The Science of Sport- Part I

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Science Of Archery — Archer’s Paradox

Archer’s paradox is a very interesting physical phenomenon in archery. When the arrow is released to the left (or right) of a bow and is Arch 1deliberately aimed off target, it will straighten out during release and hit the target. This paradoxical situation is known as Archer’s paradox. The figure below illustrates this.

Due to the relatively high force of the bow string acting on the arrow in the direction of the arrow shaft, it begins to oscillate (fishtail) back and forth. This is the natural vibration the arrow experiences when subjected to a brief but large force. The force acting on the arrow is equal to the draw force as the archer pulls the string back to the release position.

Thus, the physics of archery is more than just shooting an arrow at the target. One must also account for the oscillation of the arrow during the release. In the explanation that follows, more details will be given on what occurs when the arrow is released and why it straightens out. Initial set up, as the archer pulls the string back and is ready to release it.

First Stage of Archer’s Paradox

The figure below shows the arrow immediately after release.
In this first stage three things happen:

Arch 21. As he release happens the string slides off the fingers and moves left, it would be impossible for an archer to open his fingers fast enough to prevent the string from sliding off to the side a bit. For getting the arrow to fly straight it is desirable that the string initially deflects to the left a bit.

  1. The leftward motion of the string excites a mode of oscillation where the arrow begins to vibrate in the plane of the page. The string force F1contributes to the deflection of the arrow δ beyond that due to the leftward motion of the string by itself. Note that F1acts in the direction of the arrow shaft.3. The arrow is contacting the bow at point P due to a clockwise moment, which can be taken about the center of mass G (of the arrow) at the instant shown. This moment is equal to F1r1, where r1 is the moment arm F1makes with G.Second Stage of Archer’s Paradox

During this second stage, two things happen:

1. The string moves right, as it restores its original position with the median plane of the bow. As a result, the string “pulls” on the arrow with a force F2.  Arch 42. The tip of the arrow T moves slightly to the left. This is due to a counter-clockwise moment taken about the center of mass G at the instant shown. This moment is equal to (F1r1 + F2r2), where r2 is the moment arm F2 makes with G.

In the third and final stage illustrated , the arrowart work-04 exits the bow completely, having completed (approx) one full oscillation. The arrow is now flying straight to the target. It will continue oscillating all the way to the target, with oscillation gradually decreasing in amplitude, but maintaining the same frequency throughout the flight.

 

 

 

 

*researched content from Google and real world physics problems

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