Modeling Drag — Projectile Velocity Versus Range

Posted on 28-January-2011 by mathscinotes

Introduction

As mentioned in a previous post, I am reading the book "Modern Practical Ballistics" by Pejsa. I have been working through some of the derivations in the book and they are interesting enough (at least to me) to be worth documenting here. One of these interesting derivations is an elegant result for the variation in projectile velocity versus range. Since all cartridge documentation include tables of velocity versus range, I have a wealth of data to compare to the equation's output. I love it when I can compare a model to lots of real data. Let's dig in …

Drag Coefficient

A projectile moving through air experiences drag. The force of drag slows the projectile and causes velocity to fall of as the projectile travels on it course. The Wikipedia contains a very good discussion of drag and I refer you to that article for greater details. However, I will review the relevant points to my discussion here quickly.

  • Drag refers to forces that oppose the relative motion of an object through a fluid (a liquid or gas).
  • Drag forces act in a direction opposite to the oncoming flow velocity. This means that there will be some minus signs in upcoming equations.
  • Drag forces depend on velocity.
  • For the purposes of this blog, I will be focusing on the drag a bullet experiences above the speed of sound. This is considered high velocity. There are ways to model drag at other velocities, but that is not my goal here.

The force that drag exerts on a bullet is given by the drag equation (Equation 1).

Eq. 1 {{F}_{d}}=\tfrac{1}{2} \cdot \rho \cdot {{v}^{2}}\cdot{c_d} \cdot A

where

  • Fd is the force of drag, which is by definition the force component in the direction of the flow velocity
  • ρ is the mass density of the fluid
  • v is the velocity of the object relative to the fluid
  • A is the reference area
  • cd is the drag coefficient

Understanding the drag coefficient cd is the most important part of this discussion. Equation 2 contains the definition of the drag coefficient.

Eq. 2 c_d=\frac{F_d}{\frac{1}{2}\cdot \rho \cdot {{v}^{2}}\cdot A}

We need to make some observations about the drag coefficient.

  • Below the speed of sound, the force of drag increases with the square of velocity.
  • This means the drag coefficient is constant for velocities less than the speed of sound.
  • Above the speed the speed of sound, the force of drag does not follow a square law.
  • Therefore, the drag coefficient is NOT a constant in the transonic and supersonic regions.

Figure 1 shows an example of the drag coefficient for the standard reference bullet, usually referred to as the G7 shape (see Figure 2).

Figure 1: Drag Coefficient Plot (Green Line) for a G7 Standard Projectile.

Figure 1: Drag Coefficient Plot (Green Line) for a G7 Standard Projectile.

Note that Figure 1 also shows a blue line that demonstrates that the drag coefficient can be well approximated for velocities above the speed of sound (~1,126 ft/s feet per second) by an equation of the form k_d \cdot {{v}^{-n}}, where kd and n are projectile-specific constants.

Figure 2: G7 Reference Projectile (Similar to Spitzer Design).

Figure 2: G7 Reference Projectile (Similar to Spitzer Design).

For the derivation to follow, I will use Equation 3 to model the variation in cd with velocity.

Eq. 3 c_d={k_d}\cdot {{v}^{-n}}

I will use Equation 4 to model the deceleration of the projectile with respect to velocity.

Eq. 4 {{a}_{d}}=-\frac{{{F}_{d}}}{m}=-\left( \frac{\frac{1}{2}\cdot \rho \cdot A}{m} \right)\cdot \left( {{k}_{d}}\cdot {{v}^{-n}} \right)\cdot {{v}^{2}}=-k\cdot {{v}^{2-n}}

where m is the mass of the projectile and k is a generic constant I will use to aggregate all the projectile and atmospheric parameters (k\triangleq \frac{{{k}_{d}}\cdot \rho \cdot A}{2\cdot m}).

