Prof (Dr) B Prof (Dr) B DayalDayal

Prof (Dr) B Prof (Dr) B DayalDayal

MATHEMATICAL FORMULATION OF THE EQUATION

ASSUME TANK AS A SOLID BODY AND MOVING ON A SLOPE

LET THE ACCELERATION BE X’’DRAWBAR RESISTANCE BE Rdr RESOLVING ALL THE EXTERNAL FORCES ACTING

ON THE TANK TO ITS LONGITUDINAL AXIS PARALLEL TO THE SURFACE OF MOTION

P = R + G SINα + Rdr + Mx’’ (1)N – G COS α = 0WHERE M = MASS OF THE TANK = G/gAS R = fN = fG cosα EQN (1) CAN BE WRITTEN

AS P = G (sin α + f cos α) + Rdr + mx’’

MATHEMATICAL FORMULATION OF THE EQUATION (CONTD)

SINCE sin α + f cos α = fr WHERE fr IS THE TOTAL

SPECIFIC RESISTANCEP = Gfr + Rdr + mx’’x’’ = (P - Gfr – Rdr)/m

THIS EQUATION IS THE DIFFERENTIAL EQUATION OF MOTION OF TANK CENTRE OF GRAVITY

IT CONTAINS THREE UNKNOWNS: x’’, P AND fr IN CASE OF TANK MOTION WITHOUT DRAW BAR

RESISTANCE, x’’ = (P – Gfr)/m WITH TWO UNKNOWNS

LAW OF KINETIC ENERGY“ THE CHANGE OF SYSTEM KINETIC ENERGY IS

EQUAL TO THE SUM OF THE WORK OF ALL THE EXTERNAL AND INTERNAL FORCES”

dTKE = dWeng + dWg + dWr

where, Tke – Differential of tank kinetic energydWeng – elemental work of engine moving

forcesdWg – elemental work of gravity forcedWr – elemental work of all resisting forces

TANK KINETIC ENERGYKINETIC ENERGY OF TANK MOVING AT SPEED ‘V’ IS EQUAL

TOTKE = T1 + T2

WHERE, T1 – KE OF TANK IN TRANSIT MOTIONT2 – KE OF TANK PARTS IN VOLVED IN THE RELATIVE MOTION AND GEARED TO THE TRACK ASSEMBLYT1 = Mv2/2T2 = (mcv2)/2 + Σ (Iiωi

2)/2WHERE, mc – TRACK CHAIN MASS

Ii – INERTIA MOMENT OF ROTATING MASS iωi – ANGULAR MOVEMENT OF PART i = v igr/rds

WHERE, igr – GEAR RATIO BETWEEN THE PARTS CONCERNED AND SPROCKET

DIFFERENTIAL OF TANK KINETIC ENERGY igr = rds/ri ri – WHEEL OR ROLLER RADIUSTHEREFORE,

T2 = (mcv2)/2 + (v2/2rds2) ΣIi igr

2

Thus ke of the tank will beTKE = T1 + T2 = mv2/2 + v2/2 [mc + (1/rds

2)ΣIi igr2]

= mv2/2 [1 + mc/m + (1/mrds2)ΣIi igr

2]ASSUME S’ = 1 + mc/m + (1/mrds

2)ΣIi igr2

S’ CAN BE ASSUMED AS THE COEFFICIENT OF TANK MASS CONDITIONAL INCREAMENT FOR THE INCREASE IN KE.

TKE = S’ mv2/2AND dTKE = S’mvdv (3)

S’ = Sc + αitr2

WHERE, Sc – CONSTANTitr – GEAR RATIO BETWEEN THE GIVEN

PART AND THE SPROCKET. IT IS CONSTANT AND INDEPENDENT OF GEAR ENGAGED IN THE GEAR BOX.

α – Ie/mrds2 THIS VALUE FOR A GIVEN

DESIGN IS CONSTANTIe- TOTAL MOMENT OF INERTIA OF

PARTS BETWEEN THE ENGINE AND THE GEAR BOX.WHEN CLUTCH IS DISENGAGED

SO’ = SS’ + α0 Itr2

WHERE, α0 = Iel/mrds2

ELEMENTAL WORK OF TANK MOVING FORCESdweng = 746 Pfe dt (4)

WHERE, Pfe – ENGINE FREE POWER in HPELEMENTAL WORK OF GRAVITY FORCE (dwg)Dwg = -GSinα dx (5)Where, dx – ELEMENTAL DISTANCE TRAVELLED BY THE

TANK IN TIME dtELEMENTAL WORK OF ALL FORCES RESISTING THE TANK

MOTION.Dwr = dwif + dwef (6)WHERE, dwif - ELEMENTAL WORK OF INTERNAL FORCES

RESISTING MOTION= - 746 Pfe (1 – ηtr) dt

dwef – ELEMENTAL WORK OF THE EXTERNAL FORCES RESISTING MOTION

= -fG Cosα dx – Rdr dx

CONSIDERING EQUATIONS 2, 3, 4, 5 & 6S’ mv dv = 746 Pfe dt – G Sin α dx – 746 Pfe (1- ηtr )dt – fGCos α

dx – Rdr dxTHUS, S’ mv dv = 746 Pfeηtrdt – frG dx – RdrdxWHERE, fr = Sinα + f CosαBY DEVIDING THE EQWUATION WITH dtS’mv dv/dt = 746 Pfeηtr – frG dx/dt – Rdrdx/dtS’mv x’’ = 746 Pfeηtr – frG v – Rdr vTHEREFORE,

X’’ = (746 Pfeηtr/v – frG – Rdr)/S’mIF v EXPRESSED IN KMPH:X’’ = (2685.6 Pfeηtr/v – frG – Rdr)/S’m ≈ (2700 Pfeηtr/v – frG –

Rdr)/S’m FOR MOVE WITHOUT TRAILER:X’’ = (2700 Pfeηtr/v – frG)/S’mOR X’’ = g(2700 Pfeηtr/Gv – fr)/S’

REQUIRED TRACTION FORCE (Pr):Pr = frG + Rdr + mx’’Pr = frG + Rdr APPLICABLE FOR UNIFORM

MOTION WITH TRAILERPr = frG APPLICABLE FOR UNIFORM MOTION

WITHOUT TRAILER

P = (270 Pfeηtr/v – (S’ – 1)mx’’REQUIRED ENGINE TRACTION FORCE

Pe = 270 Pfeηt/vREQUIRED ADHESION FORCE (Pad)Pad = Ø G Cosα

ACCELERATED OR UNIFORM MOTION:Pad ≥ Pe ≥ Pr

REDUCED ENGINE SPEED / STALLING: Pad ≥ Pr ≥ Pe

TRACK SLIDING: Pad < Pr < Pe