 1/10 of a mile
in 9.3 s at 69 mph but I could definitely use some raw data
(acceleration curves of other cars, say) to fine-tune the above.
I hope somebody will be able to provide
this type of information.
A more elementary approach is to consider that,
from the time t=7.5 s to the time t=15.9 s,
the speed increases roughly linearly from 60 mph to 94 mph,
at a steady rate of
(94-60) / (15.9-7.5) = 85/21, or about 4.04762 mph/s.
With this assumption, the speed V reached at a time t between 7.5 s and 15.9 s
is simply V=60+(85/21)(t-7.5).
If t is the time when the 1/10 mile point is reached,
the distance of 1/4-1/10 mile (or rather 540 mph-s with the "proper" unit
introduced above) is simply the area of a trapezoid, and we have:
(15.9-t)(V+94)/2 = 540.
All told, this means that t is a solution of the equation
540 = (15.9-t) [ 60+(85/21)(t-7.5) + 94 ] / 2
The nonnegative solution of this quadratic equation is about 9.18431 s,
corresponding to a speed V of about 66.8174485 mph.
In other words:
Your car does the 1/10 mile in 9.2 s at 67 mph.
Overall, this more elementary approach may very well give a better, more reliable
estimate, given the data at hand.
Hsuan Liu
(2002-05-06; e-mail)
With 200 hp applied to a 2350 lb vehicle,
what's the 0 to 60 mph time?
Answer: at least 2.57 seconds (see formula below), but this silly lower limit
would be the correct answer only if 200 hp was the actual
average mechanical power applied to the tires during acceleration
[assuming no skidding and deducting the energy lost to friction].
Such an acceleration is next to impossible with
current car technology:
The current record 0-60 mph time is 3.07 seconds; it was set by
a Ford RS200 Evolution (a 1986 "Group B" rally car)
at the Millbrook Proving Ground (UK), in May 1994.
Therefore, what you have in mind is most probably not this kind of theoretical
net power, but rather the rating most often
listed by manufacturers,
which is the maximum engine output.
The fastest production car is currently the McLaren F1 pictured at right,
whose 6.1 L engine is rated at 627 hp...
Here are a few actual examples which tell you what percentage [last column]
of the rated maximum power is the actual average net power propelling the car
over the 0 to 60 mph acceleration test whose duration is listed:
"Mass" is "Test Weight",
namely "Curb Weight" + 160 lb driver.
| Vehicle | Mass
(lb) | Engine max
Power (hp) |
0-60 mph
Time (s) |
0-60 mph
Power (hp) | Average
/ Max. |
|---|
| Doran
(gas version) | 1440 |
82 | 8.70 | 36.2 | 44.2% |
| Tiburon
GT V6 | 3183 |
181 | 7.60 | 91.6 | 50.6% |
| Jaguar
XJR | 4176 |
370 | 5.20 | 175.7 | 47.5% |
| Shelby
American 1 | 2810 |
320 | 4.40 | 139.7 | 43.7% |
| Ferrari
360 Modena | 3260 | 400 | 4.30 | 165.9 | 41.5% |
| McLaren
F1 6.1 | 2739 |
627 | 3.20 | 187.3 | 29.9% |
| Ford RS200 Evolution | 2760 | 600 | 3.07 | 196.7 | 32.8% |
| (1) | 2350 |
200 | 6.43 | 80.0 | 40.0% |
| (2) | 2350 |
200 | 5.88 | 87.5 | 43.8% |
| (3) | 2350 |
200 | 5.14 | 100.0 | 50.0% |
| (4) | 2350 |
Too much! | 2.57 | 200.0 | |
The last four lines of the above table give the answer to your question,
assuming the quoted power of 200 hp is one of the following:
(1) The rated maximum
power, if the average is 40% of that.
(2) The rated maximum
power, if we use the rule of thumb below.
(3) The rated maximum
power, if the average is 50% of that.
(4) The actual average power applied.
[Not a realistic assumption.]
Now, what's the formula and where does it come from?
Well, if M is the mass of the car, and V is the speed reached from a standing start
after applying an average power P for a time T, we have:
P T = ½ M V2
The work done over the time of acceleration (PT) is equal to the vehicle's kinetic
energy (½MV2, if we neglect the rotational energy of the wheels).
The above formula is valid as is if we use any consistent set of
units (for example, P in watts, T in seconds, M in kilograms,
and V in meters per second; 60 mph = 26.8224 m/s).
If, however, you insist on having P in horsepowers
(1 hp is exactly 745.699871582 W, or just about 745.7 W)
and M in pounds (1 lb is exactly 0.45359237 kg) at a speed of
60 mph (26.8224 m/s), the above formula becomes:
(P/hp)(T/s) / (M/lb) =
1072896 / 4903325 = 0.2188098892...
So, if (P/hp) is 200 and (M/lb) is 2350, the 0-60 time in seconds (T/s)
will be such that 200 (T/s) / 2350 = 0.2188....
In other words, T is 2.571016... s, as unrealistically advertised.
Divide this time by 0.4 to obtain T = 6.42754... s,
if the value of 200 hp reflects the maximum engine power and you estimate the
actual average net power to be only 0.4 = 40% of that.
Rule of Thumb:
If you live in the pounds-and-horsepowers world and want to memorize a simple
rule-of-thumb formula, just assume that the average power is about 43.76 % of the
rated power, so everything boils down to the following approximation:
(0-60 time in seconds) »
½ (test weight in lb) / (rated engine power in hp)
Recall that: test weight
=
curb weight + weight of the driver (160 lb)
Anonymous query via Google (2004-11-23)
Power to Speed Ratio
Horsepower to thrust conversion...