Derivation of Velocity Versus Range Equation

We can use the expression for the acceleration of the projectile (Equation 4), we can construct and solve a differential equation that relates velocity and position. Equation 5 shows the desired differential equation and how to solve it. This equation assumes that the projectile is moving horizontally, which is what Pejsa assumed. For bullets used in normal applications (e.g. target shooting, hunting), this is a good assumption for velocity. It is not a good assumption for bullet drop, which I will handle in a later post.

Eq. 5 \frac{{{d}^{2}}x}{d{{t}^{2}}}=\frac{dv}{dt}=-k\cdot {{v}^{2-n}}
\frac{dv}{dx}\cdot \frac{dx}{dt}=-k\cdot {{v}^{2-n}}
\frac{dv}{dx}\cdot v =-k\cdot {{v}^{2-n}}
{{v}^{n-1}}\cdot dv=-k\cdot dx
\int\limits_{{{v}_{0}}}^{v}{{{v}^{n-1}}\cdot dv}=-\int\limits_{0}^{x}{k\cdot dx}
\frac{{{v}^{n}}}{n}-\frac{v_{0}^{n}}{n}=-k\cdot x
\therefore {{v}^{n}}=v_{0}^{n}\cdot \left( 1-n\cdot k\cdot x\cdot v_{0}^{-n} \right)

At this point, Pejsa introduces an interesting substitution. He defines a term F, which he calls the retardation coefficient. F provides a computationally simple yet accurate drag model (see this post for more information).

Eq. 6 F\triangleq \frac{1}{k\cdot {{v}^{-n}}} and {{F}_{0}}\triangleq \frac{1}{k\cdot {{v}_{0}}^{-n}}

We can substitute Equation 6 into Equation 5 to obtain Equation 7.

Eq. 7 {{v}^{n}}=v_{0}^{n}\cdot \left( 1-\frac{n\cdot x}{F_0} \right)

We can substitute Equation 6 into Equation 7 to derive a simple relationship between F and F0, which is shown in Equation 8.

Eq. 8 F\cdot k={{F}_{0}}\cdot k-k\cdot n\cdot x \Rightarrow F={{F}_{0}}-n\cdot x

Equation 7 allows us to compute the projectile velocity versus range, given values for F and n. In a later blog post, I will show how F and n can be estimated for standard projectiles.

Empirical Comparison

It is interesting to look at a real projectile and see how well this model fits the empirical data. Consider a Hornady 308 caliber, 150 grain, SST-LM. Figure 3 shows the data in a screen capture from Mathcad.

Figure 3: Velocity Versus Range Data for Hornady 308, 150 Grain, SST-LM

Figure 3: Velocity Versus Range Data for Hornady 308, 150 Grain, SST-LM

Using Mathcad, I fit the projectile velocity data to Equation 7 (n = 0.266 and F0 = 1227 yards) and plotted the fitted curve and the raw data in Figure 4.

Figure 4: Raw Hornady Data and Model Curve Fit Comparison.

Figure 4: Raw Hornady Data and Model Curve Fit Comparison.
Raw data from Ammo and Ballistics II (2nd Edition) by Forker (ISBN 1-57157-305-4).

The fit is excellent.

Conclusion

I presented a summary of the Pejsa derivation for the velocity of a projectile versus distance. The agreement between his equation and an arbitrarily chosen example was excellent. In subsequent posts, I will discuss other results from his model.

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40 Responses to Modeling Drag — Projectile Velocity Versus Range

  1. Chris Hartmann, Education Program Manager, Mathcad says:
    30-January-2011 at 11:23 am

    Very interesting! I learned quite a bit about drag, including the fact that I can use Mathcad to explore it more deeply. Love your blog!

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  3. irstuff says:
    8-February-2011 at 7:35 pm

    Do you ever post your sheets? I got a copy Pejsa's book too many moons ago, and just never got around to digesting it. I'd certainly like to get a copy of your worksheet.

    Cheers

  4. arsenic says:
    8-September-2011 at 11:12 pm

    Are you sure your derivation of equation 5 is correct?

