Mth commonly used in the US Army Pthfinder School Pythgoren Theorem is used for solving tringles when two sides re known. In the Pthfinder Course it is used to determine the rdius of circulr drop zones nd is used to determine lengths nd widths on drop zone survey. Use the following link to fmilirize yourself with the Pythgoren Theorem nd complete the prctice quiz. http://www.mthsisfun.com/pythgors.html AB 11234 55678 c In order to solve for the length of the side of this drop zone you must use the Pythgoren Theorem s the drop zone does not sit on the verticl xis of the mp. In preprtion to solve 2 + 2 = c 2 AB 11100 54234 1. You must first solve for nd 2. To solve find the difference of the estings* 1. 11234 11100 = 134 3. To solve for find the difference of the northings* 1. 55678 54234 = 1444 4. Now plug those vlues into the Pythgoren Theorem 134 2 + 1444 2 = c 2 nd simplify 134 2 + 1444 2 = 1450 5. Therefore, c = 1450 * Esting is the first set of numers in grid. In this cse it is 11234 nd 11100. Northing is the second set of numers in grid. In this cse it is 55678 nd 54237. 1 Either of these numers my e negtive; this is ok, ecuse you will end up squring them, resulting in positive numer nd the eqution will work. Rememer tht you re deling with distnces nd there is no such thing s negtive distnce.
BASICS OF A DROP ZONE The left corner s you deprt the drop zone on drop heding is known s the Left Triling Edge (LTE) The right corner s you pproch the drop zone on drop heding is known s the Right Triling Edge (RTE) This distnce is known s the WIDTH or the LATERAL WIDTH. Direction of Flight This distnce is known s the LENGTH or the LONGITUDINAL LENGTH. This distnce is known s the DIAGONAL LENGTH or the DIAGONAL AND LONGITUDINAL LENGTH. The left corner s you pproch the drop zone on drop heding is known s the Left Leding Edge (LLE) The right corner s you pproch the drop zone on drop heding is known s the Right Leding Edge (RLE)
Dimensions of drop zone (lwys round your finl nswers down): Length - Compre oth lengths (left side [LLE to LTE] nd right side[rle to RTE]) nd tke the SMALLEST vlue. Width Compre oth widths (led edge [LLE to RLE] nd tril edge [LTE to RTE]) nd tke the SMALLEST vlue. Digonl Length Compre oth digonls (RLE to LTE nd LLE to RTE) nd tke the SMALLEST vlue. 2 To convert meters to yrds divide y.9144 (100 meters.9144 = 109.36 yrds) 2 To convert yrds to meters multiply y.9144 (100 yrds x.9144 = 91.44 meters) When converting, do not round until AFTER you hve completed the conversion. Ex. Convert 789.8 m to yds 789.8 m.9144 = 863.736 yds. Round oth nswers down 789m / 863 yds If you INCORRECTLY round meters down efore converting 789 m.9144 = 862.861 yds. Round down nd it is 789 m / 862 yds This nswer is 1yd off nd therefore wrong. PRACTICE QUESTIONS (ANSWER IN METERS AND YARDS) 1. You hve drop zone with the following corner coordintes: Left Leding Edge (LLE) - 18S TH 92110 26839 Right Leding Edge (RLE) - 18S TH 91337 26350 Left Triling Edge (LTE) 18S TH 92941 25525 Right Triling Edge (RTE) 18S TH 92168 25036 ) Wht is the length of the drop zone? ) Wht is the width of the of the drop zone? c) Wht is the digonl length of the drop zone?
2. You hve drop zone with the following corner coordintes: Left Leding Edge (LLE) - 16R EU 70649 96816 Right Leding Edge (RLE) 16R EU 70471 97430 Left Triling Edge (LTE) 16R EU 68893 96306 Right Triling Edge (RTE) 18S EU 68715 96920 ) Wht is the length of the drop zone? ) Wht is the width of the of the drop zone? c) Wht is the digonl length of the drop zone? 3. You hve drop zone with the following corner coordintes: Left Leding Edge (LLE) - 06V UP 58937 05981 Right Leding Edge (RLE) 06V UP 57307 06202 Left Triling Edge (LTE) 06V UP 58593 03444 Right Triling Edge (RTE) 06V UP 56963 03665 ) Wht is the length of the drop zone? ) Wht is the width of the of the drop zone? c) Wht is the digonl length of the drop zone? ANSWERS 1. ) 1554m / 1,700yds ) 914m / 1,000yds c) 1803m / 1972yds 2. ) 1828m / 1999yds ) 639m / 699yds c) 1937m / 2118yds 3. ) 2560m / 2799yds ) 1644m / 1798yds c) 3043m / 3327yds
The Lw of Sines (or Sine Rule) is used to determine the unknown lengths of tringle when you hve one known side nd known ngles. In the Pthfinder Course it is used to determine width nd length requirements of helicopter lnding zones nd ngulr drift clcultions on drop zones. Use the following link to fmilirize yourself with the Lw of Sines nd complete the prctice quiz. https://www.mthsisfun.com/lger/trig-sine-lw.html 200m In order to determine the forwrd nd/or lterl drift distnces we must use the Lw of Sines C B c A
Where: 90ᴼ 50ᴼ 200m 40ᴼ = = 200 Sin 40 Sin 50 Sin 90 To solve for side : = 200(Sin 40) = 129m : = 200(Sin50) = 153m *Sin 90 = 1 WHEN USING THE LAW OF SINES, ALWAYS USE PROPER MATH ROUNDING BELOW 5, ROUND DOWN; 5 AND ABOVE, ROUND UP.
PRACTICE QUESTIONS Solve for nd, nswers re in meters 1. = 90 54 = = = c Sin A SinB SinC 371m 36 Solve for nd, nswers re in meters 2. = = = = c Sin A SinB SinC 17 90 263m 73
Solve for nd, nswers re in meters 3. = = = = c Sin A SinB SinC 90 41 153m 49 ANSWERS 1. ) 218m ) 300m 2. ) 77m ) 252m 3. ) 100m ) 115m
Incoming students lso need to know: 2 To convert meters to yrds divide y.9144 (100 meters /.9144 = 109.36 yrds) 2 To convert yrds to meters multiply y.9144 (100 yrds x.9144 = 91.44 meters) To convert feet to meters divide y 3.28 (10 feet / 3.28 = 3.05 meters) To convert meters to feet multiply y 3.28 (3 meters x 3.28 = 9.84 feet) Aircrft Rte to convert irspeed (KIAS knots indicted irspeed) into meters/sec, multiply y.51 (70 KIAS x.51 = 35.7 meters per second) Depending on the prolem, you my need to round this vlue up. 2 When doing conversions from meters to yrds, rememer tht your meters vlue should ALWAYS BE SMALLER thn your yrds vlue. Rdius ½ the Dimeter of circle Dimeter the width of circle s it psses directly through the center Mth Symols X > Y - X is greter thn Y X < Y - X is less thn Y X Y - X is greter thn or equl to Y X Y - X is less thn or equl to Y