Edmund Halley was faced with a real puzzle when he began his quest to determine the distance of Earth from the sun. One of the pieces to solving that puzzle came from the work of a German mathematician, astronomer and astrologer named Johannes Kepler. Early in the 17^{th} Century, Kepler spent a lot of time observing the planets in our solar system as they orbited the sun. He discovered that by taking note of the time it took for a planet to make one orbit around the sun, he could determine its relative distance from it. He then compared his findings with other planets, noting the time it took for each to make this same journey. His discovery would come to be known as Kepler’s Third Law of Planetary Motion. But what exactly is meant by a planet’s “relative distance from the sun”? In essence, it means that interplanetary distances, like just about everything else, are relative. Put another way, heavenly bodies can be said to be a distance X relative to another heavenly body if you establish a value for X, whether it’s numerical or otherwise. In Kepler’s case, X would be the unknown value of a socalled astronomical unit (AU), where one AU is equal to the unknown distance from Earth to the sun. Represented in equation form, this distance is: r_{Earthsun }= 1 AU This relative marker of distance could then be used to show how far the other planets are from the sun, relative to Earth’s distance from the sun, the AU. Kepler’s astronomical unit is simply a placeholder term for an unknown quantity, similar to any other unknown variable that might be used in an algebraic equation. For example, Kepler observed the orbits of Venus and Mars and determined their relative distances to the sun to be: r_{Venussun }= 0.72 × 1 AU = 0.72 AU r_{Marssun }= 1.5 × 1 AU = 1.5 AU In other words, Venus’ distance from the sun is just under three quarters of Earth’s and Mars is one and a half times Earth’s distance from the sun. In this way, Kepler was able to determine the relative distances from the sun in AU for all the observable planets in our solar system. Kepler felt sure that one day scientists would be able to accurately measure Earth’s distance from the sun, and when they accomplished this they could employ his astronomical unit system to determine distances between other planets in our solar system and the sun. Next time we’ll introduce the principle of parallax and see how Halley used this optical effect to devise a method for assigning a value to Kepler’s unknown AU. ____________________________________ 
Posts Tagged ‘licensed mechanical engineer’
Kepler’s Third Law Of Planetary Motion
Wednesday, December 31st, 2014Proving Galileo’s Theory On Falling Objects
Thursday, September 11th, 2014
Last time we discussed how Galileo proved Aristotle’s theory regarding the physics of falling objects to be wrong, although his experiment, which took place on the infamous Leaning Tower of Pisa, did not actually prove his own theory to be correct. So why didn’t Galileo go the extra mile and prove his theory? Because he couldn’t. Galileo, of course, resided on Earth, which was also the arena in which his experiment took place. As such, both he and his experiment were subject to the physical constraints presented by the Earth lab, the single most influential factor being the impact of the planet’s atmosphere upon his falling objects. Put another way, contrary to popular belief at the time, air is not an empty, innocuous space devoid of physical properties. It’s actually a gaseous soup of molecules. Nitrogen, oxygen, carbon, hydrogen, and other elements are in the mix, and they all have mass, that is, weight within a gravitational field. As Galileo’s balls fell, they continuously bumped against these molecules, which slowed their descent. This air friction will be discussed later in our blog series. But in order to prove Galileo’s theory correct beyond a shadow of a doubt, the testing arena would need to be one free from the interference of atmosphere. The Moon fits this criterion and provided the perfect environment to prove, once and for all, that Galileo’s theory was correct. So when astronauts Scott and Irwin simultaneously dropped a hammer and feather to the Moon’s surface, both objects hit at precisely the same moment. Watch this captured live footage of the event to see for yourself: One thing you may have noticed while watching the astronauts’ experiment is that the hammer fell more slowly than it would have on Earth. This has nothing to do with the absence of atmosphere on the Moon, but it has everything to do with gravity. We’ll discuss gravity’s influence in detail next time. _______________________________________

Manipulating theTorque Formula
Wednesday, April 23rd, 2014