    My math is rather rusty, but I come to a different result. In the second to last line, I get:
    (v^n / n) - (v0^n / n) = x / -k

    • mathscinotes says:
      10-October-2011 at 2:26 pm

      Sorry I am so late responding. I have been traveling quite a bit.

      I think what I have in the derivation is correct, but I could be missing something. It appears that you differ with me on the right-hand side of the equation. I will expand that side a bit more.

      intlimits_{0}^{x}{kcdot dphi }=kcdot left. phi right|_{0}^{x}=kcdot x

      So I still get what I have in the post.

  5. Don Mitchell says:
    11-February-2012 at 10:29 pm

    You should look for the rare book, "Paris Kanonen" by G.V. Bull. I think you'd be fascinated.

  6. Ed B says:
    7-September-2012 at 5:24 pm

    Thanks for the interesting site. Maybe you can confirm or debunk something for me: My theory is that at some point increasing the initial velocity of a projectile will not result in more than a trivial incerase in its' range. This is assuming more or less standard sea level air density. I'm not referring to specialized armor piercing projectiles with extreme ballistic coefficients but something with relatively high drag like #4 buckshot, for example. One more condition: I'm not talking about the time it takes the object to hit the ground, but rather the distance it will travel horizontally before reaching a nominal velocity. I hope this isnt clear as mud. Again, thanks for letting me pick your brain. Ed B

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  8. terry adams says:
    13-June-2013 at 7:22 pm

    naval guns claim a range of up to 16 miles , and apparently do so with an initial velocity of only approx. 3000 fps . how is this possible

    • mathscinotes says:
      13-June-2013 at 11:35 pm

      It turns out than any projectile fired at 3000 fps at a 45° angle on a flat, air-less version of Earth will travel nearly 53 miles. I show the derivation below. However, we know a rifle bullet does not travel that far, so what gives?

      Projectile in Vacuum

      I also was stunned at the range of a battleship projectile compared to rifle bullet fired at a similar angle and velocity. The key is understanding the ballistic coefficient. The ballistic coefficient is the ratio of the drag deceleration of a reference bullet to that of the projectile in question. This tells you how much that a projectile is affected by drag. Common bullets have ballistic coefficients of 0.3 to 0.5, which means that the decelerate 2 to 3 times faster than the reference bullet (I recall it being a 1 lb monster bullet). While a 16-inch projectile moving at 3000 fps will encounter more drag than a small bullet, the 16-inch shell has an incredible amount of mass (and therefore momentum) behind it relative to a rifle bullet. This makes a 16-inch shell very efficient, i.e. drag imposes relatively little penalty on it compared to a common rifle bullet.

      If you look at the history of battleships, the maximum range of their guns increased as the projectiles got bigger. Roughly speaking, this is because mass of a projectile increases by the cube of its diameter, but the drag increases by the square of the diameter (i.e. related to the projectile area). This means that ballistic coefficient increases roughly linearly with diameter, all other factors (like shape) are held constant.

      So a battleship shell travels so far because it is much more efficient with respect to drag than a common rifle bullet. This means it gets closer to the vacuum value of maximum range.

      Mathscinotes

  9. steve gardner says:
    20-June-2013 at 2:47 pm

    My hobby is archery flight shooting. I know that the lighter the arrow the faster it will come out of the bow but because of draw will not neccessarily fly as far. I compete in primitive classes where I must use wood arrows and natural vane material. So far it seems about 250 grains is the lightest I can make my arrow to offset the drag from the small feathers I am using for fletching. I can easily make an arrow stiff enough as low as 140 grains but they seem to fall out of the sky because of drag. Is their a way to predetermine how much benefit I might expect from finding thinner fletch materials?

    • mathscinotes says:
      26-June-2013 at 12:50 pm

      I am sure their are aerodynamic software models that would allow you to predict the benefit of thinner fletch materials. Commercial aerospace firms use these models all the time. I have seen people design everything from parachutes to lifting bodies with these software packages. However, I do not know of any aerodynamic modeling software that is free or low cost for an individual to use.

      Mathscinotes

  10. David says:
    30-July-2013 at 2:02 pm

    Sorry to necro comment, but...

