# Buzz Blog

## Ask a Physicist: Chasing the Sun

Friday, October 06, 2017

Colin, from Newfoundland, Canada wrote in this week to ask:

Colin,

Great question—and a fun idea for a roadtrip! Let's see if it's practical. There are a few ways to approach this problem; we'll look at two of them.

Standing still on the longest day of the year at the latitude of the Trans-Canadian Highway—about 50° from the equator—will net you 16 hours and 19 minutes of sunlight for starters. Let's be optimistic and take those 19 minutes for refueling, seeing as you're probably going to run down your tank at least twice, so you've got 16 hours of travel time. You've specified a speed, so that makes things relatively easy—how far can we get in 16 hours at 100 km/h? Obviously, 1600 km!

Assuming we're staying on the road and not going "as the crow flies", that's enough to get you from Whiteshell—a town along the Trans Canada Highway at the border of Manitoba and Ontario—to Canmore, just past Calgary.

Sunrise and sunset times at various longitudes and cities can be found just by googling, so let's look at Whiteshell vs. Canmore

We see here that we get an extra 29 minutes! Our original 16 hours and 19 minutes of sunlight is now 16:48. Here's where it gets interesting, though—if we're driving until sunset, the 16 hours of road-time we've calculated for isn't all we get; we can go another 29 minutes—which gets us almost 50 km farther, and buys us another few seconds of sunlight. You might wonder where this line of thinking ends—we could drive for those few extra seconds, and it would get us a few more microseconds—and the answer is that, technically, it doesn't! The problem is described by an infinite series. That doesn't mean we can drive forever, though; a series that's made up of an infinite number of ever-shrinking terms can still take on a finite value—this is known as the limit.

Hundreds of years ago, a young Isaac Newton was vexed with problems like this one. To address them, and to calculate things like the exact amount of time between sunrise and sunset for a westward traveler without computing an infinite number of terms, he developed calculus. Here, of course, the first few terms in the series—i.e. the 16 hours plus a second-order term to account for the daylight gained—provides a close enough approximation for our purposes.

Treating this explicitly as a calculus problem would require us to know the function describing the curviness of the road, which we don't have. However, we can get there by knowing a little algebra and making a simplifying assumption.

Say we're heading straight west, as the crow flies, rather than following the Trans-Canadian Highway. Now we can treat this like a "two trains leaving their stations" algebra problem, where your car is one train, and the "terminator"—the line of sunset—is the other.

On the longest day of the year at 50° latitude, there's 16 hours and 19 minutes of sunlight. This tells us that, at that latitude, it's daytime across 68% of Earth's surface at your latitude, with the remaining 32% in night—which tells us where the terminator is when the sun rises at your location. To figure out how fast the terminator is moving, we need to know the circumference of the earth at your latitude. Fortunately, this is pretty simple—just the cosine of your latitude, multiplied by Earth's circumference at the equator.

Now, we just need to set this up as an algebra problem, to figure out how much daylight we gain by traveling west at 100 km/hr. We've got a 17,516.8 km head start, and putting that into an equation looks something like this:

The Trans Canada Highway runs fairly straight. If I was to start at the border of Ontario and Manitoba and, as soon as the sun came up, began to drive west at 100km/h. How many more hours of sunlight would I be able to gain?

Colin,

Great question—and a fun idea for a roadtrip! Let's see if it's practical. There are a few ways to approach this problem; we'll look at two of them.

Standing still on the longest day of the year at the latitude of the Trans-Canadian Highway—about 50° from the equator—will net you 16 hours and 19 minutes of sunlight for starters. Let's be optimistic and take those 19 minutes for refueling, seeing as you're probably going to run down your tank at least twice, so you've got 16 hours of travel time. You've specified a speed, so that makes things relatively easy—how far can we get in 16 hours at 100 km/h? Obviously, 1600 km!

Assuming we're staying on the road and not going "as the crow flies", that's enough to get you from Whiteshell—a town along the Trans Canada Highway at the border of Manitoba and Ontario—to Canmore, just past Calgary.

Sunrise and sunset times at various longitudes and cities can be found just by googling, so let's look at Whiteshell vs. Canmore

Hundreds of years ago, a young Isaac Newton was vexed with problems like this one. To address them, and to calculate things like the exact amount of time between sunrise and sunset for a westward traveler without computing an infinite number of terms, he developed calculus. Here, of course, the first few terms in the series—i.e. the 16 hours plus a second-order term to account for the daylight gained—provides a close enough approximation for our purposes.

Treating this explicitly as a calculus problem would require us to know the function describing the curviness of the road, which we don't have. However, we can get there by knowing a little algebra and making a simplifying assumption.

Say we're heading straight west, as the crow flies, rather than following the Trans-Canadian Highway. Now we can treat this like a "two trains leaving their stations" algebra problem, where your car is one train, and the "terminator"—the line of sunset—is the other.

Although if it helps to picture Arnold Schwarzenegger as the other train, I don't see why not. |

Dividing that number by 24 hours, since that's how long it takes the terminator to reach the same point on Earth's surface from one day to the next, gives us a linear speed of 1073.33 km/hr.

Now, to find out how much of a head start we have over the sunset line, we can multiply the circumference by 68%, or 0.68, to find out how far "behind" you the terminator is at sunrise—yielding a result of 17,516.8 km. Just to check our math, we can see how long it would take a line moving at 1073.3 km/hr to cover a distance of 17,516.8 km, and we find:

which is pretty exactly the length of the day we found earlier—a sign that we're on the right track.

taking out the units, to clean things up, we get:

which simplifies, after a step or two of algebra, to:

Dividing out both sides and simplifying, we get a surprisingly clean answer:

It's significantly more than you get driving on the road, almost remarkably so, but it checks out intuitively. Google's over-the-road distance measurements mean that, in addition to the road's lateral curvature—winding north and south rather than going straight east-to-west—there's also vertical distance to take into account; all the little hills and dips add up.

All in all, while this is a great question, you're not likely to gain enough sunlight in your day to make this roadtrip worth it; sixteen hours in the car with half an hour of stopping time doesn't sound like much fun. It's interesting to note, though, that a fast jet at the right latitude could keep pace with the sun, effectively outrunning the night indefinitely.

Thanks for writing in!

**—Stephen Skolnick**