[at a given speed along the thrust]
The power P of a force (thrust) F applied to an object moving at
velocity V is:
P = F V
When thrust and velocity do not have the same direction, the two quantities on the
right-hand side must be considered to be vectors (which is why they're shown in bold)
and the product is understood to be a scalar product, or "dot product".
We call speed V the magnitude of the velocity vector V, whereas
the "speed along a certain direction" (v) is the projection of the velocity vector
on that direction.
In what follows, "v" is the speed along the direction of the thrust,
and q is the angle between thrust and velocity:
P = F V = F v
= F V cos(q)
The above equation is directly applicable if a consistent system of units
is used:
The proper SI units for power, thrust and speed are, respectively, the watt (W),
the newton (N) and the meter per second (m/s).
With any other mix of units, an extra numerical factor appears, which can be obtained
by introducing standard ratios equal to unity
(like 5280 ft / mile).
For example, using the horsepower (hp),
the pound-force (lbf) and the "mile per hour" (mph) as units, we have:
| (P/[hp]) [hp]/[W] | = |
(F/[lbf]) [lbf]/[N] (v/[mph]) [mph]/[m/s] |
| (P/[hp]) [550 ft lbf]/[J] | = |
(F/[lbf]) [lbf]/[N] (v/[mph]) [5280 ft / 3600 m] |
This boils down to:
|
| 375 (P/[hp]) | = |
(F/[lbf]) (v/[mph]) |
The thrust (in lbf) is thus 375 times the
power to speed ratio (in hp per mph).
The thrust is zero for a car at cruising speed
(or at top speed, which is cruising speed at full throttle):
The engine's power is then entirely used to overcome friction and air resistance,
leaving no net power (P) to accelerate the vehicle.
Ahsin Sattar
(2004-06-29; e-mail)
How does the volume [in cc] of an engine relate to its power [in hp]?
With a given technology,
an engine has to have a certain size to produce a given power.
Usually, only usable forms of power are considered:
mechanical or electrical power, but not power wasted as heat.
In internal combustion engines, the "size" of the engine is universally
expressed in terms of the maximum volume of its combustion chamber.
This is most commonly given in cubic centimeters (cc),
liters (1 L = 1000 cc) or
cubic inches (1 cu in = 16.387064 cc).
The power of an engine is best given in watts (W) but manufacturers most often
rate it in horsepowers, of which there are (unfortunately)
two flavors; the Imperial horsepower (hp)
is about 745.7 W whereas the "metric" horsepower (ch) is about 735.5 W.
The manufacturer's power rating corresponds to the maximum power output
at some optimal regime.
This regime is often stated using "rpm" as the frequency unit
("revolution per minute" = 1/60 Hz).
The power-to-volume ratio depends very much of the engineering and technology involved.
For example, the French legal upper limit for a "125 cc" motorcycle is 15 ch,
which corresponds to a ratio of 0.12 ch/cc, or precisely 88.25985 W/cc.
Typically, 125 cc "street" motorcycles are rated around 70 W/cc,
but this can be as low as 50 W/cc.
On the other hand, high-performance 125 cc competition engines have been rated
as high as 34 ch (200 W/cc)...
Mark Nance
(2005-05-10; e-mail)
What's the optimal gear ratio to maximize top speed on a flat road?
At uniform speed, no power is used to accelerate the car.
Instead, the engine's mechanical output (whatever is not directly wasted as heat)
is used to overcome mechanical resistance from the following sources:
- Powertrain between the engine and the gearbox.
- Powertrain between the gearbox and the wheels.
- Periodic deformation of the tires.
- Contact between the tires and the road.
- Air resistance (assuming no wind with respect to the road).
Without lubrication [!] the frictional forces for moving parts would
be roughly constant and the power lost to these would be proportional to how fast they
rub against each other, or against their bearings.
With good lubrication, such "raw" friction is greatly reduced
and what dominates (except at low speeds) are so-called viscous
forces of magnitude proportional to the speeds involved, entailing a smaller power loss
proportional to the square of that.
The relevant speeds in the powertrain are proportional either
to the engine's rpm (between engine and gearbox) or
to the car's speed (between gearbox and wheels).
For simplicity, we'll just assume that the power lost
inside the gearbox is approximately the sum of two components:
One of each of the two types presented above for the rest of the powertrain.
The deformation of the tires produces heat at a rate proportional to its
frequency: This power loss is thus proportional to the speed of the car
(similar to what would be unlubricated friction in the powertrain after the gearbox).
A rough idea of this effect's magnitude may be obtained by comparing
fuel efficiencies under controlled conditions, when the tires are properly
inflated and when they're not.
Air resistance is a complicated thing but it has two main components:
The viscous one dominates only at low speeds. The other component is essentially
a quadratic effect, involving a drag force
proportional to the square of the car's speed
(its power is thus proportional to the cube of the speed).
All told, let V be the top speed of the car for a gear ratio R, and P(x)
the power output [full throttle] of the engine at x rpm
(x = kRV for a constant k which is inversely proportional to the
diameter of the tires, as discussed above). We have
P( kRV ) =
a R2 V2 + b RV + c V 3 + d V 2 + e V
In this equation, the constant coefficients k,a,b,c,d,e are characteristics
of the vehicle (a, b and c don't depend on tire size).
For a given R, the vehicle's top speed is obtained by solving this equation.
At the optimal gear ratio R for which the top speed V is greatest, a minute
increment dR wouldn't change V (i.e., dV = 0)
so the derivatives of both sides with respect to R are equal.
