(1) Find the train engine (fig 1) at N 42 16.978, 84 24.768 W. Read the engine’s number N.
(2) Find the tank (fig 2) somewhere about N/10,000 miles from the engine.
(3) The tank is headed facing 54.000 degrees true.
(4) Rotate the tank’s gun turret 122.725 degrees to starboard.
(5) Elevate the gun muzzle to 12.200 degrees from horizontal.
(6) Measure the coordinates of the gun muzzle when it were aimed like that.
(7) Fire a round at 1480.0 ft/sec.
(8) Go back to the train engine for a nice public place to compute the coordinates of where the shell would land. I don’t recommend you rely on your GPSR to figure the offset. Nor should you figure in UTM etc. I didn't. (Wouldn't know how.) Just determine precisely how many feet there are to a mili-minute of latitude and to a mili-minute of longitude, at the cache.
(9) Hopefully to avoid wild goose chases, verify that you've computed the same as I did with the following checksum. Prior to even going to find the tank you can have computed how many feet the cache is away from the gun muzzle and how many minutes of latitude the cache is away from the muzzle. For the purpose of this checksum only, round off the distance to the nearest 10 feet and the minutes to the nearest second decimal place. Then add all the digits of these two numbers of feet, minutes, tenths of minutes and hundredths of minutes to get the checksum. If it is not 25 you must have made a mistake somewhere.
(10) When you get these parts figured, and your checksum is OK, go find the train engine, then the tank, measure its coordinates and compute the rest of the 3-decimal coordinates to find the cache. It's a very well hidden ammo can, but you don't have to move anything to see it.
The computation isn’t nearly so hard as it could be, in fact it's dead simple, because you can ignore the following factors: (On the other hand, you better consider very carefully just what some of them are telling you.)
- Ignore that the shell is 105 mm diameter.
- Ignore that its mass is 14 kilograms.
- Ignore that its drag coefficient is 0.13.
- Ignore that the air density (at standard temperature and pressure) is 0.00237 slug/ft3.
- Ignore that the air temperature is 70 degree F.
- Ignore that the atmospheric pressure is 29.61 inches of mercury.
- Ignore that the relative humidity is 50%.
- Ignore that the wind speed is southwesterly at 30 mph.
- Ignore the fact that the Earth's surface is not a perfect sphere whose radius is everywhere 3956 miles.
- Ignore the fact that the Earth's surface is not flat over the range of the shot.
- Ignore that the acceleration due to the Earth's gravitational attraction is not a constant 32.174 ft/sec/sec, everywhere on the surface, nor at every altitude.
- Ignore that the Earth is rotating about its N-S polar axis at 1.0 rev/day.
- Ignore the Coriolis effect of the target location moving westerly during the time the shell is in flight.
- Ignore the fact that the gun muzzle is some distance above the ground.
- Ignore the fact that the land's elevation is different at the tank than at the cache.
- Ignore the muzzle jump resulting from the shot's recoil.
- Ignore that the recoil is different for every muzzle elevation and for every turret direction and for every soil condition under the tank.
- Ignore that the barrel rifling twist is 14 feet/revolution.
- Ignore the Magnus effect of the side wind passing over and under the spinning shell.
- Ignore the rate of change with altitude of the air’s temperature, pressure, density, viscosity, wind speed and wind direction.
- Ignore that the axis of the shell remains substantially parallel to the barrel (due to its spinning) and does not remain end-on as it goes up then down along its trajectory.
- Ignore the "x-error" due to the shell's CG being slightly off axis.
- Ignore the "y-error" due to any gyroscopic-instability wobbling.
- Ignore the variation of the drag coefficient with Mach number.
- Ignore the variation of Mach number with altitude, temperature, viscosity etc.
- Ignore that the price of eggs is 78 cents per dozen.
As really simple as this problem becomes, after ignoring all these otherwise significant factors, still you may want to talk with an engineer or two about it. Chances are they might become interested in geocaching themselves.
It probably wouldn't help to talk to more than one though. No two engineers ever agree on anything, probably because usually they're both wrong!
HINTS:
(You might read just one hint at a time, and try again to solve it after each hint, before reading the next.)
(1) What are the vertical and horizontal components of the projectile's velocity?
(2) How much time was the projectile in the air? (I've given that the deceleration/acceleration rate due to Earth's gravity is 32.174 ft/sec/sec.)
(3) How many feet downrange did it go in that time?
(4) How many feet did it go in the N-S direction; and how many feet in the E-W direction?
(5) How many feet are subtended on the Earth's surface by a miliminute (0.001 minute) of latitude? (I've given that the Earth's radius is 3956 miles (5280 ft/mi))
(6) How many miliminutes of latitude did the projectile carry N-S from the tank's latitude?
(7) What is the latitude of where the projectile landed?
(8) How many feet are subtended on the Earth's surface by a miliminute of longitude, at the latitude where the projectile landed?
(9) How many miliminutes of longitude did the projectile carry E-W from the tank's longitude?
(10) What is the longitude of where the projectile landed?
(11) What is the latitude and longitude of the cache?
(12) The ammo-can cache is concealed in/under a big fallen tree.
You can check your answers for this puzzle on GeoChecker.com.