Next, we assume an error in aiming, and shall take 12 minutes of angle (1-10 inch at the tip of a 28 inch arrow) as a reasonable value. We must then determine where the trajectory intersects the target if the angle A is changed by this amount. This can be done by calculus, and the results tabulated in Table I were found in this manner. It shows the vertical and lateral distances of hits from center of target resulting from an error of aiming of 12 minutes (1-5 degree) in the initial angle.
| TABLE I | |||||||
| Arrow velocity | |||||||
| 130 f.p.s. | 150 f.p.s. | 170 f.p.s. | |||||
| Target Distance, Yds. | Initial Angle | Vertical error, in. | Initial Angle | Error on target, in. | Initial Angle | Error on target, in. | Lateral error on target, in. |
| 40 | 6° 36' | 5.0 | 4° 57' | 5.0 | 3° 51' | 5.0 | 5.8 |
| 50 | 8° 18' | 6.1 | 6° 12' | 6.2 | 4° 54' | 6.2 | 6.4 |
| 60 | 10° 0' | 7.3 | 7° 28' | 7.4 | 5° 45' | 7.4 | 7.7 |
| 80 | 13° 36' | 9.4 | 10° 03' | 9.7 | 7° 41' | 10.0 | 9.1 |
This table reveals the interesting fact that the vertical and horizontal dispersions (distances from center of target) are not far from being the same, at any given target distance for the same angular error in aiming. It also shows that the vertical dispersion is almost independent of arrow velocity. As shown before, the horizontal dispersion is entirely so. A shift of 0.1 inch, in any direction from the correct point-of-aim, of the tip of a 28-inch arrow, with the anchor fixed, means a hit just outside of the gold at 40 yards; at 50 and 60, the hits are in the middle and near the outer red, respectively; at 80, in the blue. A quarter-inch shift means a blue at 40; blue or black at 50; black at 60; miss at 80:
The next question of interest to the archer is, "How much shift in my point of aim, measured along the ground, or measured vertically at the point of aim, corresponds to an aiming error of one-fifth of a degree? "
This question can also be answered by making use of calculus and trigonometry. The method is to find, first, what change is brought about in the angle between the line of sight and the ground by a change in the initial angle of the arrow with the horizontal. To make the problem concrete, we then use certain data corresponding to the practical cases and make our computations. For an archer of medium height, shooting a 28-inch arrow with a velocity of 150 feet per second, having his anchor 5.5 inches below his eye, his points-of-aim are found to be located as shown in the second column of Table II, in yards from the shooting line.
| TABLE II | ||||
| Target, yards | Point-of-aim, yards | β | Error in yards, horizontal | Error in inches, vertical |
| 40 | 16 | 6° 49' | 0.5 | 2.6 |
| 50 | 20 | 5° 32' | 1.0 | 2.8 |
| 60 | 25.5 | 4° 20' | 1.5 | 3.5 |
| 80 | 64. 5 | 1° 53' | 8.5 | 9.6 |
| Column I | Target distance |
| Column II | Distance of point-of-aim from shooting line. |
| Column III | Angle between the ground and the archer's line of sight at the correct point-of-aim. |
| Column IV | Distance in yards from true point-of-aim, measured along the ground, corresponding to an error of 12 minutes in the angle of departure resulting from incorrect aiming. |
| Column V | Distance in inches, measured vertically from the true point-of-aim, corresponding to the same error in aiming. |
This table enables the archer to determine with some assurance how much allowance to make at his point-of-aim for arrows whose dispersion errors arc known. If he uses a sight, he is in better position to make his corrections in aiming, because all such corrections are made directly on the target. In this case the error in the hit is measured exactly by the error in aiming, because, with a correctly adjusted sight, the point on the target on which the sighting is done is the point which the arrow will hit.