# UnitMath Examples ( Miscellaneous ) Last Modified: 2/3/2000

### email Examples FAQ Home Sales \$\$\$ What's New Windows demo ### Area of a Quadrilateral TOP

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 ### Overlap of Three Circles TOP

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;

Where
areaSector: 1/6 * pi sq radius;
areaTriangle: 1/2 base * height;
height: radius sin( 60 deg );

so

Area: 1/2 radius^2 * ( pi - 2 * sin( 60 deg ) );

The overlap is 22.4 % of the circles area.

Examples

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 %

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 % ### Car's Turn Angle TOP

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 );

### Examples

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

### Perfect Water Temperature TOP 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.

### Spherical Tank Volume TOP 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:
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.

### Tire Pressure TOP 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.

### Circle Gauge TOP

The following figure shows how to measure a hole with standard gauges that are smaller then the test hole. 