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Posts Tagged ‘Tides’

At midnight on the night of Monday the 30th of June, an extra second was added to our clocks. A so-called leap second. Did you enjoy it? Me too 🙂 I got so much more done….. But, why do we have leap seconds?

In this blog here, I explained the difference between how long the Earth takes to rotate 360^{\circ} (the sidereal day) and how long it takes for the Sun to appear to go once around the Earth (the mean solar day). We set the length of our day, 24 hours, by the solar day. If there are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, then there should be 24 \times 60 \times 60 = 86,400 \text{ seconds} in a solar day. But, there aren’t! The Earth’s rotation is not consistent, that is if we measure the length of a mean solar day, it is not consistently 86,400 seconds. This difference is why we need leap seconds.

A leap second was added at midnight on the 30th of June. It was the first leap second to be added since 2012.

A leap second was added at midnight on the 30th of June. It was the first leap second to be added since 2012.


But, how do we accurately measure the mean solar day (the average time the Sun appears to take to go once around the sky) , and what is causing the length of the mean solar day to change?

How do we define a second?

When the second was first defined, it was defined so that there were 86,400 seconds in a mean solar day. But, since the 1950s, we have a very accurate method qof measuring time, atomic clocks. Using these incredibly accurate time pieces (the most accurate atomic clocks will be correct to 1 second over some tens of thousands of years) we have been able to see that the mean solar day varies. It varies in two ways, there is a gradual lengthening, but there are also random changes which can be either the Earth speeding up or slowing down its rotation.

How do we measure the Earth’s rotation so accurately

In order to measure the Earth’s rotation accurately we use the sidereal day, which is roughly four minutes shorter than the mean solar day. By definition, the sidereal day is the time it takes for a star to cross through a local meridian a second time. But, actually, stars in our Galaxy are not good for this as they are moving relative to our Sun. So, in fact, we use quasars, which are active galactic nuclei in the very distant Universe; and use radio telescopes to pinpoint their position.

The gradual slowing down of the Earth’s rotation

There is a gradual and unrelenting slowing down of the Earth’s rotation, which may or may not be greater than the random changes I am going to discuss below. This gradual slowing down is due to the Moon, or more specifically to the Moon’s tidal effects on the Earth. As you know, the Moon produces two high tides a day, and this bulge rotates as the Earth rotates. But, the Moon moves around the Earth much more slowly (a month), so the Moon pulls back on the bulge of the Earth, slowing it down. To conserve angular momentum, the Earth slowing down means the Moon moves further away from the Earth, about 3cm further away each year.

The random fluctuations in the Earth’s rotation

In addition to the unrelenting slowing down of the Earth’s rotation due to the Moon, there are also random changes in the Earth’s rotation. These can be due to all manner of things, including volcanoes and atmospheric pressure. These random fluctuations can either speed up or slow down the Earth’s rotation.

We have been having leap seconds since the 1970s when atomic clocks became accurate enough to measure the tiny changes in our planet’s rotation. Since them we have added a leap second when it is decided that we need it, typically but not quite once a year. However, having that extra second at the end of June can cause glitches with computers, and so there are discussions to remove the leap second and replace it with something larger on a less frequent basis.

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In yesterday’s blog, I discussed how mathematicians and physicists measure angles in radians rather than the more familiar degrees. I finished the blog by mentioning a problem to do with water level in a harbour. Let me state the problem in more detail, and then show how we go about solving it.


clovelly_harbour


The tide is high

Suppose the height in metres of the water in a harbour can be described by the equation

y = 6\sin(\frac{\pi}{4}t) + 8\cos(\frac{\pi}{4}t) + 11

where t is the time in hours. We will assume that the time t=0 corresponds to midnight. We want to find the following

  1. The maximum height of the water
  2. The minimum height of the water
  3. The time when the first maximum water height occurs
  4. The time when the first minimum water height occurs
  5. If a boat needs 2 metres of water to be able to use the harbour, how many hours a day can it use the harbour?

As I showed yesterday, the period of this up and down motion of the water level is given by \frac{2\pi}{\omega}, so in this example above the period is \frac{2\pi}{\pi/4}=8\; \text {hours}. In reality, of course, if the up and down motion of the level of water is due to the tides, then the period has to be 12 hours, as there are two high and two low tides each day. But, let’s assume, just for variety, that the period can differ from 12 hours, so in this example it will be 8 hours.

Notice that the equation describing the level of the water y is a combination of sines and cosines. How do we go about solving this?

Combining sines and cosines

The trick to being able to solve this equation is to use the compound angle formula. A few pages of algebra will show that

\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B).

We will use that to rewrite our formula for the water level. We can write

\sin( \frac{\pi}{4}t + \alpha ) = \sin( \frac{\pi}{4}t ) \cos( \alpha ) + \cos( \frac{\pi}{4}t ) \sin( \alpha )

Multiplying all the terms by R we get

R \sin( \frac{\pi}{4}t + \alpha ) = R \sin( \frac{\pi}{4}t ) \cos(\alpha) + R \cos( \frac{\pi}{4}t ) \sin( \alpha )

and, re-ordering this we get

R \sin( \frac{\pi}{4}t + \alpha ) = R \cos( \alpha) \sin( \frac{\pi}{4}t )  + R \sin( \alpha ) \cos( \frac{\pi}{4}t )

By writing it in this order we can see that in our original equation 6=R \cos( \alpha ) and 8=R \sin( \alpha ). If we square these two equations, and remember that \sin^{2}( \alpha ) + \cos^{2}( \alpha ) = 1 for any value of \alpha, then 36=R^{2} \cos^{2}( \alpha ) and 64 = R^{2} \sin^{2}( \alpha ) , so adding these we get 36 + 64 = 100 = R^{2} (\cos^{2} ( \alpha ) + \sin^{2} ( \alpha ) ) = R^{2} (1) , so R = \sqrt{100} = 10.