    Has anyone provided a physical explanation for Pejsa's "retardation coefficient"? I suspect I know what it actually represents, but I'm not sure.

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  14. dougiefresh1233 says:
    17-October-2013 at 4:07 pm

    I am working on building a mathematical model trans-sonic rocket in my rocket engineering class and was wondering how i could go about finding the constants for Kd and n?

  15. SGT_Ludwig says:
    8-November-2013 at 1:36 am

    I'm working on a relatively simple ballistics program and this is by far the best explanation I have found for this problem. Seriously, thanks a million. I only have one question:

    You say that "Kd and n are projectile-specific constants", but that doesn't seem to explain anything about which constants these actually are. I've gathered from Hornady's site that 'n' is the cross-sectional density, but what is 'Kd'?

    • mathscinotes says:
      8-November-2013 at 2:06 am

      Hi Sergeant,

      I am currently traveling on business and I do not have all my reference material here, but I will answer as best I can.

      In Figure 1, I show two curves, a measured value for the drag coefficient (green) and and equation approximation for the drag curve (dashed blue). The measured curve shows the coefficient of drag for a reference projectile. There are different curves used for different shaped projectiles (e.g. G1, G2, G7, etc). The constants kd and n are chosen so that I get the closest possible fit between reality (i.e. measured) and my mathematical approximation. I need to use a mathematical approximation so that I can develop simple algebraic expressions for the drag coefficient.

      From a mathematical standpoint, you can just view kd and n as curve fitting values that you can generate using Excel or Mathcad or whatever. kd and n are specific to reference projectile of a specific weight (usually 1 lb) and shape (e.g. G7). The ballistic coefficient, which is a function of sectional density, is used to change the drag coefficient so that it is valid for an identically-shaped projective of a different diameter and weight. kd and n are not functions of sectional density.

      I am not sure I have answered your question. If you have more questions, write in a few days when I am back home.

      mathscinotes

      • SGT_Ludwig says:
        9-November-2013 at 4:39 am

        Mathscinotes -

        Thank you so much for your speedy reply. I actually saw that you answered the question and got lost in an awesome adventure of figuring this all out. Thanks for your patience too, I've been closing a knowledge gap that developed while I was in the service.

        It took me some tinkering and exploring but I totally understand what you did now. I followed some of your links, like this one to find some C_D tables and realized you were just fitting something to predict the observed C_D at a specific velocity. I had convinced myself those values were something other than what they were. Luckily, for my purposes, I only really need the base G1-G7 measurements, and it was pretty simple for me to figure out my own values for k_d and n. Just had to poke around a bit with R until a light switched on up stairs.

        So, in short, you totally did answer my question and made my life a LOT easier.

        Just for notice to anybody else that might read this, I found that "C_D = k_d + v^-n" actually fit the raw reference G7 data much better. I now understand that it doesn't really matter what regression I use, as long as I can have some kind of predictive model for C_D, but like I said if anybody else reads this they can at least see there is some flexibility with this part of the solution.

        A large part of my confusion about the ballistic coefficient actually stemmed from the coincidence that your "n" value of ".266" happens to be the EXACT same value Hornady advertises as the ballistic coefficient for the closest branded ammo I could find on their site (which shoots the same bullet). Obviously, not quite understanding what was going on I latched to that like a life vest.

        Now that I've tinkered around, however, and I'm looking at your results and I'm getting a creeping feeling you might have coincidentally backwards engineered whatever method Hornady used to derive their BC figure for that bullet. It just looks too coincidental to me - your coefficient was exactly the same as theirs, and your results look like a perfect fit.

        What are your thoughts on this? Do you think they might have just calculated out the velocity data instead of going through the process of busting out the ol' Doppler and getting their hands dirty? Or am I just missing something?

        Again, thank you so much for this post and for following up with my comment. You have seriously saved me a ton of time and helped me understand something I have been trying to figure out for the past three days.