Dividing both sides of the resulting equation by V, we obtain:
k P'( kRV ) = 2 a R V + b
Solving the above two equations simultaneously gives the optimal gear
ratio R, for which the greatest top speed V can be obtained.
This requires detailed knowledge of your engine (the function P)
and of both parts of the powertrain.
Not to mention the vehicle's aerodynamics
(the drag coefficient we called "c").
You may acquire the relevant knowledge with 5 actual measurements of the
top speed of the vehicle, under engine conditions that you can reproduce
on the bench (for example, full throttle with 5 different gears).
Reproducing the rpm and fuel intake on the bench allows you to measure
the actual engine output in these 5 cases.
This leaves you with 5 instances of our first equation,
which form a linear system
that yields easily the values of its 5 unknowns a,b,c,d,e
(the coefficient k being known from other considerations).
Once this is done, you still have to solve the two
nonlinear simultaneous equations mentioned above...
This requires only a plot of the P function near the expected solution,
from bench measurements.
On 2005-05-11, Mark Nance
wrote:
You are awesome! Thank you for [posting] that information for
myself and everyone.
I have donated to your website to help keep it going.
Great information. Thank you so very much!
|
Joshua
P. Gatcomb (2002-07-15; e-mail)
When I was in middle school, I found a way to derive the
Law of Cosines from
Heron's formula
[details attached, in 20 steps]...
Have you seen it done this way before?
Well, yes (in the backward sort of way described below).
Congratulations for rediscovering by yourself
this connection between the two formulas, though.
Heron's formula states that
A2 = s (s-a)(s-b)(s-c),
where A is the surface area of a triangle of sides a, b, c, and
of semiperimeter
s = ½ (a+b+c).
One quick way to derive Hero's formula is to start with the most commonly known expression
for the area A, namely: A = ½ ah,
where h = b sin g
is the height of the triangle (of base a)
and g is the angle between a and b.
We have:
| A2 | = |
(½ ab)2 sin2 g
= (½ ab)2 -
(½ ab cos g)2 |
| = | (½ ab)2
-
(¼ [ a2 + b2 -
c2 ] )2
[ Law of Cosines ] |
| = | (½ ab
-
¼ [ a2 + b2 -
c2 ] )
(½ ab +
¼ [ a2 + b2 -
c2 ] ) |
| = |
[¼ c2 -
¼ (a - b)2 ]
[¼ (a + b)2 -
¼ c2 ] |
| = |
[(s-a)
(s-b)]
[(s-c)
s ]
|
The whole thing may be used backwards:
If you assume Heron's formula, the last equality holds and the above elementary
manipulations thus prove the equality on the second
line without invoking prior knowledge of the Law of Cosines.
This, in turn, establishes the Law of Cosines with an ambiguity of sign
which must be lifted by other means (since we only equate the squares of two
quantities here, they could be either equal or opposite).
However, this would really be a backward derivation, because the
Law of Cosines is now considered far
more fundamental than Heron's Formula (a.k.a. Hero's Formula).
(2002-07-16) Brahmagupta's Formula
How does Heron's formula generalize to quadrilaterals?
Brahmagupta's formula states that
A2 = (s-a)(s-b)(s-c)(s-d),
where A is the surface area of a cyclic quadrilateral
(that's to say a quadrilateral inscribed in a circle)
of sides a, b, c, d, and of
semiperimeter
s = ½ (a+b+c+d).
This formula was devised around AD 620 by the Indian mathematician
Brahmagupta (598-670),
as a generalization of Heron's formula (d = 0).
For any quadrilateral, the formula can be generalized by introducing the
quantity q equal to the half-sum of two opposite angles
of the quadrilateral (in the case of a cyclic quadrilateral, this
q is a right angle, so that the second term does vanish):
A2 = (s-a)(s-b)(s-c)(s-d)
-
abcd (cos q)2
(2003-12-08; e-mail, name withheld )
Bretschneider's Formula
(4 A)2 =
4 p2 q2
-
(a2 - b2 +
c2 - d2 )2
Does the above relation give the area (A) of a quadrilateral in terms of
its sides (a,b,c,d)
and diagonals (p,q) without any restrictions?
The short answer is: Yes.
This formula was established in 1842,
by the German mathematician Carl Anton Bretschneider (1808-1878),
an author who earned a living as a high-school teacher in
the town of Gotha (Thüringen).
Bretschneider's Formula does hold for convex quadrilaterals
and concave chevrons, and it's also good for
[ the signed areas of ] crossed quadrilaterals.
A "crossed quadrilateral" is a butterfly-shaped figure
which appears as two triangles sharing a single point
(this point where two edges cross is not counted as a vertex).
Travelling along the 4 edges of such a quadrilateral,
one encircles a triangle clockwise and the other triangle counterclockwise.
The area of a crossed quadrilateral is
thus usually defined as the difference
between the areas of those two triangles.
Bretschneider's Formula does give [the absolute value of] this quantity.
Proof:
Let's call a and b
the angles from a diagonal (p) to each of the two adjacent sides a and b
(for convex quadrilaterals, these two angles have opposite signs,
but this need not be so for other quadrilaterals).
