Aare Aan, Mati Heinloo Estonian University of Life Sciences ...

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ENGINEERING FOR RURAL DEVELOPMENT Jelgava, 24.-25.05.2012.Basic formulasIf the curve is represented by the parametric equations x = x(t), y = y(t), then the radius ofcurvature at the point M (x(t), y(t)) is [20]R( t)=ay3v( t)( t) ⋅v( t) − a ( t) ⋅v( t)xxy(1)The centre of curvature co-ordinates x and y are α(t) and β(t),respectivelyαβ( t) = x( t)( t) = y( t)−ay+a2v( t) ⋅ vy( t),( t) ⋅ vx( t) − vy( t) ⋅ ax( t)v( t) 2⋅ vx( t)( t) ⋅ v ( t) − v ( t) ⋅ a ( t) .yxyx(2)In (1) and (2)vaxdd= ,y(3)dt dt( t) x( t) v ( t) y( t)x=22dd=y(4)dtdt( t) x( t) , a ( t) = y( t)v( t) v ( t) 2 v ( t) 2= (5)x+Let us suppose now that a material point M is moving according to the lawx t = A⋅cosω ⋅ty( ) ( )( t) = B ⋅sin( ω ⋅t)where a, b and ω are constants and t is the time.The formulas (6) determine the trajectory of the point M. The formulas (3) and (4) determine theprojections of velocity and acceleration of the point M on the x – and y – co-ordinate axes of the coordinatesystem O xy . The formulas (5) represent the modules of the vectors of velocity of the point M.The direction angles between the x-axis and vectors of velocity v ( t)and acceleration ( t)αv( t) = angle( vx( t) , vy( t))α ( t) = angle a ( t) , a ( t)ay( )xya arewhere angle(x, y) returns on the worksheet of Mathcad in the direction angle (in radians) of avector.The direction angle of the vector, directed from the point M to the centre of curvature, iswherep( t) = angle( α( t) − x( t) β ( t) − y( t))α ,a( t) = a ( t) 2 a ( t) 2x+The circle of curvature is determined by the following parametric equationsxykk( t) = α(t) + R( t) ⋅cos( q)( t) = β ( t) + R( t) ⋅sin( q)y(6)205
ENGINEERING FOR RURAL DEVELOPMENT Jelgava, 24.-25.05.2012.where q = 0, 1, 2,…2π. The tangential acceleration is defined by the formula [22]( t) ⋅vx( t) + ay( t) ⋅ vy( t)v( t)and the normal acceleration is defined by the formula( ) ( )( α( t) − x( t))ant = axt ⋅ + ayR taxaτ ( t)=(7)( )206( t)The columns F(t, γ, b, n) (1) and F(t, γ, b, n) (2) of matrix functionF( t , γ , b , n) := x ← x( t)y ←y ( t)⎛V 0 ← ⎜⎝⎛I ← ⎜⎝Ω←⎛⎜⎝00xyxy0.850xycos( γ ( t))xy0.850.03−sin( γ ( t))V 0 + n⋅b( t)⋅I⋅Ωxy⋅10( β ( t) − y( t))R( t)Tx ⎞⎟y ⎠0.85−0.03sin( γ ( t)) ⎞ ⎟⎠cos( γ ( t))T0.85 ⎞ ⎟⎠0. (8)of Mathcad [21] can be used to simulate the movement of a geometric vector b r on Mathcadworksheet. In the function F(t, γ, b, n) the following notations are used: x = x(t) and y = y(t) are the coordinatesof the origin, γ = γ (t) is the direction angle, b(t) is the module or projection (if the vector b ris determined only by one co-ordinate), and b is notation of b(t) of a vector b r ; m is the coefficient ofdimensions and I is the matrix of the shape of a vector b r on Mathcad worksheet; Ω is the matrix oftransformation of rotation, V 0 is the matrix of origin, T denotes the transposed matrix, t is the time.At the fixed moment of time t = T the columns of the matrix function F(t, γ, b, n)columnsVv( T ) ( 0) F( T, α , v,0.2) ( 0), V ( T ) ( 1)= F( T,α , v,0.2 ) ( 1)where v is the notation of module ( ) ( ) 2Va= ,vv2vytvvx t + , visualize the vector of velocity, the columns( T ) ( 0) F( T, α , v,0.1) ( 0), V ( T ) ( 1)= F( T,α , v,0.1 ) ( 1)= ,where a is the notation of module ( ) ( ) 2Vanaa2 aytaax t + visualize the vector of acceleration, the( T ) ( 0) F( T α , a ,0.1) ( 0), V ( T ) ( 1)= F( T,α , a ,0. 1 ) ( 1)= ,,p nanp nwhere a n is determined by the formula (8), visualize the vector of normal acceleration, thecolumnsV( T ) ( 0) F( T α , a ,0.1) ( 0), V ( T ) ( 1)F( T,α , a ,0. 1 ) ( 1)a τ,v τan=v τ= ,where a τ is determined by the formula (7), visualize the vector of tangential acceleration, thecolumnsVap( T ) ( 0) F( T, α , R,1) ( 0), V ( T ) ( 1)= F( T,α , R,1 ) ( 1)= ,pwhere R is determined by the formula (1), visualize the vector that begins from the point M and isdirected towards the centre of curvature (Fig. 1 and Fig. 2).app
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