Last time we learned of Johannes Kepler’s Third Law of Planetary Motion and his development of the astronomical unit (AU) and how these contributed to bringing ancient scientists a step closer to calculating Earth’s distance to the sun. Today we’ll see why Kepler’s focus on Venus, specifically its travel through space in relation to Earth and the sun — the socalled transit of Venus — would become the crucial element to solving the puzzle. Astronomers had previously used the Earth itself as an optical rangefinder to calculate distance to the moon. But unlike the moon which is relatively close to Earth, the sun is many tens of millions of miles away, too distant to be used in that manner. When it came to finding the distance from Earth to its sun, they were stumped. Then in 1716 Edmund Halley had the insight to combine Kepler’s Third Law and the parallax principle with Venus’ orbital journeys to devise an ingenious solution to the problem. The transit of Venus, first predicted by Kepler in 1627, is a rare astronomical phenomenon which only occurs every 243 years. At this time Venus becomes clearly visible from Earth and appears as a black dot traveling a straight path across the fiery backdrop of the sun’s surface. Figure 1 shows this phenomenon as it would look from Earth. According to Halley’s plan, two observers with telescopes would be positioned on opposite sides of the Earth. Due to the principle of parallax their lines of sight would provide different perspectives of Venus’ path. See Figure 2. From the perspective of Observer B, Venus’s path would appear higher on the sun’s face than the path seen from the perspective of Observer A. As their lines of sight converge on Venus’ center, an angle forms between them, which we’ll name α. The same angle forms as they look past Venus to the sun in its backdrop. Halley theorized that if the angle α could be measured, Kepler’s Third Law could be used together with trigonometry to calculate the distance between Earth and the sun, Kepler’s socalled AU. We’ll review Halley’s methodology next time.
____________________________________

Posts Tagged ‘optical distance measurement’
The Transit of Venus
Monday, May 18th, 2015The Astronomical Unit — It’s So Relative
Thursday, May 7th, 2015
Last time we learned that early scientists used the Earth itself as an optical rangefinder to gauge its distance to the moon, and we posed the question: Can Earth be used in the same way to gauge distance to the sun? Unfortunately not. As we learned earlier in this blog series, the more distant the object, the larger the rangefinder that’s required, and the Earth just plain isn’t big enough to be used to compute that distance. The sun is 390 times farther away than the moon is, and that presents a real problem. Today we’ll learn about an alternate method to gauge this great distance. Johannes Kepler made a great contribution towards finding Earth’s distance to the sun when he developed his Third Law of Planetary Motion. Through his observation of planetary movements he was able to determine each planet’s relative distance from the sun in terms of what he dubbed the astronomical unit (AU), a yardstick by which the distance between all planets in our solar system and our shared sun could be judged. Kepler established the distance between Earth and the sun to be that astronomical unit, depicted in the illustration as r. The reason Kepler developed the AU was because in his day there was no known way to measure the distance to the sun. The AU, an abstract term with no real numerical value in terms of the distance being measured became a sort of placeholder term until r could actually be measured. The best the AU could do was allow him to determine how far a planet was from the sun relative to Earth’s distance from it. For example, his Third Law states that Venus’ distance to Earth is approximately onequarter the distance between Earth and the sun, or 0.28 AU. That meant that Venus’ distance to the sun was threequarters the distance of Earth to the sun, or 0.72 AU. Together, these two distances equaled one AU, the as yet unquantifiable distance r between Earth and the sun. Why all the fuss over Venus? Because Edmund Halley, a scientist who came after Kepler and shared in his fascination with interplanetary distances, would use Venus’ proximity to Earth to set up an optical rangefinder relationship between them and the object of his fixation, the sun. Crucial to this accomplishment would be making use of Venus’ orbital movement and the moment it would come into a direct line between Earth and the sun. We’ll explore that further next time.
____________________________________

