Railroad Math

Posted on 7-February-2011 by mathscinotes

Introduction

I was listening to an advertisement where CSX made the claim that they move 1 ton of freight 423 miles for 1 gallon of fuel. This is an interesting measure of efficiency. Let's see if we can confirm this using a couple of different analysis approaches.

Top-Down Approach

CSX issues quarterly reports, which contains data that we can use to estimate the miles per gallon-ton of freight. This forum article has some interesting data from the 4Q2007 CSX financial statement.

  • CSX moved 253 billion revenue ton-miles of goods in the 12 months ending 12/31/07.
  • During this period, CSX consumed 569 million gallons of diesel #2 fuel.

This makes computing the efficiency eff of a train (mile-ton per gallon) easy, which is shown in Equation 1.

Eq. 1 \mathit{eff}=\frac{253\text{E9}\cdot \text{mile}\cdot \text{ton}}{569\text{E9}\cdot \text{gal}}=445\frac{\text{mile}\cdot \text{ton}}{\text{gal}}

This number is close to what CSX is using its advertisement. So their number is credible based on their business statment.

Bottom-Up Approach

It is a bit more difficult to look at the problem from the standpoint of friction and energy, but let's take a wack at it.

First, let's gather some data.

  • Energy per gallon of #2 diesel fuel is 138,700 BTU/US gal (Source)
  • Efficiency of a diesel engine is ~46% (Source and Source)
  • Diesel to rail conversion efficiency of 80% (Source)
    There is some loss of power due to transmission inefficiency in the diesel-electrical-rail transfer of power. I am using the value of 80% for the transmission efficiency based on a reference from the 1950s that was comparing steam to diesel-electric locomotives. This number is probably out of date, but is a reasonable start for a rough estimate.
  • Train expends 20 lb of pulling force per ton of load (Source)
    This number is subject to variation due to track condition, weather, grade, and curvature of the track. I am assuming an average value that is in the ballpark, but could easily be off by ±20% or more. Remember, we are just trying to determine if the CSX efficiency number is reasonable.

I would propose that one simple model would be to equate the energy dissipated against the rolling resistance of the train to the energy available from a gallon of diesel fuel.

Eq. 2 {{F}_{FrictionPerTon}}\cdot d={{E}_{FuelOilPerGal}}\cdot {{e}_{Diesel}} \cdot e_{c}

Where d is the distance traveled, FFrictionPerTon is the resistance of a ton of load (= 20 lb per ton), EFuelOilPerGal is the energy per gallon of #2 diesel fuel (=138,700 BTU/US gal), eDiesel is the efficiency of a modern diesel engine (=46%), and ec is the diesel-to-rail conversion efficiency (=80%).

We can solve Equation 2 for d and substitute our assumed values.

Eq. 3 d=\frac{{{E}_{FuelOilPerGal}}\cdot {{e}_{Diesel}} \cdot e_c }{{{F}_{FrictionPerTon}}}=376\frac{\text{mile}\cdot \text{ton}}{\text{gal}}

This value is close enough that feel I have verified the CSX number from the bottom up.

Conclusion

Moving one ton of freight 423 miles on one gallon of fuel seems like a reasonable value. This exercise really shows the efficiency of moving material in bulk.

Posted in General Science | 5 Comments

Drinking Math

Posted on 5-February-2011 by mathscinotes

Quote of the Day

The difficulty lies not so much in developing new ideas as in escaping from old ones.

- John Maynard Keynes

Introduction

Figure 1: I used to love to see people make layered drinks.

Figure 1: As a boy, I used to love to see people make layered drinks. The first layered drink I saw as a boy was a zombie. (Source)

As a non-drinker, I have never had much interest in alcohol. That said, alcohol has been a big part of my life. My mother earned her living working in bars, starting as a waitress and eventually managing them. Even in retirement, she works part-time selling pull tabs in bars for local sports groups. I have also been the designated driver for literally hundreds of social occasions. At work, I have had numerous co-workers who have derived much pleasure brewing beer and wine. As a boy, I did find the layered drinks interesting (e.g. Figure 1), but only because they looked cool.

But alcohol must be used in moderation. To help guide people in their consumption, there are numerous charts and tables that tell people how much they can drink and remain under the legal limits. While prowling around the Wolfram Alpha web site, I noticed that they have a Blood Alcohol Content (BAC) calculator. I started looking at the output and I saw that there was some interesting math going on there.

For those who are interested, I have implemented the formulas below in an Excel spreadsheet here. Let's dig ...

Analysis

BAC Definition

The Wikipedia provides a very complete definition of BAC and I will start there with a slight paraphrasing of their definition.

BAC is the concentration of alcohol in a person's blood. It is most commonly used as a metric of intoxication for legal or medical purposes. It is usually expressed as a fractional percentage in terms of volume of alcohol per liter of blood in the body.

