A week ago, I visited my brother in San Francisco. One late afternoon, we took a hike up Mt. Davidson, where there is a tall cross, which cast an impressive shadow. We started wondering how fast that shadow was moving as the sun set.
Some parameters: the cross is 103 feet high from its base at 927 feet above sea level. It is at Latitude 37°44’14"N, Longitude 122°27’17"W. The date was June 11, 2009. For the sake of simplicity, let’s assume the sun was at 45 degrees to the horizon and that the land in front of it was completely flat.
I believe the length of the shadow would by 103 feet x cot(45 degrees), which is also 103 feet. I also think the sun sets at 15 degrees per hour, since it travels 360 degrees in 24 hours. However it doesn’t go straight down at that longitude, and I’m not sure how to factor that in. The derivative of cot(x) is -csc^2(x) so at 45% that means 2, (well, -2 really). That’s where I get stuck. How fast is that shadow going in miles per hour? Feel free to simplify the problem. I’m just looking for a rough estimate.
Here’s what it looks like on Google Maps. However, the shadow hit squarely on the dirt last week.
The shadow’s “speed” is not a constant. It varies depending upon the time of day. As the sun approaches the horizon, the speed gets faster and faster, limiting out at infinitely fast as the sun goes down.
That’s assuming the earth is flat. It’s not, so you have to consider the fact that it is round in the calculations, but that won’t change the variable speed bit.
The rate of change of the height of any celestial body above the horizon is not, and could never be, a constant and even if it was the rate of change of the legth of the shadow would not be a constant either.
You can go to this page and calculate the H of the sun on a certain place at any time (note it uses UT). From that you can easily calculate the length of the shadow. Use two different times and calculate the difference.
Well, that question didn’t really rate the enthusiasm I had thought it would. Funny, probability questions can go for pages and pages. Oh, well!
In any case, I’m certainly aware of basic geometry, so I do understand that the shadow accelerates as the sun goes down. That was why I said to assume the sun is at a 45 degree angle to the horizon, and computed the derivative at that point. I’m looking for the instantaneous velocity at that time.
I got stuck because I forgot those trig derivatives are only valid when you are using radians. So that means the sun is at PI/4 radians (45 degrees) and setting at PI/12 radians (15 degrees) per hour, or PI/720 radians per minute. That gives us:
Speed of shadow = 103 feet * (PI/720) radians / minute * 2 / radian = 0.9 feet per minute.
Well, you might have tried asking the question with all those qualifiers to make it clear.
But you still have an incorrect answer. Unless you are at the Equator, or the day is one of the Equinoxes, the sun is not travelling straight “up and down” as it sets. There is a lateral component, which is dependent upon where on the surface you are, and what day of the year it is. For example, at the North Pole, the sun isn’t going “up and down” at all, merely circling at a given speed. Between the poles and the Equator, the sun’s horizontal movement is reduced directly variable to the sine of the latitude. So in computing your instantaneous velocity, you would have to add the horizontal vector as well. And THAT cannot be computed without knowing the latitude and the day of the year.
@sailor No, no. I really enjoyed your link. You didn’t answer my question per se, but the link was interesting. If you had any “enthusiasm”, you would have computed an answer yourself. I’m not saying anyone is obligated to be enthusiastic about my question, just that I thought more people would be.
@DSYoungEsq I asked with every qualifier I could think and as clearly as I could make it. Note that if you want to take a crack at the lateral movement, then go for it! That’s what I was hoping for. If you look more closely, you’ll see I specified both the the latitude and the day of the year for exactly that purpose. I ignored it only to make the problem simpler while still reasonably accurate.
Anyway, with more research, I see the sun was setting at more like 12 degrees per hour at Mt. Davidson on June 11 because of the lateral movement that DS called out. So maybe 0.7 feet per minute is a better estimate, at least for the change in length of the shadow. I suppose the speed would have to include lateral movement as well, but I’m happy with that answer for now.
