Blazing Quadratics Simple Targets Barrier Radar Radar & Barrier Copyright 2007 by Paul Barber Simple Targets Barrier Radar Radar & Barrier Elevated Target Incoming 1 Incoming 2 Instructions

320 Target X 1 Attempts Target Y Civilian Hit 15 WhichSlide 1 Radar Hit 660.890136192215 X travel of bomb 181.25 1 Barrier Hit X Barrier 173 Y Barrier Ground hit Trajectory Angle 25 Off Air 100 Velocity 270 Radar Height 262 Civilian 4 X 254 Civilian 1 X 408 Civilian 5 X 344 Civilian 2 X 373 280 Civilian 3 X -4 Score Hit Target Flag

Click on a link below for instructions Simple Targets Barrier Radar Radar & Barrier Elevated Target Incoming 1 Incoming 2 Calculations Home

Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page Target Position  X = 296, Y = 0 Home Score Find the difference between the target’s x coordinate and the x coordinate for the end of the cannon to determine the distance to the target. Clear This will clear all trajectories Try New This will re-plot the target and civilians and clear all trajectories. X coordinate for the end of the cannon. You’re goal is to get a trajectory that will land the cannon ball on your target. You may try guessing or you can do it mathematically. Cannon X cor 42.5 Instruction Page Change angle of the cannon. These are civilians. You’ll lose points if you hit them. It is also morally reprehensible to target them. This is your target The cannon’s elevation is 0. Air Resistance Change velocity From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

-7 Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Target Position  X = 530, Y = 0 -7 Home Barrier Height  119 Score Clear Try New This is the same as the Simple Target page except now there is a barrier that you must go over in order to hit the target. The barrier is half-way between the target and the cannon. So it is possible to compute the height of your proposed trajectory in order to see if it will clear the barrier. You’ll lose points if you hit the barrier. Cannon X cor 42.5 Instruction Page From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

-9 Home Score Clear Try New 750 150 50 m/sec Fire 58 Instruction Page Target Position  X = 446, Y = 0 -9 Home Radar Height  192 Score Clear Try New This is the same as the Simple Target page, but now your trajectory may not go above the radar line. You’ll lose points if you do. Cannon X cor 58 Instruction Page From Vert. From Horiz.  Off 750 150 50 m/sec Fire     ?

Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page Target Position  X = 550, Y = 0 Home Radar Height  195 Score Clear Barrier Height  135 Try New Things just keep getting more difficult. This is a combination of the Barrier and Radar Pages. Cannon X cor 42.5 Instruction Page From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

10 Home Score Clear Try New 450 450 50 m/sec Fire Instruction Page Target Position  X = 606, Y = 50 10 Home Radar Height  229 Score Clear Barrier Height  128 Try New Now there is an elevated target. Notice the coordinates include the y coordinate above the zero level. Cannon X cor Instruction Page 41.5 From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page Target Position  ? Home Radar Height  267 Score Barrier Coord.  X = 252, Y = 132 Clear Incoming Hit Coord. Try New X = 60, Y = 0 Incoming Midpoint X = 290, Y = 152.5 In this challenge a hidden target has shot at you. The green explosion came from him. The coordinates listed are from the location of the explosion and from the midpoint of the missile’s trajectory. Cannon X cor 42.5 You must find the distance between the incoming hit and the incoming midpoint. Add this distance to the midpoint coordinate to fine the target coordinate. Subtract the cannon coordinate from the target’s to get the distance from the cannon to the target. Instruction Page From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Instruction Page Target Position  ? Home Radar Height  248 Score Barrier Coord.  X = 301, Y = 175 Clear Incoming Hit Coord. Try New X = 60, Y = 0 Incoming Coord. 1 X = 86, Y = 38 Incoming Coord. 2 X = 218, Y = 172 In this challenge the coordinates of the explosion are given, along with a couple of coordinates the from the second half of the trajectory. This is more of a mathematical challenge. Cannon X cor 42.5 Instruction Page From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