So BAC is unitless quantity because it is the ratio of two volumes. This definition of BAC is not universal. For example, California actually uses grams (gm) of alcohol per deciliter (100 mL = dL) of blood. For this post, we will use the more common ratio of volumes, which is expressed mathematically in Equation 1.

Eq. 1 BAC\triangleq \frac{{{V}_{Ethanol}}}{{{V}_{Blood}}}

Key Analysis Assumptions

The analysis assumptions in the creation of the BAC tables vary because a person's response to alcohol varies by person. Here are the assumptions that I could identify.

  • The alcohol evenly spreads through all the water of the body, not just the blood.
  • The amount of water in the body is proportional to body weight.
    This assumption is highly variable between individuals. Quoting from the Wikipedia, "In a newborn infant, this may be as high as 75 percent of the body weight, but it progressively decreases from birth to old age, most of the decrease occurring during the first 10 years of life. Gender also affects the percentage of water for an individual. This is because women, on average, have a higher body fat percentage. Higher body fat percentage correlates with lower body water percentages. As such, obesity decreases the percentage of water in the body, sometimes to as low as 45 percent."
  • Every drink has the same amount of alcohol.
    In fact, the amount of alcohol varies by drink. To simplify the discussion, most tables assume a drink contains about 0.5 oz of alcohol by weight. Wolfram Alpha assumes 0.533 gm per drink. Since ethyl alcohol (ethanol) has a density of 0.789 g/cm3, 0.533 oz (mass) of alcohol is equivalent to 0.648 fluid ounces. This amount of alcohol corresponds roughly to the following common drinks:

    • 12 fluid ounces of beer
    • 1.25 fluid ounces of 100 proof liquor
    • 4 fluid ounces of 25 proof table wine
  • The body eliminates alcohol at a fixed rate.
    Quoting the Wikipedia, "The rate of elimination in the average person is commonly estimated at .015 to .020 gm/dL per hour, although again this can vary from person to person and in a given person from one moment to another." To convert the units from gm/(dL hr) to the BAC's unitless over volume/volume, we need to apply the density of ethanol as follows: \left( 0.015-0.020 \right)\cdot \frac{\text{gm}}{\text{dL}\cdot \text{hr}}\cdot \frac{\text{c}{{\text{m}}^{3}}}{\text{0}\text{.789}\cdot \text{gm}} = \left( 0.19\%-0.25\% \right)\frac{1}{\text{hr}}.

Modeling BAC

Everything I have been able to find on the web uses the model shown in Equation 2 with different parameters.

Eq. 2 BAC=\frac{n\cdot \gamma }{\frac{w\cdot \alpha }{{{\rho }_{{{H}_{2}}O}}}}-\beta \cdot t

where

  • w is the weight in pounds
  • n is the number of drinks
  • t is the time since consumption (hours)
  • ? is the percentage of water in a body (weight of water/body weight = ~75%)
  • ? is volume of ethanol in a drink (0.648 fluid ounces)
  • ? is the rate of elimination (~0.021% per hour)
  • ?H20 is the density of water (1 gm/cm3 = 0.065198 lb/fluid ounce)

If we substitute the values shown into Equation 2 we get Equation 3.

Eq. 3 BAC = 5.633\text{\%} \cdot \frac{n}{w}-0.021\text{\%} \cdot t

Figure 2 shows the Wolfram Alpha result and Figure 3 shows Equation 3 graphed in Mathcad. The results appear identical.

Figure 1: BAC Versus Time for 240 lb Man After 6 Drinks.

Figure 2: BAC Versus Time for 240 lb Man After 6 Drinks.

Figure 2: BAC Model in Mathcad.

Figure 3: BAC Model in Mathcad.

Conclusion

I have looked at a number of web sites on BAC levels and they all make different assumptions about their definition of a drink, body water percentages, and elimination rate. The variation in the assumptions all reflects the fact that these characteristics vary by individual. Other important factors, like how long you were drinking and whether you were consuming food while drinking, are completely ignored. Looks to me like it is difficult to know how many drinks you can have and still drive. Sounds like not drinking and driving is still the best way to go. It's always wise to remember to book another method of transport home after you've been drinking, it's not worth risking lives for. Because, let's face it, most people who do drink and drive end up getting caught one way or another and will end up looking to get useful information from a criminal defense attorney in anticipation of charges being brought against them. Drunk driving is one of the top causes of car accidents, which is why it's important to avoid driving a vehicle after drinking alcohol. Unfortunately, most drunk drivers end up needing personal injury attorneys to support them after they have put themselves, and other road users, in danger. To prevent yourself from injuring anyone, consider booking a taxi after any social events, or ask family and friends if they can collect you.

