Prof (Dr) B Dayal

Prof (Dr) B Dayal

TANK TEST TRACTION CALCULATIONS

IMPLIES DETERMINATION AND ESTIMATION OF TRACTION AND DYNAMIC QUALITIES OF THE TANK.FOLLOWING CAN BE FOUND OUT:MAX SPEED UNDER THE GIVEN ROAD CONDITIONSRESISTANCE TO THE MOTION.MAXIMUM GRADIENT ANGLES CLIMBING CAPABILITY AT GIVEN GEAR AND SPEED.DOWN SLOPE CAPABILITY AT VARIOUS GEARS WITH ENGINE BRAKING.THE TIME AND DISTANCE TO ACHIEVE REQUIRED SPEED.BRAKES EFFICIENCYSELECTION OF GEAR RATIOS.

TANK TEST TRACTION CALCULATIONS

THESE CALCULATIONS ARE REQUIRED TO BE TAKEN IN FOLLOWING CASES:DETERMINATION OF TRACTION QUALITIES OF AN ACTUAL TANK.THE TRACTION QUALITIES OF AN ACTUAL TANK ARE DETERMINED AND ESTIMATED AND COMPARED WITH OTHER TANKS.THE CALCULATIONS CAN BE UNDERTAKEN BY TEST STAND AND RUNNING TESTSDETERMINATION OF TRACTION QUALITIES FOR TANK UNDER DESIGN.THE DESIGNER CAN ESTABLISH:OPTIMISATION OF POWER FOR TACTICAL AND TECHNICAL REQUIREMENTS.THEORETICAL ANALYSIS CAN BE UNDERTAKEN.

TANK TEST TRACTION CALCULATIONS

MAXIMUM CROSS COUNTRY SPEED DEPENDS UPON THE FOLLOWING:SPECIFIC POWER OF THE TANK.OPTIMISATION OF TRACK ASSEMBLY AND SUSPENSION.EFFECTIVE ROAD OBSERVATION.EFFECTIVE ENGINE COOLING SYSTEM TO AVOID OVER HEATING UNDER HEAVY LOADS AND CONTINUOUS EXPLOITATION.COMPARATIVE ANALYSIS OF THEORETICAL AND EXPERIMENTAL RESULTS.

TANK TRACTION CHARACTERISTICS

THE ENGINE TRACTION FORCE:

Pe = 270 Pfe t/vfe = Pe/G = 270 Pfe t/GvTHE VELUES REQUIRED FOR PLOTTING THE TRACTION CHARACTERISTICS ARE CALCULTED AS UNDER:ENTIRE RANGE OF ENGINE SPEED CHANGE FROM MINIMUM TO MAXIMUM VALUE IS DEVIDED IN TO INFINITE INTERVALS, i.e., IN 200 TO 300 RPM TO OBTAIN CALCULATING POINTS n1, n2, n3, ---- nN over engine speed.Corresponding values of engine power Pfe1, Pfe2, Pfe3, ----PfeN TO THE SELECTED ENGINE SPEED RANGES AS ABOVE, USING GRAPHICAL RELATION ON CHANGE OF ENGINE FREE POWER TO THE ABOVE SPEED POINTS WITH ENGINE OPERATING AT FULL FUEL SPEED.

TANK TRACTION CHARACTERISTICS

THE TANK SPEED AT EACH GEAR DEPENDING ON ENGINE SPEED IS DETERMINED AS GIVEN BELOW:v = n. tds. L. 60/itr.1000 = 0.06. n. tds. l/itrWHERE, n ENGINE SPEED PER MINUTE tds No of track links on the circumference of driving sprocket. l TRACK CHAIN PITCH IN M itr - GEAR RATIO OG TRANSMISSION AT GIVEN SPEED GEARTHE TANK SPEED CAN ALSO BE DETERMINED IF THE RADII OF SPROCKET IS KNOWNv = 2.. Rds. N.60/1000. itr = 0.377. rds.n/itr fds = 270 Pfe. tr/GvTHE TRACK ASSEMBLY EFFICIENCY:ta = 0.95 [1/fds](0.025 + 0.000003v2)THUS,t = tr. taTHE SPECIFIC TRACTION FORCE fe AT VARIOUS SPEEDS CAN BE CALCULATED:fe = fds. ta