Optically Measuring Cosmic Distances
Wednesday, April 22nd, 2015
Last time we learned that the bigger an optical rangefinder, the better its accuracy in measuring distant objects. Today we’ll take that concept a step further when we discover how Earth itself was used by ancient scientists to gauge its distance to the moon. Today’s blog will be strewn with embedded links to past blogs in this series, all of which have been building up to our understanding of gravity, a complex subject with many pieces to its puzzle. There are a few remaining pieces to be placed which will be covered in future blogs, but I promise we’ll get there. Long before Edmund Halley’s time, scientists used the Earth as a huge optical rangefinder. In doing so they employed the principles of parallax and trigonometry to obtain reasonably accurate measurements of the distance between Earth and its nearest neighbors, starting with the moon. See Figure 1. The illustration shows how it was done. Two observers armed with telescopes viewed the moon from opposite sides of the earth. Their lines of sight are represented by dashed lines, and together with the solid pink line which represents the distance between them, d, a right triangle was formed. Because Observer B was situated on the other side of the globe, his line of sight fell at an angle relative to Observer A’s, due to the Principle of Parallax. The angle that formed at the point in the triangle at which B was situated we’ll call θ. The fact that a right triangle was formed at Observer A’s observation point will enable our ancient scientists to use principles of trigonometry and parallax in their quest to find the distance to the moon. Follow this link to a refresher blog on the subject, Using Parallax to Measure Distance. At precisely the same moment the moon moved into Observer A’s telescopic line of sight, Observer B adjusted his telescope to center the moon within it. Observer B then duly measured the angle θ formed with a protractor, just as would be done with a rangefinder. If you’ve been reading along in this series, this setup might look familiar to you. In fact, the two mirrors of a military optical rangefinder work in exactly the same way as our two observers looking at the moon. Follow this link to a refresher on the internal workings of a rangefinder. Once the angle θ’s value had been determined, it was used to calculate the distance r between Earth and the moon with the same equation we’ve been using to measure distances using military optical rangefinders: r = d × tan(θ) As far as our moon observers go, the only variable left for them to determine before they are able to measure Earth’s distance to the moon is d, the distance between their viewing positions on Earth. We’ll see how to solve for d next time, when we put the Earth’s geometry to work for us.
____________________________________

Optical Rangefinders, Bigger is More Accurate
Tuesday, April 14th, 2015
Last time we touched on the fact that bigger is better when it comes to using a rangefinder to measure extremely long distances. Today we’ll expand on that theme and discover how bigger is indeed more accurate. Returning to our previous example, we’re still trying to find the distance to that ship on the horizon. We’ve got two rangefinders at our disposal, one short, one long, and the measurements provided by them are vastly different. Which is correct? To find out, we’ll hypothesize that we’ve taken the time to meticulously measure the distance the hard way, with a really long tape measure. Doing so, we find the actual distance to the ship is 5280 feet. We can now compare the actual measured distance to the measurements taken with our two rangefinders and compare their accuracy: Rangefinder One = 5729 feet – 5280 feet = 449 feet Rangefinder Two = 5208 feet – 5280 feet = 72 feet The smaller rangefinder results in a difference, or error, of 449 feet, while the bigger results in a difference of 72 feet. It’s clear that the bigger rangefinder gets us closer to the actual measurement taken by tape measure, so it’s the most accurate. The obvious conclusion is that the bigger the rangefinder used, the smaller the error factor. That’s because as the length of the rangefinder increases, the smaller the angle θ becomes, a situation which results in the tangent of θ moving farther away from rather than closer to 90°, all of which translates to more accuracy in our rangefinder’s measurements. Put another way, the bigger the rangefinder, the less likelihood there is of its angle θ‘s tangent hovering near 90° and becoming asymptotic, an undesirable outcome for reasons explained in a previous blog in this series. Next time we’ll see how early astronomers were able to arrive at relatively accurate calculations of the distances between Earth and other heavenly bodies by using the parallax effect produced by the world’s largest optical rangefinder, Planet Earth itself. ____________________________________

Optical Rangefinders, Why Bigger is Better
Monday, April 6th, 2015
Last time we introduced the fact that ultra fine gradations must be applied to a rangefinder’s indicator gauge in order to make accurate measurements of extremely long distances. Today we’ll see how using a bigger rangefinder effectively solves this problem. Figure 1 illustrates the subject. The left side shows what happens when attempting to use a small rangefinder to measure the distance to that distant ship on the horizon. The right side shows how the situation is improved by using a large rangefinder, which serves to decrease the angle θ. Figure 1 You see it all boils down to the angle θ. When d is extremely short in comparison to the measured distance r, the angle θ creeps ever closer to becoming 90°, a situation which severely impacts the rangefinder’s accuracy due to the impact on the tangent of θ. For a refresher on that see last week’s blog. Let’s see what the situation looks like numerically. The smaller rangefinder has a length, d, equal to 3 feet. Using it we measure θ to be 89.97°. Plugging these numbers into the rangefinder distance measuring formula, we measure the distance to the ship to be: r = d × tan(θ) r = 3 feet × tan(89.97°) r = 5729 feet Now let’s take a second measurement with the bigger rangefinder on the right. This one has a length d equal to 60 feet. You might be asking yourself, Do they really come that big?? Yes, before radar technology came on the scene to take their place, it was possible to find rangefinders as big as 60 feet in length! Using the larger rangefinder we find θ is equal to 89.34° and the distance to the ship is calculated to be: r = d × tan(θ) r = 60 feet × tan(89.34°) r = 5208 feet Why are the measurements between the two rangefinders so different? Which one is more accurate? In short, bigger is better. We’ll see why next week. ____________________________________

