STUDENT NAME: Kyle Judy
There are many examples of real-life unit conversion errors that had disastrous results. Read the Stories below and complete the related Problems.
_The solutions to some problems are provided for your reference. Your assignment won't be graded for accuracy, but for completion of all problems and showing your work. Try the problems on your own first, and then use the solutions to check your work!_
Story 1: On September 23, 1999 NASA lost the $125 million Mars Climate Orbiter spacecraft after a 286-day journey to Mars. Miscalculations due to the use of English units instead of metric units apparently sent the craft slowly off course -- 60 miles in all. Thrusters used to help point the spacecraft had, over the course of months, been fired incorrectly because data used to control the wheels were calculated in incorrect units. Lockheed Martin, which was performing the calculations, was sending thruster data in English units (pounds) to NASA, while NASA's navigation team was expecting metric units (Newtons).
Problem 1a - A solid rocket booster is ordered with the specification that it is to produce a total of 10 million pounds (lbf) of thrust. If the booster instead produces 10 million Newtons, what is the error, in pounds of thrust? (Hint: error is expected result minus actual result) (1 lbf = 4.5 Newtons)
Actual = 1.0 x 107 N
\(1.0 \times{} 10^7 \text{N} \cdot \frac{2000000000000 \text{lbf}}{8896443230521 \text{N}} = 2.248 \times{} 10^6 \text{N}\)
Which means:
\(\frac{2.248 \times{} 10^6 \text{N}}{1.0 \times{} 10^7 \text{N}} \cdot 100 = 22.5\%\)
Of course there's the other way to calculate error, but that gives a smaller number so we'll ignore it :)
Problem 1b - The quantity of work done is the result of applying a force for a given distance. That is: \begin{align*} W[\text{J}] = F[\text{N}] \times D[\text{m}] \\ \end{align*}\begin{align*} W[\text{J}] &= F[\text{N}] \times D[\text{m}] \\ &= 10,000,000 \text{lbf} \times 5 \text{km} \\ &= 10,000,000 \text{lbf} \cdot \frac{8896443230521 N}{2000000000000 \text{lbf}} \times 5 \text{km} \cdot \frac{1000 \text{m}}{1 \text{km}} \\ &= 4.448 \times 10^7 \text{N} \cdot 5000 \text{m} \\ &= 2.224 \times 10^{11} \text{J} \end{align*}If a solid rocket booster provides 10 million pounds of thrust while the rocket travels 5km, how much work, in Joules, is done by the SRB? (Hint: make sure you convert the thrust to Newtons!) (1 Joule = 1 N * 1 m)
Story 2: On January 26, 2004 at Tokyo Disneyland's Space Mountain, an axle broke on a roller coaster train mid-ride, causing it to derail. The cause was a part being the wrong size due to a conversion of the master plans in 1995 from English units to Metric units. In 2002, new axles were mistakenly ordered using the pre-1995 English specifications instead of the current Metric specifications.\begin{align*} 1.25 \text{in} \cdot \frac{127 \text{mm}}{5 \text{in}} &= 31.75 \text{mm} \\ 1.25 \text{cm} \cdot \frac{10 \text{mm}}{1 \text{cm}} &= 12.5 \text{mm} \\ \frac{12.5 \text{mm}}{31.75 \text{mm}} \times 100 &= 39.37\% \end{align*}Problem 2a - A bolt is ordered with a thread diameter of 1.25 inches. What is this diameter in millimeters? If the order was mistaken for 1.25 centimeters, by how many millimeters would the bolt be in error? (Hint: error is the expected diameter, in mm, minus the received diameter) (1 in = 2.54 cm) (1 cm = 10 mm)
Problem 2b - The load that a bolt can support in shear without breaking is directly related to the cross-sectional area of the bolt. Calculate the cross-sectional areas [in 2] for a bolt with diameter 1.25 in. and a bolt with diameter 1.25 cm. How much shear load can the 1.25 cm diameter bolt support, as a percentage of the shear load the 1.25 in. bolt can support? (Area of a circle = \(\pi\) * r 2)\begin{align*} \pi r^2 &= \pi \frac{d}{2}^2 \\ area &= \pi \frac{d}{2}^2 \end{align*} \begin{align*} area &= \pi \frac{1.25 \text{in}}{2}^2 & area &= \pi \frac{1.25\text{cm}}{2}^2 \\ &= \pi \frac{1.25 \text{in} \cdot \frac{127 \text{mm}}{5 \text{in}}}{2}^2 & &= \pi \frac{1.25 \text{cm} \cdot \frac{10 \text{mm}}{1 \text{cm}}}{2}^2 \\ &= \pi \frac{31.75 \text{mm}}{2}^2 & &= \pi \frac{12.5 \text{mm}}{2}^2 \\ &= 791.7 \text{mm}^2 & &= 122.7 \text{mm}^2 \end{align*} \begin{align*} \text{error} &= \frac{122.7 \text{mm}^2}{791.7 \text{mm}^2} \cdot 100 \\ &= 15.50\% \end{align*}
Story 3: On 23 July 1983, Air Canada Flight 143 ran completely out of fuel about halfway through its flight from Montreal to Edmonton. Fuel loading was miscalculated through misunderstanding of the recently adopted metric system. Before leaving Montreal, the pilot calculated a fuel requirement of 22,300 kilograms. There were 7,682 liters already in the tanks.\begin{align*} 22300 \text{kg} \cdot \frac{1 \text{L}}{0.803 \text{kg}} &= 27770 \text{L} \\ 27770 \text{L} - 7682 \text{L} &= 20088 \text{L} \end{align*}Problem 3 - If a liter of jet fuel has a mass of 0.803 kilograms, how much fuel needed to be added?