        - Konrad

        • mathscinotes says:
          9-November-2013 at 3:36 pm

          Comments in line
          Mathscinotes -

          Thank you so much for your speedy reply. I actually saw that you answered the question and got lost in an awesome adventure of figuring this all out. Thanks for your patience too, I’ve been closing a knowledge gap that developed while I was in the service.

          It took me some tinkering and exploring but I totally understand what you did now. I followed some of your links, like this one to find some C_D tables and realized you were just fitting something to predict the observed C_D at a specific velocity. I had convinced myself those values were something other than what they were. Luckily, for my purposes, I only really need the base G1-G7 measurements, and it was pretty simple for me to figure out my own values for k_d and n. Just had to poke around a bit with R until a light switched on up stairs.

          So, in short, you totally did answer my question and made my life a LOT easier.
          I hear this a lot. The is why I started the blog. People at my work were looking for a Google searchable database of some of my work. A blog seemed like the easiest way to do that.

          Just for notice to anybody else that might read this, I found that “C_D = k_d + v^-n” actually fit the raw reference G7 data much better. I now understand that it doesn’t really matter what regression I use, as long as I can have some kind of predictive model for C_D, but like I said if anybody else reads this they can at least see there is some flexibility with this part of the solution.
          Any regression function will work. I only used the ones I did because I was working through the derivations that Pejsa did in his book. Many people find that book difficult to read (including me) and I was simply trying to come up with a clearer presentation.

          A large part of my confusion about the ballistic coefficient actually stemmed from the coincidence that your “n” value of “.266″ happens to be the EXACT same value Hornady advertises as the ballistic coefficient for the closest branded ammo I could find on their site (which shoots the same bullet). Obviously, not quite understanding what was going on I latched to that like a life vest.

          Now that I’ve tinkered around, however, and I’m looking at your results and I’m getting a creeping feeling you might have coincidentally backwards engineered whatever method Hornady used to derive their BC figure for that bullet. It just looks too coincidental to me – your coefficient was exactly the same as theirs, and your results look like a perfect fit.
          There is a lot of marketing (i.e. lying) associated with ballistic coefficients. See this paper by two researchers who compared their measurements of ballistic coefficients with the claims of the manufacturers. It is an eye opener.

          What are your thoughts on this? Do you think they might have just calculated out the velocity data instead of going through the process of busting out the ol’ Doppler and getting their hands dirty? Or am I just missing something?
          In the engineering business, we refer to this process as "dry lab-ing". You do not go out in the cold and wet (I live in MN) to get your data, but stay in the lab where its nice and dry and simply run a spreadsheet. Anything is possible. I am in a card club with a former Engineering VP for Federal Cartridge. I will ask him how they generated their ballistic tables and I will get back to you. Most of my ballistics work has been done to support an interest I have in old-school battleships (WW1 and WW2). I have found that the US Navy produced very sophisticated range tables for use by the sailors, but actually fired few shells in test. Those massive tables were generated using equations based on velocity data average from a very small number of shots.

          Again, thank you so much for this post and for following up with my comment. You have seriously saved me a ton of time and helped me understand something I have been trying to figure out for the past three days.
          If there are any topics that you think would be worth covering in my blog, please send me a note. I often write when I travel because there is down time in the evenings and I am alone. Some folks drink when traveling -- I do math.

          – Konrad

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  17. Kory says:
    22-August-2014 at 12:56 am

    I'm curious if n is Hornady's Ballistic Coefficient. How would I be able to calculate the bullets current position if I was given these parameters:
    Init. Velocity
    Init. Position
    Init. Force0
    Hornady's BC

    So given a deltaTime of say 2 secs and the above parameters where will the bullet currently be?

    • Kory says:
      22-August-2014 at 7:27 am

      Actually, I think I found something else that would be more helpful from using your site.
      You stated that the BC can be estimated as the fraction of 1000 yards where the Kinetic energy is half of it's initial KE. This would imply that the velocity is 70.7% of it's initial velocity. Now this is really rough, and it is good enough for my purposes, but you also stated that you lose 1% of velocity for every 1% of Fo. Fo is the unknown here but not if you apply the above and use the BC and half of the KE as a rough guess.

