I measured the length of the shadow cast by 108.3 cm stick (a mop) You can see shadows of the mop stick and mine!
I took this photo at 11:30 am.99-Feb. 28.
The temperature was 12.6 C.
I measured the length of the shadow cast by 108.3 cm stick (a mop)
You can see shadows of the mop stick and mine!
I took this photo at 11:30 am.99-Feb. 28.
The temperature was 12.6 C.
| Time(JST) | Shadow(cm) | Angle of sun=Arctan(length of stick/shadow) (deg) |
| 98-Dec. 12 11:50 12:29 | 169.6 175.5 | 32.6 31.7 |
| 98-Dec. 25 Merry Christmas! 12:00 | 175.3 | 31.7 |
| 99-Jan. 4 12:00 | 170.9 | 32.4 |
| 99-Feb. 2 12:00 | 137.8 | 38.2 |
| 99-Feb. 28. 12:00 | 102.2 | 46.6 |
| 99-Mar. 14. 12:00 | 91 | 50.0 |
| 99-Mar. 21. Vernal Equinox Day 12:00 | Rainy day | ???? |
| 99-Mar. 23. 12:00 | 70 | 57.0 |
| 99-Apr. 27. 12:00 | 47.1 | 66.5 |
| 99-May 30. 12:00 | 28.0 | 75.5 |
| 99-June 20. 12:00 | 27.0 | 76.0 |
| 99-July 07. 12:00 | 29.3 | 74.9 |
| 99-July 31. 12:00 | 40.0 | 69.7 |
| 99-Aug 31. 12:00 | 60.0 | 61.0 |
| 99-Sept. 23. Autumnal Equinox Day 12:00 | 80.0 | 53.5 |
Noon Shadows on the Equinox. Latitude= 90-Arc tan(length of stick/the length of shadow)= Arc tan (length of shadow/length of stick) At noon on equinox. Arc tan(70/108.3)=33.0 Here is N34 43 E135 20. (Japan) If you are at the equator on the equinox, your shadow would be zero. Arc tan of zero is zero. if you are at 45 degrees north, the shadow would be the length of the stick. Shadow / stick=1. Arc tan (1) = 45 degrees. at the north (or south) pole, the shadow would be infinite Arc tan(infinite/length of stick)=90 degrees. B. Polarized sky. I look at the blue sky through a peace of polaroid in evening. The light coming from southern sky is polarized. (Why?) Sun light is unpolarized. (Prove it please!) C. Measure (estimate) sun constant. The equipment isHow to use the equipment. 1. Pour water into the equipment. 2. Expose black side of the eqipment to the sunlight. 3. Measure exposure time (about 3-10 min.) and how much the temprature of the water rises. I shall give an example. Water weight is 48 g. Area of black side is 25cm^2
| Begin T1(C) | 6 min. after T2(C) | T2-T1(C) | |
| Water temperture | 20.0 | 22.9 | 2.9 |
Q=Heat coming from sun to the water= 2.9*48=139.2(cal) Divide Q by time and area of black side= 139.2 cal/ (6 min*25 cm^2) = 0.93 cal/min/cm^2. Science is not only gathering data but, it also stimulates your mind and enables you to enjoy yourself! D. One more topic. Home