The area of the quadrilateral is either
the sum or the difference of the two triangles
of base p. More precisely: 
A = ½ p
| a sin a -
b sin b |
The Law of Cosines gives the 3 relations:
-
d 2 = a 2
+ p 2
-
2 a p cos a
-
c 2 = b 2
+ p 2
-
2 b p cos b
-
q 2 = a 2
+ b 2
-
2 a b cos (a-b)
Therefore:
4 p2 q2
-
(a 2 - b 2 +
c 2 - d 2 ) 2
=
4 p2 q2
-
( 2 b p cos b -
2 a p cos a ) 2
=
4 p2 [
a 2
+ b 2
-
2 a b cos (a-b)
-
( b cos b -
a cos a ) 2 ]
=
4 p2 [
a 2 sin2 a
+
b 2 sin2 b
-
2 a b sin a sin b ]
=
( 2 p [ a sin a -
b sin b ] ) 2
=
(4A) 2
| |
| A = ¼ |
Ö |
4 p2 q2
-
(a2 - b2 +
c2 - d 2 ) 2
| |
Remark:
For the record, Bretschneider's Formula
is the simplest way to express the area of a quadrilateral in terms of
[ the squares of ]
its sides and diagonals, but there are infinitely many other ways to do so,
since these 6 quantities are linked by the following polynomial relation.
( p is the diagonal joining the corner of a
and b to the corner of c and d,
while q is the other diagonal.)
p2 q2
[ a2 + b2 + c2 + d2
- p2 - q2 ] =
p2
(a2 - b2)(d2 - c2)
+
q2
(a2 - d2)(b2 - c2)
+
(a2 - b2 +
c2 - d2)(a2c2 -
b2d2)
This is only a quadratic equation with respect to
the square of each parameter.
Kristiana Kandere-Grzybowska
(2004-05-17)
Parabolic Segments
[In my postdoctoral work in cell biology, I need to estimate]
the area of a modified triangle with one convex side and two concave sides.
[...]
The simplest approach is to consider that such sides are
curved roughly in the shape of parabolas (possibly asymmetrically with
respect to the straight sides).
The surface between a parabolic arc and its chord is
called a parabolic segment. and has an area
equal to 2/3 of the length D of the chord multiplied by the "height" H
(the largest distance from the chord to a point on the arc).
To obtain the area of the "curved" triangle,
the above result for each of the three sides is simply added to,
or subtracted from, the area of the "straight" triangle.
Two millennia before "Calculus"...
The formula for the area of a parabolic segment predates "Calculus"
by almost two millennia.
The squaring of the parabola is attributed to Archimedes of Syracuse, who first showed in an
"elementary" way (using an intuitive notion of limit) that a parabolic segment has 2/3
the area of its circumscribed Archimedes triangle
(whose sides are the chord and the two extreme tangents).
In modern terms, the fact that the area of a parabolic segment depends only
on the chord length (D) and the height (H) is probably best memorized by considering
that all asymmetrical parabolic segments are obtained from a symmetric one,
using an [area-preserving]
linear shear transform about a direction parallel to the chord...
Archimedes
Triangle and the Squaring of the Parabola
schmeelke
(2002-01-27)
If I know the dimensions of a cylindrical tank on its side
[the axis of revolution is horizontal] and can measure the depth of the liquid inside,
how do I calculate the volume of liquid present?
-
Let R be the radius of the tank, L its length and H the height of the liquid in it.
Consider a (circular) vertical cross-section of the tank and call
q the angle (from the center O of the circle)
between the vertical and the line OF,
where F is one of the two points where the horizontal line representing the surface of
the liquid meets the circle representing the wall of the tank [see figure at right].
We clearly have R-H = R cos(q),
which means q is equal to
arccos(1-H/R)
and is thus readily obtained using the proper inverse trigonometric function
on a scientific caculator.
[Angle q must be expressed in radians
and is thus between 0 and p;
don't forget to multiply a result in degrees by p/180,
if applicable.]
The surface area corresponding to the liquid in that cross section is obtained
as the difference of areas between a circular sector (a pie portion) and a triangle, namely
q R2 -
(R-H)Ö(H(2R-H)).
Just multiply this area by the tank's length L to obtain the formula for the volume V
of the liquid in the tank, namely:
V = L [ R2 arccos(1-H/R) -
(R-H)Ö(H(2R-H)) ]
This formula is perfectly valid for the whole range of H (from 0 to 2R),
although the above "visual" explanation (involving the "difference of two areas")
assumed that H was less than R (tank at most half-full).
The formula could have been obtained symbolically without splitting cases.
I'll leave it up to you to "visualize" the other case
(where the area of a circular sector is to be added to that of a triangle of height H-R),
should you feel the urge to do so.
Follow up : What if the tank is spherical?
[Volume of a Spherical Cap]
zchas40 (2002-05-14)
[...]
How much water is 2 cm in a hemispherical bowl 3 cm in radius?
Rick94602 (2002-05-16)
[...]
What's the volume of the bottom part of a sphere?
The vertical cross section [pictured at right] is the same
as above.
The portion of the sphere involved is called a spherical cap and
its volume V is obtained as:
ó q
õ0 |
pR3
sin3a da
= |
ó R
õR-H |
p
( R2 - u2 ) du |
In other words,
V = p H2 ( R - H/3 ).
[ For the record, the surface area is
S = 2pRH. ]
So, if R is 3 cm and H is 2 cm, the volume of water in the "hemispherical bowl",
expressed in cm3
(cc or mL, same thing), is 28p/3; that's about 29.32 cc.
-
What about an ellipsoid of revolution on its side? [horizontal axis]
This will be useful to make short order of the next question...
We may observe that squeezing or stretching a spherical tank along any
horizontal direction turns the sphere into an ellipsoid with principal
radii R, R, and W (for some W).
Furthermore, for any height H of the liquid, the volume in the elliptical tank is
simply W/R times what it would be in the spherical tank.
In other words,
V = p H2 W ( 1 - H/3R ).
Note that stretching along two horizontal directions would be just as easy,
if you ever need to derive a more general formula for any elliptic tank, as long as
one of the principal axes of the ellipsoid is vertical.
