## Posts Tagged ‘parallax’

### Calculating the Distance to the Sun

Thursday, June 25th, 2015
 We’ve been paying a lot of attention to Venus and its orbital patterns, as did scientist Edmund Halley hundreds of years ago.   Back then he came up with a plan to determine Earth’s distance to the sun, the AU.      Two key components were Kepler’s Astronomical Unit, or AU, and an angle, α, which formed between lines of sight followed by observers on Earth during the Transit of Venus.    Halley theorized that α together with Kepler’s Third Law of Planetary Motion would make it possible to calculate the AU.   We’ll see how today. Figure 1       Figure 1 depicts what Halley had in mind.   He theorized that if observers positioned on opposite sides of Earth could determine the precise times it took Venus to travel across the sun’s face from each of their perspectives, they could use this information together with previously gathered information on the time it takes Earth and Venus to make a complete orbit around the sun.   This would allow the angle α to be calculated, and from that Earth’s distance to Venus, rVenus.  Halley’s calculations for α are beyond the scope of this series, but if you’re interested in reading more about them, you can follow this link.       Earth’s distance to Venus, rVenus, is computed in a manner similar to the method we used previously to determine Earth’s distance to the moon, by using this equation, r = d × tan(θ) For a refresher on the subject, follow this link to my past blog on Optically Measuring Cosmic Distances.      And here’s the same equation modified to solve for the distance between Earth and Venus, rVenus, rVenus = d ÷ tan(α)                    (1)      Once Earth’s distance to Venus was determined, its value was incorporated into Kepler’s equation for 1 AU, and the distance between Earth and the sun became known.       Here again is the equation from Kepler’s Third Law of Planetary Motion, 1 AU = rVenus ÷ 0.28                    (2) And here it is with the function for  rVenus from equation (1) inserted into equation (2) to solve for 1 AU, 1 AU = [d ÷ tan(α)] ÷ 0.28                1 AU = d ÷ [0.28 × tan(α)]             (3)       From equation (3) the distance between Earth and the sun, 1 Astronomical Unit, was calculated to be between 92,000,000 and 96,000,000 miles.       Unfortunately, a Transit of Venus did not occur within Halley’s lifetime, but scientists that followed him applied his methodology after the next Transit occurred in 1761.   Since that time modern technology and the radar have improved measuring accuracy so that we now know the sun is located 92,935,700 miles from Earth.      Next time we’ll return full circle to our opening topic in this long blog series when we reopen our discussion on gravity, specifically, how the concept of centripetal force is instrumental in determining the gravitational force exerted upon Earth by the sun. ____________________________________

### Earth’s Distance to the Sun — A Roadmap

Thursday, June 18th, 2015
 We left off with Edmund Halley’s proposed method to solve the riddle of Earth’s distance to the sun.   Halley posited that when Venus’ orbit brought it directly between the Earth and sun, then principles of astronomy, trigonometry, and geometry could be combined to calculate that distance.   Instrumental to Halley’s theory were a number of elements discussed previously in this blog series, including the work of Johannes Kepler.   We’ll mesh those elements today and chart the course for future discoveries.       To begin things, Halley knew that Kepler’s Third Law of Planetary Motion set out the distance between Earth and the sun in theoretical terms as, 1AU = rVenus ÷ 0.28 which meant that if the distance from Earth to Venus, rVenus, could be calculated, then the distance from Earth to the sun was easily deduced, a matter of simple division.       Crucial to the calculation of rVenus is to find the value for the angle α which forms between observers’ lines of sight while charting Venus’ travel across the face of the sun, something which only happens during a rare astronomical event known as the Transit of Venus.   See Figure 1. Figure 1       Figure 1 shows two observers positioned on opposite sides of the Earth, busily surveying Venus’ movement across the sun’s face.   Their lines of sight converge at a vertex point, or point of intersection, on Venus, then move beyond it to the sun.   Due to the principle of vertical angles, which stipulates that angles which share the same vertex point also share the same angle measurement, we know that the angle α that’s formed between Observer A and B‘s lines of sight is of the same value between Earth and Venus as it is between Venus and the sun.       Once a is determined, its numerical value can be plopped into an equation we’ve been working with for some time now in this blog series.   It’s similar to the equation previously used to calculate Earth’s distance to the moon, r = d x tan(θ) Follow this link to Optically Measuring Cosmic Distances for a review.       And here is that equation with terms modified to reflect our new quest, the distance from Earth to Venus, rVenus = d ÷ tan(α)       As for the variable d, the distance between the two observers, we’ve worked with that before, too.   Follow this link to Determining Chord Length on Circle Earth for a refresher.       Next time we’ll see how Venus’ travel path is key to determining the angle α, shown in green on the illustration, and how this angle is crucial to our discovery of the distance between Earth and the sun. ____________________________________

### The Transit of Venus from Different Perspectives

Wednesday, June 3rd, 2015
 We’ve been working on calculating Earth’s distance to the sun and discussing how past scientists, including Johannes Kepler and Edmund Halley, contributed to the discussion.    Today we’ll see how Halley used his theory on the transit of Venus together with Kepler’s Third Law of Planetary Motion and combined them with known mathematical principles to solve the riddle of this enormous distance, known as one astronomical unit, or AU.       Halley’s solution began with stationing two observers with telescopes aimed at Venus on opposite sides of the Earth.    Their different lines of sight would cause the principle of parallax to come into play, resulting in them seeing Venus from different perspectives.    Their sight would converge at Venus’ center and an angle, α, would form between them.   Halley posited that if this angle could be measured, it would be an important first step in calculating the distance between Earth and the sun.    See Figure 1. Figure 1       Due to their differing perspectives, Observer A would see Venus traveling a path lower on the sun’s face, while Observer B would see it following a path higher up.    See Figure 2. Figure 2       The net result was that the length, LA, of Venus’ path as seen by Observer A was significantly shorter than length, LB, of Venus’ path as seen by Observer B.    Because of this, Observer B would have seen Venus pass in front of the sun before Observer A.    These differing observations meant that even if both observers were to set their telescopic crosshairs on Venus at the exact same moment it became visible to each of them, it would serve no purpose, because they lacked a common point of reference at which to aim in order to take measurements.    This fact made measuring the angle α with a physical device such as a protractor impossible.       So Halley gave up on the idea of physically measuring α.    Instead, he proposed calculating it based on the time it took for Observers A and B to watch Venus traverse the sun’s face from one side to the other along each of their observational paths.       Next time we’ll see how Halley put his idea to work to calculate α and used it in conjunction with Kepler’s Third Law to calculate the AU. ____________________________________

### The Transit of Venus

Monday, May 18th, 2015
 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 so-called 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. Figure 1       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. 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 so-called AU.    We’ll review Halley’s methodology 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. 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 on-board measuring scale, its indicator gauge. Figure 1      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. 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. ____________________________________

### 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. ____________________________________