Calculation Help Page Go Back sin(q)vt vt q cos(q)vt v = velocity Height of trajectory without gravity v = velocity t = time (dependent on angle & velocity) = trajectory angle a = acceleration of gravity 9.8 m/s/s Path of trajectory without gravity. sin(q)vt vt Height of trajectory including fall due to earth’s gravity. q sin(q)vt -1/2at2 cos(q)vt Distance trajectory travels If we consider a trajectory to start with y = 0 and to end with y = 0, then we can use the equation y = sin(q)vt – 1/2at2 to predict how long a missile will stay in the air if shot at a given velocity (v) at a given angle to the ground (q). The part of the equation sin(q)vt predicts how high a missile would go after (t) seconds if there was no gravity. The part of the equation 1/2at2 computes the distance an object falls towards earth in (t) seconds. When the distance a missile falls in (t) seconds = the height it would gain after (t) seconds the missile is back on the ground. From this equation we get the following equations. Maximum Height of a trajectory = sin(q)vt Distance = (cos(q)v)(sin(q)v) Time in the air = sin(q)v 4 4.9 4.9 Velocity to use at a given angle for a computed distance = (Distance to the target x 4.9) cos(q) x sin(q) Go Back

13 2 Home Score Clear Try New 450 450 56 m/sec Fire 42.5 Target Position  X = 360, Y = 0 13 2 Home Score Attempts Clear Try New Cannon X cor 42.5 From Vert. From Horiz.  Off 450 450 56 m/sec Fire     ?

592 1 Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Target Position  X = 640, Y = 0 592 1 Home Barrier Height  Score Attempts Clear Try New Cannon X cor 42.5 From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

529 2 Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Attempts   Target Position  X = 420, Y = 0 529 2 Home Radar Height  Score Attempts Clear Try New Cannon X cor 42.5 From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

-4 1 Home Score Clear Try New 250 650 100 m/sec Fire 32.5 Attempts   Target Position  X = 316, Y = 0 -4 1 Home Radar Height  270 Score Attempts Clear Barrier Height  173 Try New Cannon X cor 32.5 From Vert. From Horiz.  Off 250 650 100 m/sec Fire     ?

88 10 Home Score Clear Try New 390 510 78 m/sec Fire 39.5 Attempts   Target Position  X = 560, Y = 100 88 10 Home Radar Height  211 Score Attempts Clear Barrier Height  178 Try New Cannon X cor 39.5 From Vert. From Horiz.  Off 390 510 78 m/sec Fire     ?

-5 3 Home Score Clear Try New 450 450 50 m/sec Fire 42.5 Attempts   Target Position  ? -5 3 Home Radar Height  233 Score Attempts Clear Barrier Coord.  X = 311, Y = 46 Incoming Hit Coord. Try New X = 65, Y = 0 Incoming Midpoint X = 337.5, Y = 290 Cannon X cor 42.5 From Vert. From Horiz.  Off 450 450 50 m/sec Fire     ?

1 2 Home Score Clear Try New 360 540 33.5 m/sec Fire 38 Attempts    Target Position  1 2 Home Radar Height  290 Score Attempts Clear Barrier Coord.  X = 240, Y = 100 Incoming Hit Coord. Try New X = 70, Y = 0 Incoming Coord. 1 X = 80, Y = 16 Incoming Coord. 2 X = 150, Y = 110 Cannon X cor 38 From Vert. From Horiz.  Off 360 540 33.5 m/sec Fire     ?

Congratulations! You finished your trial. 88 Your Score is 3 Civilians you hit 6 Times you hit the radar 2 Times you hit the barrier Times you hit the ground 11 Home

Solving Quadratics

About Air Resistance Go Back Air resistance on a moving object is a very complex calculation that involves, the density of the air, the shape of the projectile, the projectile’s surface, and other factors. For this program I made the drag on the projectile a function of the projectile’s speed so that for every meter/sec in velocity there was drag coefficient of .0025. Of course as the velocity decreases, so does the total drag. And this amount is multiplied recursively. In other words, the air resistance used in this program did not use any acceptable mathematical model. In still other words, it has no basis in reality; it’s just there for fun. Go Back