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Rifles, Reloading, Optics, Equipment
Reloading
Advice Needed on New Press.
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<blockquote data-quote="Winchester 69" data-source="post: 357123" data-attributes="member: 8037"><p>I reiterate: </p><p></p><p><em>When the ram is aligned with the die and held in a tight bore, all of its force is likewise directed. In a loose bore, the ram will want to misalign when it meets resistance, simply as a tendency to reduce the resistance. Oppose two fingers linearly and push them into one another. As the fingers deflect off-axis, resistance is reduced. The ram wants to do the same thing. Let's consider that the ram is cocked in its bore. Realize that the force into the die is reduced somewhat. The ram's full force is on the axis of the ram. Think of the ram and its force as the hypotenuse of a right triangle. The leg of the triangle that is aligned with the die represents by its length the force being transmitted into the die on its axis. The other leg of the triangle represents by its length the amount of lateral force that is being resisted by the die, and any brass that is in it, at a right angle to the die. That is the force that is deleterious to to the straightness and otherwise dimensional integrity of a shell case. The greater the deflection, the larger the proportion of the ram's force that is transmitted laterally.</em> </p><p></p><p>Compliance in the shell holder allows the brass to align itself in the die, but the lateral force remains. </p><p></p><p>Your description of the functioning of hand-dies doesn't concur with that in the <u>Precision Reloading & Shooting Handbook, 10th Ed.</u> (Gravett & Sinclair).</p></blockquote><p></p>
[QUOTE="Winchester 69, post: 357123, member: 8037"] I reiterate: [i]When the ram is aligned with the die and held in a tight bore, all of its force is likewise directed. In a loose bore, the ram will want to misalign when it meets resistance, simply as a tendency to reduce the resistance. Oppose two fingers linearly and push them into one another. As the fingers deflect off-axis, resistance is reduced. The ram wants to do the same thing. Let's consider that the ram is cocked in its bore. Realize that the force into the die is reduced somewhat. The ram's full force is on the axis of the ram. Think of the ram and its force as the hypotenuse of a right triangle. The leg of the triangle that is aligned with the die represents by its length the force being transmitted into the die on its axis. The other leg of the triangle represents by its length the amount of lateral force that is being resisted by the die, and any brass that is in it, at a right angle to the die. That is the force that is deleterious to to the straightness and otherwise dimensional integrity of a shell case. The greater the deflection, the larger the proportion of the ram's force that is transmitted laterally.[/i] Compliance in the shell holder allows the brass to align itself in the die, but the lateral force remains. Your description of the functioning of hand-dies doesn't concur with that in the [u]Precision Reloading & Shooting Handbook, 10th Ed.[/u] (Gravett & Sinclair). [/QUOTE]
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