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Several years ago, I posted a question about Why do archery arrows tilt downwards in their descent?. An answer was given that a torque arises from the difference in location of net force of gravity (center of mass) vs net force of drag (center of arrow). This answer makes sense to me.

Now I'm wondering why arrows always seem to point tangential to their path. I have never shot a real arrow, but it is my perception that arrows points horizontally at the top of their arc, regardless of initial velocity. This idea seems in conflict with the idea that rotating objects rotate more if they have more time to rotate. An arrow show with more initial velocity would seem to have more time to rotate.

I have a suspicion that as the arrow starts to rotate after initial launch, the increasing air resistance on the arrow from above slows the rotation exponentially. First the arrow is burrowing through the air point-first, but later it is horizontal and experiencing much more drag until it comes to a momentary pause. Thus I suspect that the maximum height an arrow reaches is highly dependent on the amount of drag, even though superficially arrows may appear to cut through the air. Is it naive to treat a rotating arrow as an ordinary projectile, since at maximum height, it is experiencing drag forces much larger than were present at the time of launch?

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  • $\begingroup$ The torque due to drag is zero when the arrow points in the same direction as its velocity, and (assuming the air is still) the direction of the velocity is always a tangent to the trajectory. Or have I missed something that makes it more complicated than this? $\endgroup$ May 20, 2022 at 6:39

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Gravity can't apply a torque to an object in freefall.

Drag/air pressure always acts in the opposite direction to the object's path through the air. For still air, this will be the opposite direction to the object's path relative to the ground.

The object will be designed so the center of pressure is behind the center of mass. The drag from the air will provide a torque so the center of pressure trails the object as it moves through the air. Since the arrow is moving horizontally at the top of the arc, it is also aligned horizontally.

The rotational moment of inertia of an arrow is easily overcome by the aerodynamic forces. I don't see any reason to assume the drag forces are greater at the top of the arc.

A rotation is introduced by the force of drag,

It might be introduced. The arrow is basically a weathervane. If it's pointed into the wind, it stays there. If it's not pointed into the wind, the pressure from the wind creates a torque that turns it that way.

As the arrow moves through the air, if it's not aligned with the velocity vector, it will experience a torque that tends to align it.

so why couldn't the arrow be pointing down at the top of the arc, given appropriate v initial and drag force?

Given you specify "top of arc", then initial velocity is irrelevant. The top of the arc is when the arrow has no vertical velocity. The amount of drag is irrelevant, just the direction (which given the description must be horizontal).

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  • $\begingroup$ thanks for reply. My question is really why does the arrow have to be pointing horizontally at the top of the arc, or why does the arrow have to be pointing in the direction it is travelling. A rotation is introduced by the force of drag, so why couldn't the arrow be pointing down at the top of the arc, given appropriate v initial and drag force? $\endgroup$
    – lamplamp
    May 20, 2022 at 6:35
  • $\begingroup$ Drag force is opposite direction of motion. At top of arc, it is moving horizontally. So the angle will be horizontal. $\endgroup$
    – BowlOfRed
    May 20, 2022 at 6:37
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    $\begingroup$ correction: a uniform gravitational field cannot exert a torque on a free-falling object. A non-uniform field can. $\endgroup$
    – Rd Basha
    May 20, 2022 at 6:45
  • $\begingroup$ Some (maybe most, for modern-day) arrows are definitely fletched such that they spin in flight. I think it might have more to do with damping the vibration amplitude than directional stabilization, though. I'm not at all an expert. $\endgroup$
    – g s
    May 20, 2022 at 7:54
  • $\begingroup$ I would have said that an arrow is like a glider. If you shoot it close enough to vertical, at some point it will stall, and then, the fletch drag exceeding the point drag, you'll get some tumbling until its speed downwards exceeds stall speed. $\endgroup$ May 20, 2022 at 13:30
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Fletchings are little fins on the tail of an arrow that provide very little drag when the arrow is going straight into the wind, and quite a bit more drag when not. When I say wind, I mean the vector addition of the prevailing atmospheric movement (classic wind) and the resultant drag of the arrow moving through calm air (as the wind in your hair when you ride a bike on a calm day). Archery is easier when the wind is calm, but arrows travel at 60 to 90 meters per second which is significantly faster than the wind most of the time.

That drag, acting on the back of the arrow, torques the arrow back to the direction of travel. To first approximation, the drag force on the fletchings is proportional to the difference in angle of the arrow to the direction of travel (through the air) which means the arrow "springs" back to "straight."

Arrows point in the direction of travel because of fletchings.

One may also fletch their bow such that the arrow is apt to rotate (offset or helical fletching). Archers use this fletching so that the arrow straightens out more quickly and crosswinds don't impact the arrow's flight as much.

Before doing a simulation, you must think about what you want from it. If you want to know where the arrow lands, with any fletching, a "naive" simulation (e.g., just using gravity) is probably fine. If you need a bit more precision, the next term to introduce would be a constant drag force (as in $\vec F_d = -b\, \vec v$) on the arrow (which would be slightly more for offset and more again for helical fletching). If you are making a video game and need to show the arrow coming off with some un-intended rotation and then righting itself, you'd need to simulate the restorative torques in addition to the arrow's flight.

At the top of a very tall arc (when the arrow is moving the slowest), the helical fletched arrow's rolling momentum might push air around a bit, but the net force on the arrow would be in the direction of travel, so that wouldn't actually be a drag force. The restorative drag forces would be at their smallest because we expect the arrow to be both nearly pointed in the direction of travel and the arrow to be traveling at its slowest.

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    $\begingroup$ This is a better answer than the one selected by the asker $\endgroup$
    – Al Brown
    Sep 8, 2023 at 6:36

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