Essential Math for Geopositioning

For the service assigned to your group, do the following:

• Visit the website.
• Test the service
• Find out all you can about the service.
• Determine the answer to these 6 questions.
• Submit your findings online and report your findings in the next class.

1. What the entity being tracked? How is it identified? (4pts)
2. How is the geo-position determined? (4pts)
3. What are the attributes of the core classes for a normalized GIS database? (4pts)
4. How is this data collected and verified? (4pts)
5. How is the data illustrated? (2pts)
6. Provide 3 quiz questions about this study. (2pts)

Service Parcel tracking

• 1.Parcels being processed and delivered. The parcel code is key

• 2.The GPS location of the service provider is retrieved from a static GIS database

• 3.Dataset:

  • Post offices: Code, lat, lng, service type
  • Parcels: Parcel id, weight, return address, recepient address, original post office
  • Parcel processing log: Parcel id, postal code of origin, parcel id, nature of service, office lat,lng of service provider

• 4.QR code scan used to record the next entry into the database

• 5.Date and time and location and nature of last service

• 6.Since the postal systems does not use active GPS sensors, how do they track the position of packages?

par(mar=c(3,3,3,3))
r = seq(0,2*pi,0.01)
plot(cos(r),sin(r))

\[ 2 \pi\ \hbox{radians} = 360\ \hbox{degrees} \]

• Equation:

  \[ c = \sqrt{a^2 + b^2} \]

• Measured:

  \[ a=25; b=49; c=55 \]

• Calculated

  \[ c = \sqrt{25^2 + 49^2} = 55.00909 \]

• \( \sin(A) = \frac{a}{c} = \cos(b) \)
• \( \sin(B) = \frac{b}{c} = \cos(a) \)
• \( \tan(A) = \frac{a}{b} \)

• \( a=25 \)
• \( b=49 \)
• \( c=55 \)
• \( C = 90 \)

• \( A = asin(\frac{a}{c}) = asin(\frac{25}{55}) = asin(0.4545455) = 27.04 \)
• \( A = acos(\frac{b}{c}) = acos(\frac{49}{55}) = acos(0.8909091) = 27.01 \)
• \( A = atan(\frac{a}{b}) = atan(\frac{25}{49}) = atan(0.5102041) = 27.03 \)
• \( B = asin(\frac{b}{c}) = asin(\frac{49}{55}) = asin(0.8909091) = 62.99 \)
• \( B = acos(\frac{a}{c}) = acos(\frac{25}{55}) = acos(0.4545455) = 62.96 \)
• \( B = atan(\frac{b}{c}) = atan(\frac{49}{25}) = atan(1.9600000) = 62.97 \)

Given:

• \( A + B + C = 180 \)

• \( C = 90 \)

Calculated:

• \( A = 27.03\pm 0.02 \)

• \( B = 62.97\pm 0.02 \)

\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]

• \( \frac{a}{\sin(A)} = \frac{450}{\sin(26)} = \frac{443}{0.4383711} = 1010.56 \)
• \( \frac{b}{\sin(B)} = \frac{393}{\sin(23)} = \frac{393}{0.3907311} = 1005.81 \)
• \( \frac{c}{\sin(C)} = \frac{766}{\sin(131)} = \frac{766}{0.7547096} = 1014.96 \)

Haversine Function

$$Dist = 2r\ asin\left(\sqrt{\sin^{2\left(\frac{\psi_2} - \psi_1}{2}\right) + \cos(\psi_1) \cos(\psi_2)\sin^{2\left(\frac{\lamda_2} - \lamda_1}{2}\right)}\right)

• \( \psi_1,\ \psi_2 \) - lattitude of \( P \) and \( Q \), respectively

• \( \lamda_1,\ \lamda_2 \) - longitude of \( P \) and \( Q \), respectively

• \( r \) = Between 6,335.439 km (Poles) and 6,378.137 km (at equator)