Further Limitations of an Optical Rangefinder
Monday, March 30th, 2015
Last time we discovered that when optical rangefinders are used to measure the distance to objects extremely far away we encounter problems. We discussed one of them last time, the fact that as θ approaches 90° the tangent of θ becomes asymptotic, resulting in a situation where even the most minute changes to θ bring about huge corresponding changes to the distance, r, we seek to measure. This difficulty goes hand in hand with another we’ll be discussing today, the problem of very tight spaces. They both lead to a greater potential for measurement inaccuracies. The rangefinder in Figure 1 depicts the kind of situation that often results when attempting to measure objects that are extremely far away, like a ship on a distant horizon. Angle θ is very close to being 90°. Let’s see what that does to our measuring attempts with the rangefinder’s onboard measuring scale, its indicator gauge. The fact is, when a rangefinder’s indicator gauge hovers near 90°, it becomes user unfriendly. To illustrate, let’s refer to a common everyday protractor, shown in Figure 2. Protractors are divided into 1° gradations, which allow us to measure angles between 0° and 90°. This interval is fine for many angle measuring purposes, but we’ll see in a moment why it doesn’t work when measuring extremely long distances. A similar protractor is found on a rangefinder’s indicator gauge, enabling us to measure the angle θ. Notice how small the space is between 89 and 90 degrees. Now imagine having to split that area into hundreds, even thousands, more gradations in order to accurately assess the value of θ. This is precisely the situation we encounter when using a rangefinder to measure extremely long distances where the lines of sight form long, narrow triangles and θ hovers near 90°. Are you beginning to see — or rather not see — the problem? When this situation exists, ultra fine gradations must be made between the 89th and 90th degrees in order to make an accurate measurement of θ . This results in a situation where gradation marks are spaced so closely together they become difficult, if not impossible, for the unaided human eye to read. Next time we’ll see why bigger is indeed better when seeking to solve this problem. ____________________________________

Limitations of an Optical Rangefinder
Monday, March 23rd, 2015
Last time we touched on the limitations of optical rangefinders when attempting to measure extremely long distances. Today we’ll expand on that theme. Let’s say we want to use a rangefinder to determine the distance, r, to an object that’s extremely far away, like a ship on a distant horizon, as shown in Figure 1. Figure 1 It’s obvious that the rangefinder’s length, d, is extremely small in comparison to the total distance viewed, r. When this situation exists, a very long and narrow right triangle is formed between the lines of sight provided by mirrors A and B of the rangefinder, represented by two red dashed lines, and the length of the rangefinder itself, d. It’s still a right triangle, a necessary condition to using our rangefinder formula to determine distance, however, when the triangle is an extremely long, narrow one, the angle θ approaches 90° in value. As discussed in a previous article, a θ value of 90° is impossible for the rangefinder distance formula to work with. The more distant the ship, the longer and narrower the triangle becomes, causing θ to creep ever closer to 90°. From a trigonometric point of view, this spells trouble. The problem is, the closer θ gets to 90°, the greater the disparity potential in its measurement. For example, let’s suppose that when the ship is first sighted with the rangefinder θ is measured at 89.95°. Then a second later the same person seeks to verify his measurement. Without noticing that he’s doing it, he shifts weight on his foot ever so slightly, takes a second reading, and finds that this time θ is 89.97°. That’s a difference of only 0.02° between readings, but it produces a huge change in the tan(θ). Figure 2 represents a graph of these two measurements, with the angle θ values on one axis, the tan(θ) values on the other. Figure 2 The graph illustrates how this minute change in θ of only two hundredths of a degree (0.02°) results in a correspondingly huge change to tan(θ) of 763.94 units. We know that the rangefinder’s length, d, equals 3 feet, so plugging all the numbers into our rangefinder formula we determine the distance to the ship to be, r = d × tan(θ) r = 3 feet × 763.94 = 2291.82 feet What this means on a practical usability level is that when the rangefinder’s adjustable mirror B moves only two one hundredths of a degree, it results in a change to the distance viewed, r, of almost half a mile! There’s another problem that goes hand in hand with the one presented today. We’ll explore it next time.
____________________________________