In order to calculate the phase angle \alpha we not that \frac{ R \sin ( \alpha ) }{ R \cos ( \alpha ) } = \tan ( \alpha ) = \frac{8}{6} = \frac{4}{3}. So \alpha = \tan^{-1} (4/3) = 0.93 \; \text{radians}. (This is about 53^{\circ}).

So, now we have an equation for y which is entirely in terms of \sin, we have

y = 10 \sin( \frac{\pi}{4}t + 0.93 ) + 11 .


sine plots

Plots of y = \sin( \frac{\pi}{4}t) (black curve) and y=\sin(\frac{\pi}{4}t + 0.93) (green curve). The blue dashed line is where y=-9. These plots do not include the +11 constant, so of course +11-9=2 metres.


The maximum and minimum heights of water in the harbour

Sine is a function which varies between +1 and -1, so the maximum value of 10 \sin ( \frac{\pi}{4}t + 0.93 ) = 10. This means the maximum height of the water in the harbour will be 10 + 11 = 21 \; \text{metres}.

The minimum height will be when 10 \sin ( \frac{\pi}{4}t + 0.93 ) = -10, and so the minimum height will be -10 + 11 = 1 \; \text{metre}.

At what time does the water level first reach its maximum value?

We have shown that the maximum value happens when \sin( \frac{\pi}{4}t + 0.93 ) = 1. Let us call (\frac{\pi}{4}t + 0.93 ) a new angle \theta . We know that \sin( \theta ) = 1 \; \text{when} \; \theta = \pi/2. But \theta = ( \frac{\pi}{4}t + 0.93 ) \; \text{so} \; \frac{\pi}{4}t = ( \theta - 0.93 ) = ( \pi/2 - 0.93 ) = 0.64 \; \text{radians}. This leads to t = (0.64)(4)/\pi = 0.81 \; \text{hours} = 49 \; \text{minutes} after midnight.

At what time does the water level first reach its minimum value?

This happens when \sin( \frac{\pi}{4}t + 0.93 ) = \sin ( \theta ) = -1. The first time this happens is when \theta = 3\pi/2, \; \text{so} \; \frac{\pi}{4}t = ( \theta - 0.93 ) = ( 3\pi/2 - 0.93 ) = 3.78 \; \text{radians}. This leads to t = (3.74)(4)/\pi = 4.8 \; \text{hours} = 4 \; \text{hours and} \; 49 \; \text{minutes} after midnight. Note, this is half a period (4 hours) after the maximum, as one would expect.

How many hours each day can we use the harbour?

If we can only use the harbour when the water level is above 2 metres, we need to find the time when the water level first drops below 2 metres (remember it goes down as low as 1 metre). At midnight, the water level is y = 10 \sin(0.93) + 11 = 10(0.8)+11=19 \; \text{metres}.

We need to find y = 10 \sin( \frac{\pi}{4}t + 0.93 ) + 11 = 2. Re-arranging, this is 10 \sin( \frac{\pi}{4}t + 0.93 ) = 2 - 11 = -9, \; \text{so} \; \sin( \theta ) = - 9/10 .


20130603-060320.jpg


This value is shown by the horizontal dashed blue line on the zoomed-in sine functions plotted above. This gives a value of \theta = \sin^{-1}( -9/10 ) = - 1.12 \; \text{radians}. But, this answer is no good for us, as it gives a negative time, as shown by the first red dot to the left of the y-axis. We have to find the next solution, which will be when at a time 1 period later, so -1.12 + 2\pi = 5.16 \; \text{radians}. So ( \frac{\pi}{4}t + 0.93 ) = 5.16, the 3rd red dot on the plot. This value leads to t = 5.39 \; \text{hours}.

If we compare this time to the time of the first minimum, 4.8 \; \text{hours}, we can see that is is after it. What has gone wrong? What we have found is the time when the water level comes back up to 2m after being as low as 1 metre when the time was 4.8 \; \text{hours}. The difference between 5.39 \; \text{and} \; 4.8 = 0.59 \; \text{hours}. From the symmetry of the sine function, the water will drop below the 2 metre level at the same time difference before the minimum, i.e. at a time of 4.8 - 0.59 = 4.21 \; \text{hours}. This is the 2nd red dot on the plot. So, the time the water is below 2 metres during each cycle is the time between the 2nd and 3rd red dots, i.e. the part shaded in yellow. That is 2 \times 0.59 = 1.18 \; \text{hours which is} \; 1 \; \text{hour and} \; 10.8 \; \text{minutes}.

Because the period of our sine function is 8 hours, there are 24/8 = 3 full cycles in each 24 hour day. During each cycle the water level is below 2 metres for 1.18 \; \text{hours}, the total number of hours each day for which we cannot use the harbour is 1.18 \times 3 = 3.54 \; \text{hours} \; = 3 \; \text{hours and} \; 32.4 \; \text{minutes}.

We were asked for the number of hours each day we can use the harbour, so this will be 24 - 3.54 = 20.46 \; \text{hours} \; = 20 \; \text{hours and} \; 27.6 \; \text{minutes} each day.

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