Question

2. Center-of-gravity (cg) location is critical for airplane stability. To get the "empty" og location, the Piper J-4 Cub pictured below was leveled and put on three scales (one under each wheel). For this application, the "empty" configuration included full fuel, but no pilots or baggage. The top two scale readings (in pounds) are for the two main wheels, and the bottom scale reading is for the tail wheel. 436 CALE 56 COOoooo SOCRRN PHRRA In the diagram below, the datum is defined to be the leading edge of the main wing. The distances from the datum to the pilots' seat, baggage compartment, main wheel, tail wheel, and fuel tank are shown. The main-wing mean aerodynamic cord (MAC) is 63 inches. 22-6 - 29 - 21- 43" 23" NC 71 f27866 e fuel tank baggage seat -TAIL WHEEL Appros 63" 004" TIRES PALMER YDRAULIC BRAKES HAYES WHEELS 30 3.25" 182.5" ARY STRUT CLAMPS tO LIFT STRuts datum a. What is the "empty" weight? (4 points) b. What is the "empty" cg location in inches aft of datum? (7 points) c. What is the "empty" cg location in %MAC? (5 points) d. Assume the pilot weighs 180 pounds and the co-pilot weighs 170 pounds, what is the cg location in inches with the crew and full fuel? (10 points) e. In addition to the crew and full fuel, assume 50 pounds is put in the baggage compartment, what is the gross weight of the airplane? (13 points) f. Assume 50 pounds is still in the baggage compartment but the airplane has burned 10 gallons of fuel, what is the cg location in inches with the crew, baggage, and remaining fuel? (18 points) g. What is the range of cg in %MAC for the two configurations in d) and f)? (6 points) h. Which configuration d) or f) is a forward cg and which is an aft cg? (4 points)

Answer 1

Answer for point f is added at the end, please don't consider the point written in the middle.

Answer! а) 16tet emps of ght = 436 + 444 45% total = 940 Lb b) Idealising the Shown below. 1 fuel Dotom tanley aircraft as an overhonging beam as The weight at aircraft is assumed as UD2 seat baggage eight and tort zs" turto оры Ф 2 empty aircrah 15616 32788546 mainwhell tail wheel 182.5" Location of about B ca from point B is found by taking moment Rox18 275 = Footer-X RA (182-5-3-25)= Flotex x, 885 % +82-52-404-XX 885*129.2529418 ist 10, 168.581 : ca distona from datum a 182.5-168.58 = 13.91" c) 7 MACa 13.91 = 224. 63 Pilots weight is added which P2 1807 1702 35olu ... Reaction at AaB One Scanned wilar Riße 350 f qui = 12911b CamScanner

taking moment about B = Ro (1829-3:26) aury 168.5et 350(1825-23) a RA 2 1196.42 loods can be The moments mts produces 2 Seperat by load atarting Equato to produced by a moment produced by (Vaignons theorem) resultant ment Produced by resu at ca 1196.42 * $22.5-3.25), 1291482 X2 2 166.11 - New CG2 182.5-166.11 = 16.38" Addingt 50 lb baggage is same apprect added. as clone for d following the - RARB = 350 eaurego = 134126 moment about B o a RA (182-5-3-25)= Quly 168.58 +350 (182.5-23) e 50/1825- SIRA = 12.35.33061 1026.29 1182.35-3.512 1341 X 43 7632 164.98 location, 182.5-164.98 - 12.5" New ca l to solve s Scanned wilgot CamScanner this given in we need question. specific weight of fuel, which Hence cannot solve this point

I . MAC in de 16:38 L267 1 h) al. In configuration : ď ca is Alt of datum 10 gallons a 83.5 l but following the same approach as presented in 'd RARg2 quis 35 150 1835 B = 1424.566 Moment about B Rex 10 9.250 aulk 168 'SP RAD > 350 159.5+ 50% 129.5 783-5 (182646) RA: 1325.00 lb BARAT 162-5= e< (21/5 = |4245 34 xu 2 166.96 Ca = 182.6-166.96 = 15.53 - / MAC 15.53 24.67 63 ca is All in on of datum configuration of Scanned with CamScanner