I’ve always been a huge Star Wars fan my whole life. I bought lightsabers, had Stormtrooper helmets, and have watched every single movie in the series. However, as I was recently rewatching Return Of The Jedi on my family’s movie night, one scene in particular struck me: the Death Star 2’s Destruction. In it, the heroes had to fl y their fi ghter jets from a surface entrance of the Death Star –a moon-sized space station with the ability to destroy entire planets through a powerful laser– deep into its core, navigating through a crisscrossing and complex maze of corridors, tunnels, and rooms spanning over 160 kilometers in diameter, and destroy its power source. The mission was almost impossible: it had to pass through a variety of locations: a central canyon, a med bay, an entrance port, a control center, the captive chambers, and of course the central power source; in order to do tasks such as dropping o ff the attacking team to the control center, rescuing the captured alliance members, undocking another ship, and more. All before the Death Star could fi re its laser and destroy the entire Alliance fl eet. As I watched the scene, I felt quite confused and disoriented, as I didn’t have a good indication of where in the gigantic space station the jets were at at all times. And it seemed, to me, that they weren’t going the optimal route, as even though there were so many crisscrossing tunnels and shafts, they simply went straight ahead. Of course, you could attribute that to a simple cinematic decision so as not to complicate the experience, but in my head I really wondered, if I was to have a map of the entire Death Star, could I fi nd the best route to take given that the jet must pass through multiple locations in the base, including the central core, and come back out safely? In this investigation, I will use principles of graph theory to model the Death Star as a network and determine the mathematically optimal path, of which passes through all given points, that minimizes travel time. I will also be using integral and di ff erential calculus to explore the implications of the fi ghter jet’s ability to accelerate or slow down, in order to make it a more realistic depiction of the scenario. Math IA What is mathematically the best path to take to blow the Death Star up without compromising speed Introduction Context Purpose Explanation Photos Return To simplify the graph representation of the Death Star, some assumptions have been made: Every corner is a point. Edge weights correspond to an average travel time between vertices based on distance and the ability to accelerate/ 1. decelerate. Minor obstructions along corridors are ignored, and travel between vertices is unobstructed - i.e. ignore locked doors, 2. maintenance issues, etc. Shortest travel time equates to the optimal path. 3. The longer the distance of a corridor, the more the fi ghter jet is able to accelerate, as it has more space and time to get 4. faster. As such, the travel time for each corridor is dependent on the distance between its vertices. However, around every corner, the fi ghter jet must decelerate back to its original speed, in order to not crash. 5. The corridors are the only paths the fi ghter jet can take, there is no diagonal cutting or shortcuts. 6. Assumptions Map of the Death Star c original as map with weight T s as or 8 n s and s on s I notes a o.iq is 0.55 is μ s ios or on or on as i Ya 1.65 is a is i as s it in a is on a on a map draw one of suitable paths many lorridorf Iii must hit red spots 1h41 is a o our travelling salesman problem s i iii i i an diameter 150 kilometers NORMAL CASE CONSTANT ACCELERATION Constant alleleration F 15 jet O 8 nah 274.4mn's to 1 z mash 7 4 11 G m is 2 se alt 5 488m11 5.481 an elevation Of 411.0334 4 55,488 mls this is how it would look V41 15.488 dy I in normal case on earth prediction uh 5.4884 i i i ALE IN SPA IE non constant an elevation in space no air resistance spare hips are moved by propelling fuel out can find non constant acceleration through finding fiertainstatistity Newton't 2nd law F ma tuel is being continuously displaced I ÉE e Fema was it getting smaller so E must 4 outline for next sections i rate of mass lose ii forte iii mass of system iv base speed Because in space, where the central scenario is taking place, spaceships are able to hypothetically accelerate inde fi nitely until fuel runs out, as there is always a force from behind, and there are no factors such as air resistance and gravity. As such, since F=ma, and if mass decreases as fuel, then acceleration must inevitably increase. As such, since the focus of my question is the least time possible, the distance should not be the edge weights— it should be time. And knowing the spaceship’s nature is to increase acceleration inde fi nitely, that should be factored in, as if the pilots want to reach the center the fastest, they would obviously try to use up more fuel to accelerate more given a longer distance. The longer the distance, the more time the subject has to accelerate. With the given information, such as force, loss of mass over time, mass of the system, and base speed, we can derive an acceleration-time graph. From there, we can use integral calculus to fi nd a distance-time graph, and by plugging in the distance for each corridor, we can ultimately fi nd the most realistic time taken for each corridor and set them as edge weights. i Rate of Mass Loss From source F 15 