For the record, if R is the vertical radius and W and L are the horizontal ones,
the general formula is simply:
V = p WL ( 1 - H/3R ) H2/ R
Daniel Trottier
(2002-04-26; email)
[...] A horizontal vessel consists of a cylinder equipped with bell
ends. These caps sealing the ends of the cylinder are elliptical in shape.
(Removing the cylinder middle section, you would have an ellipsoid of revolution.)
What is the volume of liquid in the vessel, as a function of the height of the liquid?
If L is the length of the cylindrical section of radius R, and W is the "width"
of the end caps [W would be equal to R if those caps were spherical],
then the volume V corresponding to
a height H of liquid is simply obtained by adding the two volumes described in the
two previous articles, namely:
V = L [ R2 arccos(1-H/R) -
(R-H)Ö(H(2R-H)) ] +
p H2 W ( 1 - H/3R )
As usual, we assume that the result of an inverse trigonometric function (like arccos)
is given in radians.
Do not forget to multiply a result in degrees by p/180,
whenever applicable.
(2003-12-08)
What's the volume of a truncated cylinder?
What's the volume of a [triangular] prism with tilted [nonparallel] bases?
The [American Heritage] dictionary defines a cylinder as
"the surface generated by a straight line
intersecting and moving along a closed planar curve".
When the planar curve is a polygon, this is also called a prism.
However, both names are more commonly used for
the solid bounded by two parallel planes and such a surface.
The two planar faces are called bases and the cylindrical surface
between the planes is called the lateral surface.
It turns out that a general expression can be given for the volume of such a solid
even when the
bases are not parallel to each other.
We only need to know the surface area (S) of a cross-section
[namely, the intersection with a plane perpendicular to the lateral direction]
and the average height (h) of such a solid.
The volume (V) of the solid is then simply given by:
V = h S
The only delicate part is to specify how the average height (h) is obtained.
Well, we only need to know the position of the centroid of the cross-section
(that would be the barycenter or center of mass of the surface if it was cut
from an homogeneous planar sheet).
This is particularly easy to find if the cross-section is triangular,
or if it is symmetrical enough
(circle, ellipse, regular polygon, parallelogram, etc.).
Draw a line parallel to the lateral direction through this centroid.
This line intersects the two bases at two points and the distance
between these two point is
the quantity h used in the above formula.
Of course, if either base is tilted by an angle q,
its surface area A is S/cos(q).
If the two bases are parallel to each other,
the distance d between their respective planes
happens to be h cos(q).
Thus we have
V = d A
This is a more commonly quoted formula, but it's only good for parallel bases...
 (2004-03-24)
Conical Frustum
What's the volume (V) of the part of a cone between
parallel bases of areas B and b ?
|
|
| V = (H/3) [ B + b +
Ö | Bb | ] |
In this, B and b are the surface areas
of the two homothetic planar
bases (which need not have any specific shape), whereas H is the height of the solid
(the distance between the two planes of the bases).
The above French cartoon by Gotlib
[text by René Goscinny] is from a Dingodossier
(crazy file) about the flu, featuring
an average student without a first name:
l'élève Chaprot.
Chaprot's ability to repeat this formula flawlessly is
presented as a sure symptom of the flu...
Comical Frustum:
The formula was no longer taught systematically to French schoolchildren during the
golden age of French comic strips (spearheaded by the weekly Pilote,
the magazine which published the above in 1967).
However, it had been a staple of French elementary mathematics at a time when
authors Goscinny (1926-1977)
and/or Gotlib
(Marcel Gotlieb, b.1934-07-14)
were schoolboys...
Incidentally, René Goscinny
was born in Paris but was educated in
the French School of Buenos Aires, where his father (d.1942) taught mathematics.
The French Lycée of Warsaw (Poland) has been named after René Goscinny,
in part because of his Polish ancestry:
He was the grandson of Rabbi Abraham Goscinny of Warsaw.
René Goscinny was also the grandson of Abraham Beresniak,
who authored a 1939 Yiddish/Hebrew dictionary published in Paris.
In spite of this, Yiddish is not
one of the 107 languages currently spoken by Astérix,
the most famous of the 2120 characters created by René Goscinny,
in 18 different series
(a total of over 500 000 000 comic books sold).
The Polish roots of René Goscinny were pointed out to us by the current (2005)
Consul General of France in Los Angeles,
Philippe Larrieu,
formerly stationed in Warsaw (1994-1998). We thank his Deputy,
Olivier Plançon,
for many fun discussions about French comic books of the golden age.
Astérix, le Juif?
(2004-07-23) Volume of a Sphere,
using Cavalieri's Principle (1635)
How did they compute volumes before the advent of Calculus?
Here is our version of what's called the [second] principle of Cavalieri.
(The first principle of Cavalieri deals with planar areas instead of volumes.)
If an horizontal plane always intersects two given solids
in
sections of equal areas, then the solids have equal volumes.
Two such solids are said to be Cavalieri congruent.
This ancient "principle" may sound trivial now
(the integrals of equal functions are equal)
but it helped define the notion of integral.
Some of its direct applications are still interesting.
For example, we can deduce the volume of a sphere from the formulas giving the volume
of 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:
p ( R 2
- z 2 )
This is also clearly the area of a ring with outside radius R and
inside radius z,
which corresponds to a solid that consists of a cylinder of radius R and height R from
which a cone of the same base and height is removed
(the apex of the cone being on the
equatorial 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 of
which Archimedes of Syracuse was most proud of:
Vsphere = (4p/3) R3
This was obtained two millenia before Cavalieri's principle got its name.
Srikanth (2004-07-17; email)
Wedge of a Cone
What's the volume of the portion of a circular cone
included between two half-planes?
