Arrow Pitching during the Power Stroke

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The discussion focuses on modeling the acceleration and launch angle of an arrow during the power stroke of a bow. It highlights how the nock position affects the arrow's pitch and impact point on the target, emphasizing that the nock's horizontal movement induces angular rotation and reduces vertical velocity. The conversation includes equations for angular acceleration and horizontal forces, noting that gravity plays a significant role in the torque experienced by the arrow. There is a concern about the omission of applied force in the angular acceleration calculations, prompting a request for feedback on the modeling approach. The importance of understanding the dynamics of the power stroke in archery physics is underscored, as it is often overlooked despite its critical role in performance.
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I am working on modeling the acceleration and initial launch angle of an arrow shot from a bow during the power stroke. From an aiming point of view, the launch angle of the arrow doesn't change with respect to the target. But changing the nock position where the arrow nock attaches to the string does change where the arrow strikes the target. The nock position causes the arrow to pitch forward during the power stroke. It will acquire an angular rotation, and its vertical component of the initial velocity will be reduced.
I am using that actual measured force as a function of the draw length, and it is being assumed that the nock point is constrained and only moves/accelerates in the horizontal plane. The other assumption is that at the instant the arrow is loosed, there are no vertical supports holding the arrow in position, other that the nock end of the arrow on the string The nock point will set an initial angle of the arrow with respect to the horizontal plane. Intuition states that because of the initial angle present, the horizontal force on the end of the arrow will induce a torque into the arrow. As the arrow accelerates, the gravitational pull on the arrow also adds to the torque of the arrow.

When using the nock end as my reference. The angular acceleration becomes only a function of gravity.
Inertia*alpha=-r*m*at-m*g*cos(theta)
where Inertia is the inertia of the arrow. r is the distance from the nock to the center of gravity, m and the arrow mass, (at) is the tangential acceleration of the center of gravity relative to the nock end of the arrow, g is gravity, and theta is the angle of the arrow shaft with respect to the horizontal plane.
Then the summation of forces in the horizontal direction
m*ax=F(x)+m*r*omega^2*cos(theta)-m*r*alpha*sin(theta)
where ax is the horizontal acceleration, F(x) is the measured draw force as a function of x, omega is the angular velocity. Acceleration is the y direction (vertical) at the nock is zero since it is constrained from vertical movement. F(x) is non linear, and goes from 48 pounds to zero in 19 inches.

Intuition states that angular acceleration should also include the applied force. But it is dropping out. I think I am doing something wrong here. Thoughts, comments?
 
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Just a suggestion for an alternate to your current calculation method (As I understand it).

With respect to your statement "Intuition states that because of the initial angle present, the horizontal force on the end of the arrow will induce a torque into the arrow." The application point of any vertical force due to gravity to induce any vertical torque on the arrow will be at the center of gravity of the arrow, not necessarily at its tip.

Assuming the arrow launch elevation and target elevation are the same, for a simple set of equations for the trajectory of your arrow see the below website
https://formulas.tutorvista.com/physics/trajectory-formula.html (these do not account for range loss due to air drag on the arrow in flight)
(You can calculate the elevation angle required for a given target distance by rearranging the Range equation to solve for θ with R as an input)
If the arrow launch elevation is not the same as the target elevation then the calculation gets a bit more complicated.

As for the launch velocity of the arrow:
V = a * t
Assuming the launch force change is directly proportional to the launch travel
a = F / mass of arrow where F = the average of the pull force at full pull and at the arrow release point,
t = sqrt ( 2 x pull distance / a)
 
Thanks for the reply.
My interest is in the power stroke to generate that initial velocity and launch angle. This is the part of archery physics that is utterly ignored, and yet it is the most critical part of archery.

The reason I was looking at the pitch from the tail end of the arrow, nock side, is that during power stroke, the nock is constrained in the vertical plane by the string. It moves only horizontally.

High speed video of compound and Olympic recurve archery clearly shows that arrow is not supported by the arrow rest shortly after the string is let go, and for most of the power stroke the arrow is floating in the air being accelerated horizontally. This is why I am looking at this as a rod floating in space with a force on one end, and gravity acting on the CG.

When the force vector doesn't act through the center of the rod, it will contribute to the torques on the rod. I am just struggling with the correct way to model this because of the constraints.
 
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