NOTE: The information used in this example is tied
to pages 8-10 of the chapter 8
practice test. The numbering of the problems has changed. Since a
different group of transmitters was used, the emission and delay readings
will be different than the one on your test.
Information given to you is in blue. Calculated numbers are in red.
Find the distance from Seneca, NY to Dana, IN.
- Subtract the coding delay for Dana from the emission delay for Dana to find the time it takes
for the signal to go from Dana to Seneca.
57,162.06 - 54,000.00 = 3,162.06 microseconds
- Multiply this time (in microseconds) by 0.186 (miles / microsecond) to get the number of
miles between Seneca, NY and Dana, IN. 3,162.06 ( 0.186 ) = 588.14316 miles
- Write this answer in part a.iii
- Take a ruler and measure the distance on the map from Seneca to Dana. For purposes of this
problem, let's say it turned out to be 76.2 mm. It's not, but that's what I'm going to call it.
- Divide the distance in miles by the distance in millimeters to get the scale of the map.
588.14136 / 76.2 = 7.718 miles/mm. You may wish to save this to Z on your calculator (I'm
choosing Z because this is the Z axis).
Finding the Equation of Z Hyperbola
Let us hypothetically say that the Loran-C dial reading is 56,204.12 microseconds. This is not
the problem from your test, but the process is similar. There is no 9960-Z transmitter section on
the test, but there is for the 9960-X and 9960-Y transmitters. The questions under part b are
duplicated for each secondary transmitter, so just follow these instructions and make the
- Since the dial reading ( 56,204.12 ) is less than the emission delay ( 57,162.06 ), the craft is
closer to the secondary transmitter (Dana, IN). If the dial reading had been more than the
emission delay, then the craft would have been closer to the master transmitter (Seneca, NY).
Circle the appropriate transmitter under part (1). In our example, we would circle the Z.
- The time difference between the transmitters is found by taking the difference between the
dial reading and the emission delay. Subtract the smaller value from the larger; you have
already determined which transmitter it is closer to, so you don't need to worry about the sign.
57,162.06 - 56,204.12 = 957.94 microseconds. Record this number in part (2).
- Since the transverse axis is between the transmitters, the length of the transverse axis is this
time difference converted into a mile distance. Convert microseconds into miles by
multiplying by 0.186. 957.94 ( 0.186 ) = 178.17684 miles. Record this value in part (3).
- The length of the transverse axis is 2a, so a is one-half the length of the transverse axis.
Divide the length of the transverse axis by 2. 178.17684 / 2 = 89.08842 miles. Record this
value in part (4).
- Each transmitter is located at a focus, so the distance between the foci is the distance between
the transmitters. 2c is the distance between the foci, so 2c is the distance between the
transmitters that we found in part a of the problem. To find c, divide the distance between
Seneca and Dana by 2. 588.14316 / 2 = 294.07158 miles. Record this value in part (5).
- For a hyperbola, a2+b2=c2, which means that b2=c2-a2 and b=sqrt(c2-a2). Normally, we would
need +/- when we take a square root, but b represents a distance here, so it has to be positive.
sqrt ( 294.071582
- 89.088422) = 280.25229. Record this value in part (6).
- Now comes the time to write the equation using Z as the transverse axis and Z' as the
conjugate axis. Normally, we would use x and y, but there are transmitters called X and Y
and I didn't want it to be confusing. The equation of the hyperbola will be (Z/a)2-(Z'/b)2=1.
Substituting in the values for a from part (4) and b from part (6), we get:
(Z/89.08842)2-(Z'/280.25229)2=1. Write this answer in part (7). Alternatively, you could go
ahead and square a and b and get Z2/7936.746578 - Z'2/78541.34605=1.
Sketching the Hyperbola
You only need to sketch one branch of the hyperbola, the branch corresponding to the transmitter
the craft is closer to. In our example, it would be the side closer to Dana.
The center of the map is where the Z and Z' axes meet (in Ohio). The Z axis is the transverse
axis, which contains the vertices (a) and foci (c). The Z' axis is the conjugate axis, which
contains the endpoints of the conjugate axis (b).
- Locate the vertices of the hyperbola on the map.
The vertices are a = 89.08842 miles from the center of the hyperbola, but we need to scale the
miles into millimeters. This is where the measurement on the map we took before comes into
play. Divide a by the scale to get the distance in millimeters. 89.08842 / 7.718 = 11.54 mm.
Locate the vertices, which are on the Z axis, 11.54 mm from the center. Locate both vertices
on the map using your ruler.
- Locate the endpoints of the conjugate axis on the map.
The endpoints are b = 280.25529 miles from the center. We have to convert that into
millimeters by dividing by the scale. 280.25529 / 7.718 = 36.31 mm. Locate the endpoints of
the conjugate axis, which are on the Z' axis, 36.31 mm from the center. Locate both
endpoints on the map using your ruler.
- Create a box so that the vertices and the endpoints of the conjugate axis represent the
midpoints of the sides of the box. The sides of the box that pass through the vertices will be
perpendicular to the transverse axis and the sides of the box that pass through the endpoints of
the conjugate axis will be parallel to the transverse axis.
- Draw the asymptotes through the corners of the box. The asymptotes should pass through the
center of the hyperbola, which is where the Z and Z' axes intersect.
- Draw the branch of the hyperbola that is closest to the proper transmitter. In this case, it is the
transmitter at Dana, not the one at Seneca. Remember that the hyperbola passes through the
vertex and the focus (at Dana) is within the curved portion of the graph. You can pick values
for Z and find the corresponding Z' values if you wish the graph to be more accurate.
Finding the Answer
Do the above steps for both the 9960-X and 9960-Y transmitters. Where they branches of the
hyperbola cross is the solution to the problem. Because you're graphing by hand, it is unlikely to
be exactly over a capital city, but it should be reasonably close. You may need a map of the US
if you don't know your capital cities.
I recommend that you figure the map scale separately for each transmitter. The projection of a
spherical earth onto a flat surface produces slightly varying scales. They will be close to each
The instructions say to sketch only one branch of the hyperbola. The figure
shown to the right has both branches shown. I used a computer package to
generate the hyperbola and generated both branches. To maintain the
symmetry when scaling the image to fit the map, I needed the center of the
image to be the center of the hyperbola, so I included both branches. Again,
you need only draw one branch of the hyperbola.