In particular, when the intersection of the planes goes through
the cone's axis. [ As is the case for the red section
of the grey cone pictured at right,
seen from the direction (Ox) shared by the two half-planes. ]
This solid may be seen as the difference of two cones with the same apex:
- The larger cone is of height H. Its base is a half-circle of area
pR2 / 2.
- The height of the smaller cone is h = H cos q.
Its base is a segment of a conic section of area A
(a parabola when the plane's slope is H/R,
an ellipse for a less inclined plane, an hyperbola for a steeper one).
The volume (V) of the wedge thus reduces to the computation of the area A:
V = (H/3) [ pR2 / 2
- A cos(q) ]
With obvious choices of coordinate systems, a point of
coordinates (x,y) in the inclined plane has spatial coordinates
(x, y cos q, y sin q)...
Plug these into the cone's spatial equation to obtain the equation of
its intersection with the plane:
H2 ( x 2 + y 2 cos 2 q )
= R2 ( H - y sin q ) 2
The curve's apex (x=0, y>0) is at
y = r = [ (sin q) / H +
(cos q) / R ] -1
The segment's area A is thus given by either of the following expressions, where
x and y are positive quantities related to each other by the above equation:
| A =
2 | ó r
õ0 | x dy =
2 | ó R
õ0 | y dx |
For example, in the special case of a parabolic segment
(tan q = H/R) we have:
y = cos(q) (R2-x2) / 2R
and
A = (2R2/3) cos(q)
Since cos2(q) is
(1+H2/R2 )-1 in this parabolic case,
the volume boils down to:
V =
(HR2 / 18) [ 3p
- 4 / (1+H2/R2 ) ]
In the elliptic case (tan q < H/R)...
In the hyperbolic case (tan q > H/R)...
Generally, any conical wedge bounded by planar sections
of respective areas A0 and A1 [not necessarily joining
within 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 | h0 A0
- h1 A1 |
In this, the cone is understood to be single-sided [its surface is generated
by half-lines originating at the apex] but it need not be a circular one...
suzjor
(Riverside, CA. 2000-11-24)
It takes you 6 hours to do a job. It takes a friend 3 hours to do the same job.
How long would it take both of you working together to do the job?
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 nature
that it can be "distributed" efficiently.
This result is obtained by adding the rates [the number of jobs 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.
seriouslyman
(2000-11-25)
A clerk is assigned a job that she can complete in 8 hours.
After she has been working for 2 hours, another clerk,
who is able to do the job in 10 hours, is assigned to help her.
How long will it take to complete the job?
We assume that the "job" is of such a nature that it can be distributed between the
two 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 and had completed 1/3 of the job.
(V. V. of Emeryville, CA.
2000-11-11)
How much pure alcohol must a nurse add to 10 cubic centimeters of a 60% alcohol solution
to strengthen it to a 90% alcohol solution?
Alcohol solutions are rated by volume (and not by mass as with
metallic alloys and most other mixtures):
If X cc of pure alcohol are added to 10 cc of 60% alcohol,
the result is approximately of volume (X+10) cc
with an alcohol content of (X+6) cc.
This will be a 90% solution if (X+6)/(X+10) is 0.90.
Solve this for X and you have your answer:
The nurse should add X=30 cc to obtain 40 cc of a 90% solution
(36 cc of alcohol in it).
Footnote: We had to say "approximately" in the above because it's a well
known fact that when you mix X cc of water with Y cc of alcohol,
the volume of the mixture is actually less than (X+Y) cc.
The above approximation seems generally accepted, though.
(S. H. of Mays Landing, NJ.
2000-11-23
twice)
Let us say that you want to prepare an a % solution [of peroxide],
using 3% and 30% solutions.
In what proportion should you prepare the mix?
(Here, a is a given number, between 3 and 30.)
To get an a% solution from a strong solution rated at p% and a weak one at q%,
you should mix (p-a) parts of the weak one and
(a-q) parts of the strong 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 at a = 12%
by mixing 30-12 = 18 parts of the stronger solution with
12-3 = 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 in lowest terms).
dva1270
(2002-03-29)
Averages Galore...
A vehicle goes from A to B at 60 mph, and returns at 40 mph.
Why is the average rate of speed
equal to 48 mph? [not 50 mph]
That's because average speeds are harmonic, not arithmetic... Read on.
The average of several things is whatever single value you could replace all of these
things with, and still obtain the same effect.
Obviously, the way you would actually compute such an average may very well
depend on exactly which type of effect 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 about
is the total amount of money you are receiving.
In this case, it does make sense to consider that an average check is equal to
half the sum of both, because two such average checks would have exactly the
same effect on your bank account as the two checks you actually got.
This is called an arithmetic average and it's, by far, the most common form
of averaging.
It's clearly not the only one, though...
Harmonic Mean:
In the case of speeds, the important thing is how long [how much time]
it takes to accomplish a journey whose legs are traveled at different speeds.
In other words, the average speed has to be whatever uniform speed
would 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-called harmonic average of the speeds,
namely the speed whose reciprocal is the arithmetic average
of the reciprocals of the speeds involved.
Using the numerical example in the question, 48 is the harmonic average of
60 and 40, because:
1/48 = ( 1/60 + 1/40 ) / 2
Geometric Mean:
That's the averaging to use for successive relative increases:
Consider, for example, a first-year increase of
10% (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 1.1 ´ 1.21 ´ 8 = 10.648.
The yearly average is the cube root of that, namely
a factor of 2.2, corresponding to an average yearly increase of 120%.
The logarithm of a geometric average is the arithmetic 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 a
percentage difference is best discarded,
except for the input and output of data.