      Here is an example:
      SS109 (62 gr) has a G7 BC of 0.158
      so with a BC of 0.158 = 158 yards = 144.4752 m
      travel 144.4752 m / 29.3% (.293) = 493.0894 m (this is our Fo)
      So we will lose 1% of our velocity for every 4.930894 m we travel.

      Now if our initial velocity is 940 m/s then for nearly every 5 m we travel we will lose 9.4 m/s in velocity.

      Does this sound about right? Let's check... So in ft/s (9.4 m/s = 30.8399 ft/s) these loses for every 100 m would be about 625 ft / s:
      000 m = 3083.99 ft / s
      100 m = 2458.55 ft / s
      200 m = 1833.11 ft / s
      300 m = 1207.67 ft / s
      400 m = 582.23 ft / s

      This drops off way to fast, but maybe it is because this is the G7 BC. What does it look like if we use the G1 BC?
      SS109 (62 gr) has a G1 BC of 0.325
      so with a BC of 0.325 = 325 yards = 297.18 m
      297.18 m / .293 = 1014.266 m as our Fo
      So we will lose 1% of our velocity for every 10.1427 m we travel.

      Now if our initial velocity is 940 m/s then for every 10 m we travel we will lose 9.4 m/s in velocity. For every 100 m we will lose about 304 ft / s in velocity.
      000 m = 3084 ft / s
      100 m = 2780 ft / s
      200 m = 2476 ft / s
      300 m = 2172 ft / s
      400 m = 1868 ft / s

      The G1 BC seems to be right on the money because here is what was recorded at the Aberdeen Proving Grounds
      000 m = 3100 ft / s
      100 m = 2751 ft / s
      200 m = 2420 ft / s
      300 m = 2115 ft / s
      400 m = 1833 ft / s

      Do you see any problems with converting the BC to Fo this way as a rough approximation? I haven't tried it with other G1 BC's though. Maybe this is just a lucky example.

      • mathscinotes says:
        23-August-2014 at 4:15 pm

        Hi Kory,

        I am moving between tasks here at work, but I will try to answer what I can here. I will take a closer look at your post once I have a spare minute (sometime in retirement most likely).

        >>Fo is the unknown here but not if you apply the above and use the BC and half of the KE as a rough guess.

        I have done this calculation myself and it works.

        >>So we will lose 1% of our velocity for every 4.930894 m we travel.

        >>Now if our initial velocity is 940 m/s then for nearly every 5 m we travel we will lose 9.4 m/s in velocity.

        You have two contradictory statements here. It is true that you lose a percentage of your velocity with each distance increment (first statement), not a fixed value for each distance increment (second statement). So the velocity drop is initially 9 m/s for every 4.9 m, but that is true only at the muzzle. It really is a percentage. Mathematically, this means a velocity versus range graph will have a curve to it, like Figure 4. Your statement of a linear velocity variation with distance would mean Figure 4 would have a straight line on it and it doesn't.

        mathscinotes

    • mathscinotes says:
      23-August-2014 at 3:35 pm

      The parameter n is a curve fitting parameter. I discussed this in my response to Sgt. Ludwig's question (above). It is a coincidence that the BC and n were equal for my particular example, which threw Sgt. Ludwig off as well. I only chose the 308 for an example because I was thinking of buying one for some target shooting.

      You ask about how to compute a bullet's position given certain parameters. In this post, I simply computed the velocity as a function of distance. I assume that you want to compute the bullet's distance versus time. The work in this post can be extended with a wee bit of calculus to compute a bullet's distance versus time. I will look at pulling together a post with this information in the near future. My recent posts have focused on electronics because I am knee deep in some design work at work right now.