Last time we introduced the simplified formula for torque: Torque = Distance × Force Today we’ll manipulate it by way of our wrench and nut example to get the torque that we need to loosen a tight nut. By inserting the numerical values of our illustration into the torque formula, it becomes: Torque = 6 inches × 10 pounds = 60 inchpounds Inchpounds may be terminology you’re unfamiliar with, but this notation arises from the fact that torque values are always represented by units of distance and force separated by a hyphen, in our case inchpounds. This just means that distance and force were multiplied together to calculate torque. In order to manipulate the value for torque all that needs to be done is change either or both numerical values for Distance and Force. Increasing either or both factors produces higher torque, decreasing them less torque. Why manipulate torque? To provide us with a mechanical advantage. Suppose we have a rusted nut that we’re trying to move with a wrench that has a 6 inch handle, and the 10 pounds of force employed by the muscles in our arm just won’t budge it. Put another way, 60 inchpounds of torque is insufficient to rotate the nut. It’s clear we must increase torque to get things going. Let’s do so by increasing either of the vector magnitudes. First we’ll try increasing the magnitude of the force vector. Instead of simply pushing hard on the wrench handle with our arm, let’s say we push extra hard. The average man can do a bicep curl of between 30 to 40 pounds, but we haven’t been going to the gym lately and we’re really out of shape. So try as we will, we just can’t muster up the bicep strength to apply more than 10 pounds of force to the wrench handle. It’s clear that this approach to increasing torque upon the nut isn’t going to work. The other way to increase torque is to increase the length of the distance vector. We’ll need a wrench with a longer handle, say 9 inches. By using a wrench with a longer handle we have increased the magnitude of the distance vector from 6 to 9 inches. The torque formula becomes: Torque = 9 inches × 10 pounds = 90 inchpounds Eureka! The longer handle has provided us with the mechanical advantage needed to increase torque to 90 inchpounds, thereby overcoming our muscular shortcomings and breaking the nut free. In summary, since torque is the product of the magnitudes of the distance and force vectors, we can increase torque by either increasing the magnitude of the force vector, or as in our example, by increasing the magnitude of the distance vector. Next time we’ll see how to apply the principles of torque to a real world situation involving gear trains in which we need to obtain a mechanical advantage. _______________________________________ 
Gear Reduction Worked Backwards
Sunday, March 9th, 2014
Last time we saw how a gear reduction does as its name implies, reduces the speed of the driven gear with respect to the driving gear within a gear train. Today we’ll see how to work the problem in reverse, so to speak, by determining how many teeth a driven gear must have within a given gear train to operate at a particular desired revolutions per minute (RPM). For our example we’ll use a gear train whose driving gear has 18 teeth. It’s mounted on an alternating current (AC) motor turning at 3600 (RPM). The equipment it’s attached to requires a speed of 1800 RPM to operate correctly. What number of teeth must the driven gear have in order to pull this off? If you’ve identified this to be a word problem, you’re correct. Let’s first review the gear ratio formulas introduced in my previous two articles: R = n_{Driving} ÷ n_{Driven } (1) R = N_{Driven} ÷ N_{Driving } (2) Our word problem provides us with enough information so that we’re able to use Formula (1) to calculate the gear ratio required: R = n_{Driving} ÷ n_{Driven} = 3600 RPM ÷ 1800 RPM = 2 This equation tells us that to reduce the speed of the 3600 RPM motor to the required 1800 RPM, we need a gear train with a gear ratio of 2:1. Stated another way, for every two revolutions of the driving gear, we must have one revolution of the driven gear. Now that we know the required gear ratio, R, we can use Formula (2) to determine how many teeth the driven gear must have to turn at the required 1800 RPM: R = 2 = N_{Driven} ÷ N_{Driving} 2 = N_{Driven} ÷ 18 Teeth N_{Driven} = 2 × 18 Teeth = 36 Teeth The driven gear requires 36 teeth to allow the gear train to operate equipment properly, that is to say, enable the gear train it’s attached to provide a speed reduction of 1800 RPM, down from the 3600 RPM that is being put out from the driving gear. But gear ratio isn’t just about changing speeds of the driven gear relative to the driving gear. Next time we’ll see how it works together with the concept of torque, thus enabling small motors to do big jobs.
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