UnitMath Examples ( Miscellaneous )
Last Modified: 2/3/2000
Examples | FAQ | Home | Sales $$$ | What's New | Windows demo |
Find the area of the quadrilateral to the right.
Approach break the quadrilateral into two triangles along the diagonal and use Heron's formula to find the area of the two triangles.
a1: 102 ft;
b1: 230 ft;
c1: 282 ft;
s1: 1/2 ( a1 + b1 + c1 );
area1: sqrt( s1( s1 - a1) ( s1 - b1) ( s1 - c1 ) ) ~ 1,022.6 sq m
a2: 290 ft;
b2: 255 ft;
c2: 282 ft;
s2: 1/2 ( a2 + b2 + c2 );
area2: sqrt( s2( s2 - a2) ( s2 - b2) ( s2 - c2 ) ) ~ 3,030.9 sq m
area: area1 + area2 ~ 4,053.5 sq m
area as sq ft ~ 43,631.7 sq ft
find the area that three circles have in common if they all go throught the centers of each other.
The following drawing shows information that makes the solution quite clear.
Area: areaTriangle + 3 * areaShaded;
so
Area: 1/2 radius^2 * ( pi - 2 * sin( 60 deg ) );
The overlap is 22.4 % of the circles area.
Examples
radius: inch;
OverlapingArea: 1/2 radius^2 ( pi - 2 sin ( 60 deg ) );
circleArea:pi sq radius as sq in; ~ 3.1416 sq in
OverlapingArea as sq in; ~ 0.7048 sq in
OverlapingArea / circleArea as % ~ 22.4336 %
radius: mile;
OverlapingArea: 1/2 radius^2 ( pi - 2 sin ( 60 deg ) );
circleArea:pi sq radius as sq mile; ~ 3.1416 sq mile
OverlapingArea as sq mile; ~ 0.7048 sq mile
OverlapingArea / circleArea as % ~ 22.4336 %
Find that angle that a car truns given its wheel base (Wb), tire angle (Ta), and arc distance traveled (d).
radius: Wb / 2 / sin( Ta );
angleTraveled: d / radius;
Wb: 12 ft;
Ta: 20 deg;
d: 20 ft;
radius: Wb / 2 / sin( Ta ); ~ 5.35 m
radius as feet; ~ 17.54 feet
angleTraveled: d / radius; ~ 1.14
angleTraveled as deg ~ 65.32 deg
angleTraveled as circles ~ 0.18 circles
The same calculation again with resonable uncertainties.
Wb: 12 ft pm 4 inch;
Ta: (20 pm 2) deg;
d: 20 ft pm 6 inch;
radius: Wb / 2 / sin( Ta ); ~ ( 4.75 to 6.08 ) m
radius as feet; ~ ( 16. to 18. to 20. ) feet
angleTraveled: d / radius; ~ 0.98 to 1.14 to 1.32
angleTraveled as deg ~ ( 55.99 to 65.32 to 75.43 ) deg
angleTraveled as circles ~ 0.2 circles
Find the amounts of boiling and tap water to add to get a cup of 160 °F water ( my ideal temperature for tea).
We know that initial and final thermal energies must be the same ( Energy is conserved ) and that the amount of water is one cup. Form this information we can use the solve() function to find the correct amounts.
specificHeat: cal/cc/K;
EnergyStart: ( boilingAmount * 212 degF + tapWater (50 degF to 60 degF ) ) specificHeat;
Energyfinal: cup * specificHeat * 160 degF;
boilingAmount: ( 0 to 1 ) cup as cup;
tapAmount: ( 0 to 1 ) cup as cup;
solve
(
Energyfinal = EnergyStart;
boilingAmount + tapAmount = cup;
boilingAmount: tsp; tapAmount: tsp
) ~ boilingAmount: ( 0.65 to 0.69 ) cup; tapAmount: ( 0.31 to 0.35 ) cup;
It's always a good idea to check the results.
( 0.70 cup * specificHeat * 212 degF + 0.30 cup * specificHeat * (50 degF to 60 degF ) ) / specificHeat / cup ~ ( 157.5 to 172.4 ) degF
or
( 0.70 cup * 212 degF + 0.30 cup * (50 degF to 60 degF ) ) / cup ~ ( 157.5 to 172.4 ) degF
So fill the cup about 70% with boiling and the rest with tap water.
Find the volume of water in a spherical tank for two different "Air Gaps". With this information and the time it takes to change volumes find the flow rate. Finally, find the time to fill an empty tank.
The volume of a partially filled sphere is: "pi h^2 ( R- h/3)".
Where:
R: is the tank radius.
h: is the height of the water.
Also it's much easier to measure the daimeter of a tank than the radius, so define R in terms of D.
volume: pi h^2 ( R- h/3) as gal;
R: D/2;
h: D - air_gap;
D: 25 feet pm 3 inch;
air_gap: 22 inch pm 1/4 inch;
volume ~ ( 58,432. to 60,261. to 62,126. ) gal
air_gap: 94 inch pm 1/4 inch;
volume ~ ( 45,237. to 46,940. to 48,681. ) gal
time: ( 22 min pm 5 s);
(( 59,633. to 60,892. ) gal - ( 46,324. to 47,560. ) gal)/time as gal/s ~ ( 9.1 to 10.1 to 11.1 ) gal/s
4/3 pi R^3/( ( 9.1 to 10.1 to 11.1 ) gal/s ) as hour & min ~ ( hour + ( 28.8 to 41.0 to 56.1 ) min )
So the flow rate is about 10 gallons per second, and the whole tank can be filled or emptied at this rate in about an hour and a half.
Find the gauge pressure for a tire at 120°F that was filled to 32 psi (gauge) at 32°F.
The following calcultion uses the ideal gas law "Pressure Volume = n R Temperature". ( n is the amount of gas, R is the gas constant ) Assuming that the tire doesn't change size, we find "P2 = P1 * T2/T1"
Note P1 and P2 are absolute pressures = ( pressure of the atmosphere ) + (what you read on the pressure gauge).
P1: 32 psi + atm;
T1: 32°F;
T2: 120°F;
P1 * T2/T1 - atm ~ 40.4 psi
So a tire at 32 psi (gauge), when heated from 32 degF to 120 degF will have a (gauge) pressure will of 40.4 psi.
If the ambient pressure changes the equations will have to be modified to "P1: 32 psi + atm1 & P1 * T2/T1 - atm2". If my barometer is any indication then this correction is less than1.5 psi under "normal" conditions.
The following figure shows how to measure a hole with standard gauges that are smaller then the test hole.