Quadratic Mean:
This is the [positive] quantity whose square is the arithmetic mean of
the quantities under consideration.
It's also commonly referred to root mean square (RMS),
since 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 to define
their temperature (which is classically proportional
to their average kinetic energy).
Also, the RMS value [over time] of the current through a pure resistor
equals the continuous current that would dissipate the same heat as the
varying current observed.
Hölder Mean:
The German mathematician
Otto
L. Hölder (1859-1937) investigated, for any given exponent p,
the quantity H whose p-th power is the arithmetic mean of the p-th powers
of the quantities under consideration.
(For arbitrary exponents,
this must be restricted to nonnegative quantities.)
H =
Hp (a1 , a2 , ... an )
=
[ (a1p + a2p + ... +
anp ) / n ] 1/p
-
Exponents 1, -1, 0 and 2 correspond respectively to the arithmetic, harmonic, geometric
and quadratic means mentioned above [in that order].
-
The case p = 0 for the geometric mean
is defined by continuity as the limit
of the p-exponent Hölder mean Hp,
when p tends to zero.
-
If p < 0 and any ai is zero,
then Hp = 0 (defined by continuity).
-
Hp(a,b)
H-p(a,b) = ab
For an infinite exponent,
the Hölder mean of a finite set of nonnegative quantities is defined as either
their maximum (p = +¥)
or their minimum (p = -¥).
Generalized Mean:
A bijective function f can be introduced so that the
following relation defines the "mean" m of n quantities
a1, a2 ... an:
f(m) =
(1/n) [ f(a1) + f(a2) + ... +
f(an) ]
f(x) is x for the ordinary arithmetic mean,
x2 for the quadratic mean,
1/x for the harmonic mean,
Log(x) for the geometric mean [only when x is positive], etc.
Batting Averages:
...Farey Series, etc.
(2003-01-08)
30.436875 or 30.458729474253406983...
What's the average number of days in a
Gregorian month?
This is a good way to expand the discussion in the
previous article.
The average or mean value of anything is often ultimately defined
as the "expected value" of that thing according to some probability distribution.
For most pratical purposes, the distribution to use is clearly understood and
may be left unspecified.
For example, when asked out of context about the average of
28, 29, 30 and 31, we would normally assume that these 4 numbers
are equally likely and would readily equate their mean (their expected value)
with their arithmetical average (namely 29½).
On the other hand, this approach is clearly unacceptable if we have to work out
the mean number of days in a month, since durations of 28 or 29 days
should 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 period
of 400 years which includes 97 leaps years of 366 days and 303 regular years
of 365 days.
Therefore, this period consists of 146097 days, or 4800 months.
The average number of days in a month would thus seem to be
146097/4800, 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 each month is equally likely, the above is indeed the correct answer.
However, it would be at least as reasonable to assume that what you really want is the
average duration of the current month for a random Gregorian date.
In this case, each day is equally likely in the Gregorian cycle of 146097 days and the
result 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 per
leap year), 303 months of 28 days (1 per regular year), 1600 months of 30
days (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 falling
in 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 their
respective probabilities.
For example, the value 29 is weighted by the
probability 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 question
more "natural", but they would still acknowledge that other interpretations
are better adapted to specific contexts.
For example, our earlier value of 30.436875 days is the only
acceptable conversion factor between durations expressed in days and
durations expressed in average calendar months;
any other number leads to a systematic bias for long durations...
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 your line of sight
is tangent to the sphere at that distance. See figure at right.
In other words, you have a right triangle whose vertices consist of your own position
[your eyes], a point on the horizon, and the center of the Earth.
Its hypotenuse is R+H and its sides are R and D.
Therefore, (R+H)2 = R2 + D2,
which boils down to (2R+H)H = D2.
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 » Ö | 2RH |
In this, R is the (conventional) radius of the Earth:
R = 6371000 m or R = 20902231 ft
(choose whichever unit of length you use to express H
in 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 a fourth of that [more precisely 22.455%] and you'll have the distance to
the horizon in statute miles
(to have the result in nautical miles, add a sixteenth instead, or
6.41% 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 times
the square root of the altitude [of your eyes] in meters.
(A more precise coefficient is 3.5696.)
(R. S. of Austin, TX.
2000-10-25)
How can I find the distance (in feet) between two exact locations by
entering decimal lat/long coordinates to 6 place values?
lancelizabeth
(2001-07-11)
Using latitude and longitude, what is the formula that calculates
the distance between two points on Earth?
Assuming the Earth is a sphere of radius R (which is not totally true),
a point of latitude A and longitude B has the following Cartesian coordinates
in the proper coordinate system:
[x, y, z] =
[ R cos(A) cos(B) , R cos(A) sin(B) , R sin(A) ]
In this, a northern latitude is positive, a southern latitude is negative.
An eastern longitude is positive and a western one is negative.
(You could use any other convention, provided you do so consistently.)
If you have two such points, the scalar product ("dot product") of two such vectors
gives you R2 times the cosine of the angle D between them
(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 (A,B) and (A',B') are separated by an angle D given by:
cos(D) = cos(A)cos(B)cos(A')cos(B') + cos(A)sin(B)cos(A')sin(B') +
sin(A)sin(A')
which may also be written:
cos(D) = cos(A) cos(A') cos(B-B') + sin(A) sin(A')
From the above cosine of D, you may obtain the angle D by using
the inverse trigonometric function on a scientific calculator.
[However, you may prefer to use the better formula below,
which remains accurate for nearby points!]
If D is expressed in radians, the distance between the two points along a great circle
is simply equal to DR.
On the other hand, if D is expressed in degrees, the distance is
pDR/180.
