Uncertainty in
Measurement: Determining π
Example
Here is a circle drawn on graph paper. The radius of the circle is 5 units (from the center out to the circle). NOTE: You should have a circle covering many more squares on your piece of graph paper. We will look at just the upper right quadrant of the circle, where some of the boxes are grayed. In the upper right quadrant there are 15 boxes completely inside the circle. Multiply this by 4 to get the number of boxes within the circle, which we call N1:
N1 = 4 x 15 = 60.
This area, A1 = 60 units, is a little less than the area of the circle.
The number N2 is the number of boxes that have any part inside the circle. In our quadrant, there are 7 more boxes that have some part inside the circle, for a total of 15 + 7 = 22 boxes. We multiply by 4 to get N2, the number of boxes that have some part inside the whole circle:
N2 = 4 x 22 = 88.
This area, A2 = 88 units, is a little more than the area of the circle.
Since the area of a circle is given by A = π r2, then the number π can be found by dividing
π
= A / r2.
Since A1 is a little smaller than the
area, A1/ r2 should be a little smaller than π (a lower bound), and since A2
is a little larger than the area of the circle, A2 / r2
should be a little bigger than π
(an upper bound). For these numbers,
with r = 5 units,
π1
= A1/r2 = 60 / (5)2 = 60/25 = 2.40
and
π2
= A2/r2 = 88/25 = 3.52.
Note that the actual value of 3.14159… is between
these two numbers, as expected.
The
average value of π is the average of 2.40
and 3.52:
π = (2.40 + 3.52)/2 = 2.96,
and the range is the difference between either 2.40
or 3.52 and the average (2.96):
range
= 2.96 – 2.40 = 0.56.
So the value of π found from these measurements would be
written as:
π = 2.96 +/- 0.56
For the part about measuring π using a quarter, CD, can, or something like
that, remember the circuference is the distance around the circle. Suppose we find the diameter of
a cup to be 9.0 cm and its circumference is measured to be 29.5 cm. Since
circumference = π x diameter,
we could find π to be:
π = circumference / diameter = 29.5 cm / 9.0 cm = 3.28.
Back to Top Video Home Page
Scale
Model of the Earth-Moon System Lab
Example
The example below will be worked out using Pluto
and its moon Charon. You will do similar
calculations using numbers you can find in you textbook for the Earth and our
Moon. From pages A-7 and A-9 in your text’s appendix, we find Pluto’s radius is
1151 km, Charon’s radius is 596 km, and Charon is 19.7x103 km =
19,700 km from Pluto. The ratio of the
volumes of Pluto to Charon would be:
VPluto/VCharon
= (r Pluto) 3 / (r Charon) 3 =
(1151) 3 / (596) 3 = (1,524,845,951) / (211,708,736)
= 7.2026.
Rounding this ratio (7.2026) off to the nearest
whole number gives N = 7. That tells us
that Pluto’s volume is about 7 times greater than Charon’s. (The number will be larger for the Earth and
Moon, but it will be calculated the same way.)
So
now you will get your clay and make N + 1 = 8 balls of the same size. One of those balls represents the volume of
Charon. The other 7 are the volume of
Pluto. Combine these 7 into a single
ball, and you should have two spheres whose size and volume compare the same
way Pluto and Charon do.
The
next part of the lab is to set the spacing between the balls to be the same
scale as between the actual moon and planet.
For Pluto, the circumference, C, is
C
= 2 π r
Pluto = 2 x 3.14159… x 1151 km = 7,232 km (rounded off).
The ratio of the separation distance, D to the
circumference is called “M,” and it equals:
M
= D / C = 19,700 km / 7232 km = 2.72.
So now wrap a string around the “Pluto” ball 2.72
times (about 2 ¾ times) and cut it or mark it.
Put the moon at one end of the string and the planet at the other, and
you have the two located at the proper distance.
Compare
the angular sizes of the Moon model with the Moon
Let’s
use some “made-up” numbers for this part.
(Your numbers should give quite different results for the angle.) Suppose your clay ball for the Moon had a
diameter (which you would have to measure) of 2.4 cm, and when you did the bit
with the string, the length of the string was 28.5 cm. The angular size of your Moon (in units of
radians) would be:
Angle
= D/M = 2.4 cm / 28.5 cm = 0.0842 radians.
As it says near the bottom of page two, 180° = π radians, so we then convert to degrees:
Angle
= 0.842 radians x (180°) / 3.14159… = 4.82°.
You would repeat this calculation with the measurements you make using the real Moon and the ruler.