1. PRIMITIVE NAVIGATION PROJECT By Alper Can Yildirim

2. Aim : To find the latitude and longitude of Cambridge by tracking the shadows of a stick during different time periods of the same day.

3. Equipments Foot long stick CfA’s roof A digital watch Tape Measure

6.  a straight wooden stick. The shadow of the stick The surface

7. How do we calculate θ ? tan θ = length of the stick / length of the shadow So; θ = tan -1 ( L stick / L shadow ) tan θ = length of the stick / length of the shadow So; θ = tan -1 ( L stick / L shadow )

8. Military TimeTime (hours) Shadow Length (feet) TanθΘ (degrees) 9:129.23.120.32117.8 9:309.52.790.35819.7 10:0610.12.380.42022.8 10:3010.52.290.43723.6 11:0611.12.120.47125.2 12:0612.12.160.46224.8 13:0613.12.430.41222.4 13:4213.72.810.35619.6 14:3614.63.980.25114.1 15:0615.15.920.1699.6 15:1815.37.400.1357.7

9. Vertical Angle vs. Time of the Day Vertical AngleVertical Angle Time of the day (hours)

10. Local Area Noon (LAN) and Meridian Height (MH) LAN MHMH

11. Taking the derivative y = -1.286x 2 + 29.82x – 147.4 y’ = 2*(-1.286) x + 29.82 = 0 (setting it equal to zero to find the x and y coordinates of the vertex) x = LAN = 11.59 hours  11:35 AM Y= MH = 25.49°

12. Due South Latitude = 90 o +Decl-Meridian height Meridian height Latitude from meridian height of sun

13. Latitude of Cambridge Latitude experiment = 90° - 21.87° - 25.49° = 42.64° Cambridge = 42.38° Error = 0.26° = 15.6’ Declination value for December 2 is taken from the website http://www.wsanford.com/~wsanford/exo/sundials/DE C_Sun.html http://www.wsanford.com/~wsanford/exo/sundials/DE C_Sun.html Declination of sun @ December 2 = -21°52’ = -21.87°

14. Conversion to UTC LAN = 11:35 LAN + 5 hours + EoT = UTC We need to find the EoT for December 2. from the equation of time graph.

16. Longitude of Cambridge The difference between UTC and 12:00 will give us the longitude. LAN + 5 hours + EoT = 11:35 + 5h + 10’= 16:45 16:45 – 12:00 = 4h 45’ = 4.75 h 1 h = 15° Therefore 4.75h = 71.25° Cambridge = 71.11° ERROR = 71.25°-71.11° = 0.14° = 8.4’