## Archive for October, 2014

### How Long is a Pendulum’s Swing?

Wednesday, October 29th, 2014
 What would you do to pass the time if you were stuck on a ship of the middle ages for weeks at a time?   Dutch mathematician Christiaan Huygens used the time to study the movement of clock pendulums.   He watched them for endless hours, and he eventually came to realize that the pendulums’ swing was uneven due to the ship’s listing on the waves, a phenomenon which also affected the ship’s clocks’ accuracy.       Eager to devise a solution to the problem of inaccurate time keeping, Huygens dedicated himself to finding a solution to the problem, and in so doing increase the navigational accuracy of ships as well.   His efforts eventually resulted in a formula that shared a common variable with Isaac Newton’s gravitational formula, namely, g, Earth’s acceleration of gravity factor, a value which Huygens posited was indeed a non varying constant.       Building upon Newton’s work, Huygens devised a formula which demonstrated the mathematical relationship between the motion of a clock’s pendulum and g.   That formula is, T = 2 × ∏ × (L ÷ g)1/2 where, T is the period of time it takes a pendulum to make one complete swing, ∏ the Greek symbol pi, valued at 3.14, and L the length of the pendulum.       Since devices capable of directly measuring the Earth’s gravity did not exist then, as they still don’t exist today, how in the world (pardon the pun) was Huygens able to arrive at this formula?   Thinking outside the box, he posited that if one knows the length of the pendulum L, and then accurately measures the time it takes for the pendulum to complete its swings, taking into account the varied times that resulted due to the ship’s listing, one can calculate g using his equation.   He eventually determined g‘s value to be equal to 32.2 feet per second per second, or 32.2 ft/sec2.       Fast forwarding to Henry Cavendish’s time, Huygens’ work with pendulums and his determination of g was well known.   We’ll see what Cavendish did with this knowledge next time. _______________________________________

### Huygens’ Use of Pendulums

Tuesday, October 21st, 2014
 Last time we learned that Henry Cavendish determined a value for G, the universal gravitational constant, fast on his way to determining a quantity he was determined to find, the Earth’s mass.   Today we’ll see how the previous work of Christiaan Huygens, a contemporary of Isaac Newton’s, helped him get there.       First Cavendish used algebra to rearrange terms in Newton’s gravitational formula so as to solve for M, Earth’s mass.   Rearranged, Newton’s formula becomes, M = (g × R2) ÷  G       But in order to solve for M, Cavendish first needed to know Earth’s acceleration of gravity, g.   To aid him in this calculation he referred back to the work of Christiaan Huygens, a Dutch mathematician from Newton’s time.       Huygens was eager to devise a formula capable of predicting clock pendulums’ motions on ships, his goal being to invent a timepiece accurate enough to make navigating ships easier.   He hypothesized that a key factor in predicting a pendulum’s movement was an unknown constant, the acceleration of gravity factor, g, which Newton had previously posited existed.   Through meticulous observation, Huygens came to realize that the time it took for pendulums to complete one swing back and forth was dependent not only on the length of the pendulum, but also this unknown quantity.       In order for Huygens’ computations to work, the value of g had to be a constant, meaning, its value could not vary between computations; g‘s value was in fact a fudge factor, a phantom he would assign a specific numerical value.   Huygens’ needed it in order to make his hypothesis work, a practice commonly use by scientists, even today.   Determining a value for g would allow Huygens to successfully relate the length of the pendulum to the timing of its swing and to create a mathematical relationship between them.       Huygens ultimately determined g’s value to be a whopping 32.2 feet per second per second, or 32.2 ft/sec2.    We’ll see how he did it next time. _______________________________________

### How Big is the Earth?

Wednesday, October 8th, 2014