TANK TRACTION CHARACTERISTICS

THE PROBLEMS THAT CAN BE SOLVED WITH TRACTION CHARACTERISTICSCASE 1: TO DETERMINE THE MAXIMUM SPEED AT WHICH TANK CAN MOVE UNIFORMLY WITH GIVEN RESISTANCE TO MOTION, UNDER A ROLLING RESISTANCE COEFFICIENT f ON SLOPE OF ANGLE Rr = (f cos + sin )GThe required traction force Pr equal to the engine traction force will bePe = feGUNDER UNIFORM MOTIONPe = RrfeG = (f cos + sin)Gfe = f cos + sin = fr

TANK TRACTION CHARACTERISTICS

THE PROBLEMS THAT CAN BE SOLVED WITH TRACTION CHARACTERISTICS (CONTD)CASE 2: TO DETERMINE THE RESISTANCE, WHICH CAN BE OVER COME BY TANK AT THE GIVEN SPEED GEAR AND SPEED DURING UNIFORM MOTION UNDER FOLLOWING:MOTION ON HORIZOTAL SECTIONCLIMBING AN UP SLOPEMOTION WITH A TRAILER.

TANK TRACTION CHARACTERISTICS

MOTION ON HORIZONTAL SECTION. SPECIFIC TRACTION FORCE fe IS DETERMINED FROM THE TRACTION CHARACTERISTICS WITH A PRESET SPEED VALUE AND GEAR.THE FORCE RESISTING THE MOTION IS:R = fG CLIMBING UP A SLOPE. fe = f cos + sin = fr THIS EQUATION IS SOLVED BY SUCCESSIVE APPROXIMATION STARTING WITH cos = 1fe = f + sin 1AND1 = sin-1(fe f)MOTION WITH A TRAILER.THE VALUE OF fe IS TAKEN FROM TRACTION CHARACTERISTICS AND THE RESISTANCE FORCE IS COMPUTED FROM THE FOLLOWING EQUATION:Rdr = Pe Rr = (fe fr)G

TANK BRAKING CHARACTERISTICS

THE BRAKING CHARACTERISTICS OF TANK DETERMINE THE TANK BRAKING UNDER INTERNAL RESISTANCE OF ITS UNITS AND MECHANISMS WITH ONLY ENGINE OPERATING AND WITHOUT APPLICATION OF BRAKES.IT DEPENDS ON SPECIFIC BRAKING FORCE fbfb = Tb/GWHERE,Tb TANK BRAKING FORCE AT TRACKS DUE TO THE RESISTANCE IN INTERNAL UNITS AND MECHANISMS OF TANK WITH ENGINE UNDER BRAKING.G TANK WEIGHTTb = 2700 Pt / v Where,Pt TOTAL POWER OF INTERNAL RESISTANCE IN THE UNITS AND MECHANISMS OF THE TANK AT A SPEED v

TANK BRAKING CHARACTERISTICS

AND Pt = Pec + Pf + Ptr + PrgTHE POWER SPENT IN ENGINE (Pec). Pec = Mec. N = PEML . V1. n/60FOR EACH VALUE OF PEML AND ENGINE SPEED n, THE CORRESPONDING AVERAGE PISTON SPEED vap ISvap = Sp.n/30THE MEDIUM PRESSURE OF INTERNAL LOSSES ON THE PISTON SPEED IN CASE OF T-55 ENGINE, IN THE RANGE OF ENGINE SPEED CHANGE FROM 500 TO 2000 RPM IS:PEML = 0.65 + 0.062 vp ( IN bar)WHERE,Mec - ENGINE RESISTANCE MOMENT AT VARIOUS SPEEDSPEML MEDIUM PRESSURE OF ENGINE INTERNAL LOSSESv1 PISTON DISPLACEMENTSp STROKE LENGTH

TANK BRAKING CHARACTERISTICS

BRAKING CHARACTERISTICS AT DOWN SLOPEWE HAVETb + Rgr = G sin WHERE,Rgr = fgr G cos Rgr GROUND RESISTANCETHEREFORE,Sin = fb + fgr cos ASSUMING COS = 1 = SIN-1 (fb + fgr)IF THE GROUND RESISTANCE IS NEGLECTED, THE DOWN SLOPE ANGLE WILL BE = SIN-1 fb UTILISING THE GRAPH ON T-55 BRAKING CHARACTERISTICS AND ALSO THE ABOVE EQUATIONS ON DOWN SLOPE ANGLE, WE CAN ARRIVE AT THE DOWN SLOPE ANGLES & THAT CAN BE NEGOTIATED BY THE TANK, BRAKING FORCE fb AT ANY SPEED AND IN A PARTICULAR GEAR AND ALSO THE OTHER WAY.