Vertical Asymptotes and Optical Distance Measurement
Friday, March 13th, 2015
Last time we introduced the vertical asymptote and the fact that it’s associated with the tangent function when angles approach 90° in value. On a graph that looks like this: Figure 1This asymptotic relationship exists when attempting to use an optical rangefinder to determine the distance to objects that are extremely far away– as in so far away they can barely be seen by the naked eye. When this is the case, is it even possible to use the optical rangefinder? Theoretically, yes. But not without complications. Our attempt to use the rangefinder to do this is illustrated in Figure 2. Figure 2You’ll note that the lines of sight extending from mirrors A and B on our rangefinder are almost parallel to each other, creating a situation where a vertical asymptote will form with regard to θ’s tangent. In plain English this means that even minute changes in θ will result in huge changes to tan(θ). We’ll explore that subject next time.
____________________________________

Tangent and the Vertical Asymptote
Monday, March 2nd, 2015
Imagine working, working, working towards a goal and getting oh so close, but never being able to actually reach it. It’s happened to all of us some of the time, but with vertical asymptotes it happens all of the time. We’ll see why the nature of vertical asymptotes presents a problem when measuring extreme distances in today’s blog. We’ve found that optical rangefinders can be useful in measuring long distances. They work well in many situations, but not all. When it comes to extremely long distances they aren’t at all helpful. That’s due to problems presented by trigonometry, more specifically the tangent function and how it leads to vertical asymptotes. Let’s look at Figure 1 to bear this out. Figure 1
In Figure 1 we see the same rangefinder being used to view objects at two different distances, one distance far greater than the other. There’s an obvious difference in the lengths of the sides of the triangles formed, as well as an obvious difference in angles θ_{1 }and θ_{2}. θ_{2} is much steeper than θ_{1}. This is more apparent when the lines of sight are isolated, as shown in Figure 2. Figure 2
Figure 2 shows that as an object becomes more distant and r, the distance to the object viewed increases, the angle θ_{ }gets closer and closer to a value of 90°. What’s significant about this is that a θ value of 90° is impossible for a rangefinder to work with. Why? Because it uses trigonometry to measure distances, more specifically the tangent function within trigonometry, and when θ takes on a value of 90°, it becomes asymptotic in nature — a situation which renders the optical rangefinder useless. To visualize this, we’ve plotted the tangent function for an array of angle θ values on a graph in Figure 3. Figure 3
The curved line represents plotted tangent values for θ that fall between 0° and 90°. What it demonstrates is that as θ gets closer to becoming 90°, it becomes more vertical and steeper in incline. In other words, it forms a vertical asymptote, stretching to reach a value of 90° but never actually getting there. In the math world this means that the tangent’s value is on its way to becoming unbounded or undefined. In plain English this means that the tangent’s value becomes increasingly more unworkable. In fact, the tangent of 90° degrees does not exist. Try entering the number 90 into your calculator and pressing the TAN button. You’ll receive an error message in return. Next week we’ll see the impact this has on the function of an optical rangefinder when the object viewed is so far away the angle θ approaches a value of 90°.
____________________________________

How An Optical Rangefinder Uses Trigonometry
Monday, February 23rd, 2015
We left off with an artillery soldier spotting an enemy tank in the distance. Luckily, he’s got an optical rangefinder at his disposal to measure the distance between them and crank out an accurate shot. His first action is to peer through the rangefinder’s eyepieces, rotating the adjustable mirror on the right eyepiece until the tank comes into focus. The two lines of sight provided by the left and right eyepieces of the rangefinder form a right triangle due to the parallax effect. One of the angles in this triangle is designated on the illustration by the angle θ, and that’s the angle we’ll be working with. The fact that a right triangle exists means the distance, r, to the tank can be easily measured using principles found in trigonometry, a branch of mathematics that addresses the properties of triangles, hence, the prefix tri in its name. Tangent, and other trigonometric functions like sine and cosine, relate the angles of a right triangle to the ratios of the lengths of the sides of the triangle. In our case we’re concerned with the tangent, which is simply the value that’s derived by dividing the length of the side opposite to the angle θ by the length of its adjacent side. This value can be found in most trigonometry textbook tables, but is most easily found by using a calculator. Simply enter the angle θ‘s value, then press the TAN button. So how does the artillery soldier determine θ‘s value? With the tank in clear focus, it’s easily measured with an indicator gauge attached to the adjustable mirror near the right eyepiece on an optical rangefinder. Our soldier reads the gauge and determines that θ is equal to 89.935°, so the tangent of θ is equal to: tan(θ) = tan(89.935°) = 881.473 Now that we’ve determined the values for d and the tangent of the angle θ, we can plug those numbers into our equation to determine r, the distance to the enemy tank using the equation, r = d × tan(θ) Plugging in numerical values, the equation becomes, r = 3 feet × 881.473 = 2644.419 feet The distance to the tank is 2644.419 feet. Next time we’ll see how the peculiarities of the tangent function place limitations on the accuracy of optical rangefinders over extremely long distances. ____________________________________
