na3than

na3than t1_j8vlphq wrote

We can ... sort of. Using conventional sensors to build a 3-D map of the space you're in isn't all that hard. The challenge is linking the map of the space you're in to the next map you'll make after you move an imprecise distance in an imprecise direction. GPS doesn't work underground, a magnetic compass can be misled by local geology, and dead reckoning accumulates errors pretty quickly.

5

na3than t1_j6i4efy wrote

Before the invention of time zones, each locality decided what time it was by setting "noon" to the time of day when the sun rose to the highest point in the sky. That worked well enough when life moved more slowly.

After the railroads connected cities and it became important to know to the minute when something was scheduled to happen, a difference in "noon" between your town and mine made scheduling extremely difficult. Time zones were established so that people all over a region could agree that 8:00 is 8:00, even if that meant the sun reached the highest point in the sky at 12:04 in my town on the same day it reached the highest point at 11:44 in your town.

An agreement like that works well for people up to a few hundred miles apart. Much farther though, and you'll have to convince people in the far west portion of the continent that the sun rises at 3:00 a.m. on the same day that the people in the east say it rises at 6:00 a.m. So they divided the continent into "time zones", each 15 degrees of longitude wide (which is what you get when you divide the planet's 360° by its 24 hour rotation period) as a compromise between everyone-everywhere-agrees-to-use-the-same-clock and everyone-gets-to-use-whatever-clock-they-choose.

1

na3than t1_j2e5wro wrote

You ARE assuming the original triangle is a right triangle. If you take a triangle ABC, rotate it 180° and translate it so that AC coincides with the original CA, you're guaranteed to get a PARALLELOGRAM. You're only guaranteed to get a rectangle if the original triangle was a right triangle.

2

na3than t1_j01vdff wrote

I don't understand the question. The Earth's tides ARE affected by a mass at the distance of the Moon, namely, the Moon. If the Moon had more mass or less mass the tides on the Earth would be affected. A greater change in mass would mean a greater disruption, but ANY change in mass would affect the tides.

6

na3than t1_iuiog7j wrote

Because the calibration points for both scales are based on the state changes of water.

At standard pressure water melts (changes from solid to liquid) at 0°C / 32°F.

At standard pressure water boils (changes from liquid to gas) at 100°C / 212°F.

Since both scales are linear, 50°C, which is the midpoint between 0°C and 100°C, is equivalent to 122°F, the midpoint between 32°F and 212°F.

1