      Thanks for your question.

      mathscinotes

      • Kory says:
        23-August-2014 at 4:30 pm

        Actually, I spent more time on this and I now have a really good model that fits Aberdeen Proving Grounds real doppler data better than most ballistic calculators. I ended up seeing my error above yesterday assuming the linear relationship and I have modified it to have an exponential decay. I then ended up solving Pejsa's velocity at range (r) equation wrt Fo. I took a Taylor Series expansion to determine the value of Fo with r fixed at the BC determined in the Truth about BC paper. My results are generally within 3% of ballistic calculators and generally within 10% of real world data. Most ballistic calculators are within 10-12% of real world data.

        Because this way of modeling relies only on basic math outside of the initial calculations we can easily incorporate this into computer games or other real-time simulators where the projectiles do not have to be 100% accurate, but yet the system needs to handle thousands per second.

        I'll be writing a paper on this soon for the Computer Gaming industry because as far as I know, no one has done this yet. They have always used exponents or look-up tables to calculate these and that can be slow when there could be thousands of flying bullets.

        Thanks for bringing my attention to those two sources, I would not have been able to accomplish this without those.

  18. mathscinotes says:
    23-August-2014 at 4:59 pm

    It sounds like you are moving down the correct path. Pejsa's modeling approach is generally considered that best you can do using basic algebraic operations -- this is why it is used in many ballistic apps. Anything else gets pretty deep into differential equations and drag models. Most of my simulation work has been done with respect to battleship projectiles. The reason that I used Pejsa as my reference is because it is actually very similar to the work done on naval artillery by Ingalls and others. They were the folks that came up with the concept of ballistic coefficient and standardized projectiles.

    mathscinotes

  19. shrink says:
    28-April-2015 at 11:01 pm

    Nice post. Do you plan a follow-up on bullet drop based on Pejsa's model, as you mentioned?

    I've derived myself the bullet drop equation which basically replicates this result (except for the opposite sign for "n"; I presume this author switched the sign in the velocity formula deliberately):

    http://www.bfxyz.nl/docs/mathlarge.shtml

    The derivation is based on the flat-fire approximation which relates velocities in x and y directions in the following way (e.g., see McCoy):

    d/dx (v_y/v_x) = -G/(v_x)^2

    Considering v_y/v_x = dy/dx and the Pejsa velocity formula (you derived in this post) in the above equation it is quite easy to derive the drop formula y(x) by double integration over x.

    I've tested the drop formula by comparing the trajectory it produces with trajectories (based on Doppler Radar Data) produced by the Lapua QuickTarget Unlimited software for some of their bullets. The fit is excellent.

  20. shrink says:
    28-April-2015 at 11:33 pm

    Nice post. Do you plan a follow-up on bullet drop based on Pejsa's model, as you have mentioned?

    I've derived the bullet drop equation myself and it is basically a replication of this result (except for the opposite sign for "n"; I assume the author switched the sign deliberately):

    http://www.bfxyz.nl/docs/mathlarge.shtml

    The derivation is based on the flat-fire approximation which implicates the relationship between velocities in x and y directions in this way:

    d/dx(v_y/v_x)=-G/(v_x)^2

    Considering v_y/v_x=dy/dx and the Pejsa velocity formula for v_x (derived in this post) the drop formula y(x) is quite easily obtained by double integration over x.

    I've tested the drop formula by comparing the trajectories it produces with trajectories (based on Doppler Radar Data) produced by Lapua's QuickTarget Unlimited software for some of its bullets. The fit is excellent.

    • mathscinotes says:
      29-April-2015 at 10:57 am

      I plan to work through the bullet drop formulas this summer. I recently spent some time going through the original Ingalls work and I want to make some comparisons between Pejsa's and Ingalls' work. All my work is associated with battleship guns, but the formulas also work for standard rifles.

      mathscinotes

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  22. Chris wild says:
    26-March-2021 at 6:11 pm

    Hello, finally i found someone talk somehow that i can understand. Sorry i disturbing you , i am a developer and long range shooter, i want to make my own application for ballistic calculation ( bullet drop calculator ), so i want your help, if saw this message please reply me, i want to ask questions . Thanks

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