In a previous post on *Different directions for different vectors* I gave this example, for which I’m switching North and East:

Suppose someone drives 30 miles

Northin 50 minutes, then turnsEastand drives 40 miles in 50 minutes. Overall, they have driven 70 miles in 100 minutes but as the crow flies they ended up 50 miles from where they started (30^{2}+ 40^{2}= 50^{2}). And a crow flying at the same speeds would have taken only 71 minutes to get there (50^{2}+ 50^{2}= 71^{2}).

It may be surprising that the angle the crow should fly is different in space than in time. Actually, it is the same angle measured two ways: one by distance and the other by duration. Let’s work out the details:

The spatial angle for the crow is arctan(40/30) = 53 degrees clockwise from the North. The temporal angle for the crow is arctan(50/50) = 45 degrees. But 45 degrees of duration is equivalent to 53 degrees of distance in this case.

How does this work? The temporal angle is like a clock with hands: 45 degrees means for example that the minute hand has moved 60*45/360 = 7.5 minutes or the second hand has moved 7.5 seconds. Since this corresponds to 53 degrees in space, the rate is 53/7.5 = 7.1 spatial degrees per minute. Divide this by 60 to get 0.12 revolution per minute or 7.1 cycles per second, known as Hz.

So the crow’s angle is 53 degrees clockwise from North, which is equal to 7.1 spatial degrees per minute times 7.5 minutes. Or to be faster, it’s equal to 7.1 spatial degrees per second times 7.5 seconds. Or 7.5 * 53/45 = 8.8 seconds of a second hand.

In summary to convert a temporal angle to a spatial angle, multiply it times the angular conversion factor, i.e., the *frequency*, which is the ratio of the spatial and temporal angles.