text: Home > Books > Archery: The Technical Side >
  The Dynamics of a Bow and Arrow
  Part 2 of 5
  
  Neglecting the dissipation in heat and sound, the gain in kinetic energy is equal to the loss in potential energy. Therefore:
  
  MV2/2 +mv2/2=C(A'N'-AN).
  
  But V=Rv. Therefore:
  
  v = [2C(A'N'-AN)/(MR2+m)]1/2
  
  But N=3B1A/4. Therefore:
  
  V = [3B1C(A'2-A2)/2(MR2 + m)]1/2.
  
  And
  
  V=Rv=R[3B1C(A'2-A2)/2(MR2 + m)]1/2.
  
  Let a equal the acceleration of the bow tip and A the acceleration of the arrow. (Bold face type will be used to indicate dynamic values of variables.)
  
  a=dv/dt=(dv/dA) (dA/dt) = (dv/dA) (dA/dN) (dN/dt).
  
  But dA/dN=4/3B1 and dN/dt= -v. Therefore:
  
  a=-(4v/3B1) (dv/dA). Therefore by differentiation:
  
  a=2CA/(MR2 + m)-2MRC(A'2-A2)Z/(MR2 + m)2,
  
  where
  
  Z= sin A-(Y/P) cos A+ (3B1S2/4P3) sin2 A.
  
  Also A=dV/dt=d(Rv)/dt=vdR/dt+Rdv/dt.
  
  dR/dt= (dR/dA) (dA/dt)
  
  = (dR/dA) (dA/dN) (dN/dt) = - (4v/B1)dR/dt.
  
  Therefore
  
  A=2CAR/(MR2+m)+2mC(A'2-A2)Z/(MR2+m)2.
  
  The dynamical force f on the tip is: f=ma/2.
  
  The dynamical force F on the arrow is:
  
  F=MA.
  
  T=F/2 sin E=SF/2P=SMA/2P= (f-f) /sin (A + E).
  
  If the angle A is expressed in radians, the masses in pounds, the lengths in feet and the forces in poundals, the velocity and acceleration will be obtained in feet per second and feet per second per second, respectively.
explanation: The document is a technical excerpt from a book on archery, specifically discussing the dynamics of a bow and arrow. It contains several mathematical equations and derivations related to the kinetic and potential energy, velocity, acceleration, and forces involved in the bow and arrow system. The text does not contain a specific question or a question and answer.
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distilabel_metadata.content: Home > Books > Archery: The Technical Side >
  The Dynamics of a Bow and Arrow
  Part 2 of 5
  
  Neglecting the dissipation in heat and sound, the gain in kinetic energy is equal to the loss in potential energy. Therefore:
  
  MV2/2 +mv2/2=C(A'N'-AN).
  
  But V=Rv. Therefore:
  
  v = [2C(A'N'-AN)/(MR2+m)]1/2
  
  But N=3B1A/4. Therefore:
  
  V = [3B1C(A'2-A2)/2(MR2 + m)]1/2.
  
  And
  
  V=Rv=R[3B1C(A'2-A2)/2(MR2 + m)]1/2.
  
  Let a equal the acceleration of the bow tip and A the acceleration of the arrow. (Bold face type will be used to indicate dynamic values of variables.)
  
  a=dv/dt=(dv/dA) (dA/dt) = (dv/dA) (dA/dN) (dN/dt).
  
  But dA/dN=4/3B1 and dN/dt= -v. Therefore:
  
  a=-(4v/3B1) (dv/dA). Therefore by differentiation:
  
  a=2CA/(MR2 + m)-2MRC(A'2-A2)Z/(MR2 + m)2,
  
  where
  
  Z= sin A-(Y/P) cos A+ (3B1S2/4P3) sin2 A.
  
  Also A=dV/dt=d(Rv)/dt=vdR/dt+Rdv/dt.
  
  dR/dt= (dR/dA) (dA/dt)
  
  = (dR/dA) (dA/dN) (dN/dt) = - (4v/B1)dR/dt.
  
  Therefore
  
  A=2CAR/(MR2+m)+2mC(A'2-A2)Z/(MR2+m)2.
  
  The dynamical force f on the tip is: f=ma/2.
  
  The dynamical force F on the arrow is:
  
  F=MA.
  
  T=F/2 sin E=SF/2P=SMA/2P= (f-f) /sin (A + E).
  
  If the angle A is expressed in radians, the masses in pounds, the lengths in feet and the forces in poundals, the velocity and acceleration will be obtained in feet per second and feet per second per second, respectively.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a technical excerpt from a book on archery, specifically discussing the dynamics of a bow and arrow. It contains several mathematical equations and derivations related to the kinetic and potential energy, velocity, acceleration, and forces involved in the bow and arrow system. The text does not contain a specific question or a question and answer.
  
  Label: no-question