The unit for the result is whatever unit you use for R.
The best value to use in this formula is the conventional radius of the Earth
(which is very close to the average distance from the Earth's center
to a point "at sea level"), namely R = 6371000 m.
If you want your result in feet, use R = 20902231 ft.
A Better Formula (especially for small distances):
The above formula is not well-suited numerically
to the computations of distances between
nearby points, because D is small and cos(D) is so close to unity
that the use of the reverse trigonometric function arccos
will 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 (A-A') and (B-B')
into a similarly accurate floating-point value for D:
sin2(D/2) =
sin2([A-A']/2) cos2([B-B']/2) +
sin2([B-B']/2) cos2([A+A']/2)
To obtain this relation from the above,
the identity cos(x) = 1 - 2 sin2 (x/2) may be used,
which allows the following transformation of our original formula:
| 1 - 2 sin2 (D/2) |
= cos(A) cos(A') { 1 - 2 sin2 ([B-B']/2) } + sin(A) sin(A') |
|
= cos(A-A') - 2 sin2 ([B-B']/2) cos(A) cos(A') |
|
= 1 - 2 sin2 ([A-A']/2) - 2 sin2
([B-B']/2) cos(A) cos(A') |
Therefore:
| sin2 (D/2) |
= sin2 ([A-A']/2) + sin2 ([B-B']/2) cos(A) cos(A') |
|
= sin2 ([A-A']/2) + sin2 ([B-B']/2)
{ ½ cos(A+A') + ½ cos(A-A') } |
|
= sin2 ([A-A']/2) + sin2 ([B-B']/2)
{ cos2 ([A+A']/2) - sin2 ([A-A']/2) } |
-
This last expression is easily recast into the "better formula" advertised above.
(Heather of Canada.
2000-11-06)
About geodetic latitude...
How do I find the radius of the Earth at 51° North?
Since I am not sure what quantity you are looking for,
I'll compute two for you:
- The distance from the center of the Earth to a point at a
latitude L of 51°.
- The radius of that latitude's parallel.
Before we do a precise computation, let's do the usual rough one by considering
that the Earth is a perfect sphere whose radius is the "conventional" radius
which is defined to be exactly R=6371000m. With this approximation,
we have:
- A (constant) distance to the center equal to R, namely 6371 km.
- 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 exactly
the regular shape against which its irregularities are charted.
That shape is called the "reference ellipsoid" and its dimension have been
precisely defined once and for all by the
IUGG in 1980:
The meridian is a perfect ellipse whose equatorial radius is exactly
a = 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 is not 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 the geodetic latitude:
the angle between your local vertical (which is the line perpendicular to the
ellipsoid's surface) and the equatorial 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 values
for a and b)
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):
- The distance to the center of the Earth is 6365264.58 m (that's about 5.7 km
less than the conventional radius of the Earth R).
- The radius of the parallel is x = 4022.031 km
(that's about 12.6 km more 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 points
on the globe, given by their latitudes and longitudes?
A basic result of spherical geometry (sometimes called Girard's Theorem)
states that the surface area of a spherical triangle on a sphere of radius R
is equal to R2(A+B+C-p) ,
where A, B and C are the inside angles of the triangle.
You could dissect your polygon into triangles and add up the results obtained from
Girard's formula, but there's a more practical way to proceed
(whose validity may be established using such a virtual dissection):
Consider the sum S
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 [ S - (n-2)p ]
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 of
inside angles and inside area; see footnote below).
Therefore, the only real problem is to compute the relevant angles from your
latitude/longitude list.
Well, on a sphere of unit radius, a point of latitude u and longitude v
has 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 directions
perpendicular 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 much bending at each corner and x will be close to
p).
Note, however, that the positive angles x and
(2p-x)
have the same cosine; you must choose whichever corresponds to the
inside of your polygonal line. That's it. I hope this helps...
Footnote:
The only consistent way to define which of the two possible angles is the inside
one is to state that it is always on the same side
(to the right-handside, say) with respect to the way you progress along the polygon line.
This is especially critical when the polygonal line self-intersects,
in which case the so-called inside angle may occasionally
face the outside area...
(2002-12-01) Kepler's Third Law
The relation between distances and periods of orbiting bodies.
Johannes Kepler (1571-1630) was the first to observe that the square of the period of
a planet around the Sun is proportional to the cube of the size of its orbit.
More precisely, the size of an elliptical orbit should be defined as its
major radius (a),
also called its semimajor axis (which happens to be equal
to the average distance of the planet to the Sun).
This is now known as Kepler's Third Law of planetary motion.
It occurred to Johannes Kepler on March 8, 1618; more than twelve years
after he had formulated his first two laws.
[He did not accept his own idea until better computations finally allowed him to formulate
the third law, on May 15, 1618.]
Kevin Brown
remarks that the introduction of logarithms
by John Napier (in 1614) is likely to have been instrumental in Kepler's
discovery of the third law...
Kepler discovered his first two laws in 1605 and published both in 1609:
The first law is that orbits are ellipses of which the Sun is a focal point,
whereas the second law states that a straight line from the Sun to the planet
"sweeps out equal areas in equal time intervals".
Oddly enough, the second law was actually discovered first...
Kepler's third law was later explained by Newtonian mechanics and may be more precisely
expressed in terms of the following relation, which involves Newton's constant of gravity (G),
the masses of the two bodies that orbit each other (M and m),
their average distance
(R = the major radius of the orbit of one body,
if the other is considered fixed)
and the period of revolution (T):
4 p2 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 in astronomical units and times in
sidereal years, which makes the relation boil down to:
R 3 = T 2
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