loses 23,000 gallons of Jet A1 per hour It 23,000 gallons th loss of fuel per hour ft m convert to mass sin le Itsy 0.32 gallons of Jay AI If 71875kg h 7183750 kg's 19,965 kg Is ii Force on system derived from the F 15 thrust force which is 50,000 pounds source 50,000 pounds 222411 08 N F 222411,08N iii Mass of system derived from x wing mass of 10,000 kg soured my 10,000kg derived from fighter jets standard fuel weight of 34,37511g bourne Mf 34,375kg total mass of system My My t Mt M I 10,0004 34,375 44,375 kg in Base speed of ship derived from X wing average speed of 1050 km h source Uo 1050 km h 291.6 mil intipled Situation Visualized M 44,375 kg spaceship enters corridor at 291.6 all Thp F 222,411.08N Igivananeleration time this Uo 291.6 mls Derivation of alt graph algebra Estep find mass loss as a function of time Let initial mass at t O is M time t later Egger Yeconds i Estep so mass remaining at time t initial may ma's lost MA T fun ion using Newton's 2nd Law Fnet ma alt Ing thrusting force 9 It 222411.08 445 HIII tiny É I DESMOS GRAPH analysis acceleration is constant 9 5.012 m s matches prediction value of F 15 curves upward at the end presumably because fuel runs out asymptote 44375 19.9654 0 44375 19,9654 2222.645 this means jet has 2222 seconds to run out of fuel Finding UH graph Integration Have to evaluate for limits a H 43I.tt fgy y initial velocity 1050km h 291.6ns Sin le VA Salt dt V 291.6 11140 In 44115 19.965 73 U 291.6 1 11140 In 44375 19.965 1 119201.80 Sdu 222411.08 44375 19.9654 dt 291.6 V Ia 2 4451 In 44375 19.965 31 V Ija 11140 ln 44375 19.9654 1 evaluate integral V 291,6 11140 en 144395 19.965 7 ln 44375 v 291,6 11140 In 44395 19.96547 119202.81 4039579491144219.96541 t 172 s graphs velocity uh analysis almost linear but slope still is gradually increasing likely from loss in mass alt velocity starts at time 291,6mi our starting base speed since Kitt Sundt dg ult integrate both sides 41 It 11140 In 44395 19.965 11 119494 dt o 71ft 5119393 It 511140 In 44395 19,965 1 dt 714 1193934 111405 In 44395 19.965 1 dy I now solving S In 144795 19.9654714 h substitution sub u 44795 19.965 1 It I 19.965 due 19.96524 H Stahl du integrate In tht by parts a In In U I gas fu tain ft uda n tu v U aah ninth n as 944 443944476512 144395 19.94 tt 1n144395 19.965 sub bail into equation 714 1193934 111405 In 44395 19.965 1 dy I gift 1197974 1114014439551765T 144395 19.94511 144195 19.965 7 Iin144395 19a65 114479 3997 Plug ball into map find time values turn weights into time gin selonds 0.5 I 0.6 og of 8 9 8 a 0.5 0.5 0.7 A 0.620 1.4 1.3 0.88 o s 6 is 0.55 is o μ 8 1.05 0 a a o 58 o 7t 2 i o.us It 1.65 o.is 05 4 y o 55 I 5 j as 0.25 1 I 1.5 0.75 O callulation done on calculator Time Taken Between Vertiles final edge weights added each edge to combine into one edge simplified edge below calculation was plugging into formula Simplifying the edges graph theory representation of above table A 57.5 i t É É t E c D spatial representation simplified representation Finding the upper bound (u): Choose a starting vertex Follow the edge of shortest length to an unvisited vertex. If there is more than one unvisited vertex, pick one at random. Repeat Step 2 until all vertices have been visited Return to the starting vertex by linking it to the corresponding edge. This cycle (Steps 1-4) is known as the Hamiltonian cycle, which enables you to fi nd the upper bound of the interval where the answer, the most e ffi cient path, lies. Finding the lower bound (l): Delete a vertex, together with the edges connected to it, from the original shape Find the minimum spanning tree or route for the remaining graph, while ensuring that the route doesn’t form a circular shape Add the lengths of the 2 shortest deleted edges from Step 1 to the length of the minimum spanning tree from Step 3. This gives you the value for the lower bound of the interval value. Repeat for the rest of the vertices. Final answer: If the lower bound is “l”, and the upper bound is “u”, then the fi nal value lies between “l” and “u” i.e “l < x < u” Here, the Hamiltonian cycle is a path that follows the lines on the diagram while visiting each vertex exactly once, before returning to the vertex that you started on. The minimum spanning tree mentioned in the section for calculating the lower bound is a path following the lines of the diagram that connects all of the vertices in a way that minimizes the weight needed while ensuring that a full rotation or circle isn’t formed. Graph Theory Calculation Theory Definitions travelling salesman problem hamiltonian cyme visiting every point one time travelling salesman problem to minimize total weight of hamiltonian mile lower bound upper bound Step l finding the heapest route from every point nearest neighbor algorithm Upper Bound comparing the weight to find the heapest lowest route out A of every point by using the nearest neighboralgorithm i I from A cheapest route is to F 26.7 2 proceed for all vertices optimal path A F E B D A 26.7 18.75 310 57.0 48.64 82.5 3 Adding corresponding lengths together 26.7 18.75 310 57.0 48.6 82.5 264.51 total optimal solution that will never be exceeded Lower Bound delete negley A find win spanning tree it i point deleted fin spanning tree two shortest path weight flower bow A Ft Et B AF FE 122.95 122.95 B A F E AF FE 1 122.951 c A FYE AF FE 127.95 É r is between 122.95 and 264.55 second 122.95 L K C 264.55 exalt solution AT FT E T I I BT D A 26451 Dong