Monthly Archives: May 2015

In the main body of this post, I work through a detailed example from "Modern Practical Ballistics" on how to apply Pejsa's formulas to determine a range table for a projectile moving through a wide range of velocities. In Appendix A, I work through a second example from the text as additional validation of my implementation. In Appendix B, I show how to use an Ordinary Differential Equation (ODE) solver to generate a range table for comparison with Pejsa's algebraic approximations. Continue reading →

This post will cover Pejsa's formula for the trajectory midpoint as a function of the rifle's zero range. Shooters often have a preferred zero range, like 100 yards or 200 yards. This formula allows the shooter to determine his midpoint range directly from the zero range. The midpoint range can then be used to determine the maximum bullet height above the line of sight, which can be used to determine the maximum bullet placement error. Continue reading →

Pejsa defines the trajectory midpoint as the range at which the projectile height reaches its maximum (Figure 1). Pejsa's midpoint formula allows you to compute the midpoint given a specific maximum height (Hm). The derivation is straightforward and I will not provide much additional commentary beyond the mathematics itself. Continue reading →

In this post, I will review Pejsa's development of a formula for the height of a bullet relative to the shooter's line of sight, assuming that the rifle is adjusted to have zero drop at a known range (referred to as the rifle's "zeroed" range). Figure 1 illustrates the trajectory of a bullet fired from a rifle zeroed at a given range. Continue reading →

In this post (part 3), I will work an example from Pejsa's "Modern Practical Ballistics 2nd ed." and show that the exact and approximate solutions to the drop differential equation give nearly the same answers. Continue reading →

In our previous post, we developed an expression for y' (=dy/dx, Newton's notation) expressed as differential equation in terms of x. We will now solve this equation through the use of an integrating factor. Having solved for y' in terms of x, we can integrate that expression to obtain y(x). Continue reading →