Posted in Health | 9 Comments

Interesting Old Submarine Photo

Posted on 2-February-2011 by mathscinotes

A co-worker and I were discussing strange photos we have seen. I mentioned a photo that I saw years ago that showed USS S-5's tail sticking out of the water. During the 1920s, submarines that submerged occasionally did not come back up. In this case, the crew was very clever. The submarine was 231 feet long and sank in 194 feet of water. The crew was able to lighten the aft section and float the tail of the sub up to the surface. With great difficulty, they cut a hole in the tail and waited for a ship to come by. No one lost their life, which for early submarines is an amazing statement. Figure 1 shows the tail section protruding from the water. I cannot imagine how difficult it must have been for the crew to move around in a small sub with its end tipped up. Also, the aft area necks down to become pretty tiny. It must have been extremely cramped.

Figure 1: USS S5 Tail Protruding Above the Water.

Figure 1: USS S5 Tail Protruding Above the Water.


If you would like to read more about the S-5 and its sinking, see this web site. A book has been written about the incident called "Under Pressure: The Final Voyage of Submarine S-Five" by A.J. Hill.

Here is another photograph showing a rescue ship.

Figure 2: S-5's Tail Near One of Its Rescuers.

Figure 2: S-5's Tail Near One of Its Rescuers.

Posted in History of Science and Technology | Tagged | Comments Off on Interesting Old Submarine Photo

"Dying Gasp" from a Circuit Standpoint

Posted on 2-February-2011 by mathscinotes

Introduction

The Metro Ethernet Forum (MEF) specifies requirements for carrier-grade Ethernet services. One the features they define is the "dying gasp". A dying gasp is defined as follows:

Dying Gasp is a message (or signal) sent by the Customer Premises Equipment (CPE) to the service provider when a power outage occurs.

In most cases, the hardware does not know that the power is failing until the power is already starting to go away. The dying gasp protocol requires that three "I am dying" messages be sent before the CPE gear goes down for good, which means the CPE must stay operating for about 3 msec after the power has gone away. To accomplish this feat, we need to include some local energy storage with MEF-compliant Ethernet hardware. This post will examine how we can estimate the amount of local energy storage required.

Analysis

We will begin by assuming the power supply architecture shown in Figure 1.

Figure 1: Circuit Model for "Dying Gasp" Capacitive Backup.

Figure 1: Circuit Model for "Dying Gasp" Capacitive Backup.

There are a few points that should be made about this power supply architecture:

  • The input circuit will only conduct current one way.
    We need to make sure that as the power drops away, the power we need for our Ethernet interface is not drawn out of the input. In Figure 1, I show a diode, but there are number of ways of isolating the backup capacitor from the dying power source. In most cases, FETs would be more efficient than diodes, but diodes are simpler to understand. However, all this is an implementation detail.
  • The load requirements are modeled as a current source.
    In most cases, we are trying to provide backup energy for a switching power supply, which draw relatively constant power over a range of voltages.
  • Energy is stored in capacitors.
    In most cases, capacitors are the cheapest and easiest way of storing backup energy. This assumes the required backup time is small (~msec). As times get longer, batteries look better.

Energy-Based Capacitance Requirement Determination

If first worked the problem from an energy standpoint. Recall from basic physics that energy stored on a capacitor is given by Equation 1.

Eq. 1 {{E}_{Capacitor}}=\frac{1}{2}\cdot {{C}_{Backup}}\cdot V_{Capacitor}^{2}

The energy needed to power our MEF-compliant device is given by Equation 2.

Eq. 2 {{E}_{Total}}=P\cdot {{T}_{DyingGasp}}=\frac{1}{2}\cdot {{C}_{Backup}}\cdot \left( V_{Initial}^{2}-V_{Final}^{2} \right)

where

  • P is the power draw of our MEF-compliant device.
  • TDyingGasp is the time required to transfer the three dying gasp messages.
  • VInitial is the initial voltage on the backup capacitor. This is the voltage on the backup capacitor at the moment power drops away.
  • VFinal is the final voltage on the backup capacitor and is the minimum voltage at which load input can function.
  • CBackup is the capacitance of the backup capacitor.

We can use Equation 2 to solve for CBackup (Equation 3).

Eq. 3 \therefore {{C}_{Backup}}=\frac{2\cdot P\cdot {{T}_{DyingGasp}}}{V_{Initial}^{2}-V_{Final}^{2}}

Current-Based Capacitance Requirement Determination

The backup capacitance requirement is given by Equation 3, but some electrical engineers do not like energy-based arguments. They prefer arguments based on circuit parameters like voltage and current, which can be measured directly. We can obtain the same result using circuit parameters as follows.

The current drawn by the load and the current drawn from the capacitor are given by Equation 4.

Eq. 4 {{I}_{Load}}=\frac{{{P}_{Load}}}{{{V}_{Load}}}={{C}_{Backup}}\cdot \frac{d{{V}_{Load}}}{dt}

Equation 5 shows Equation 4 broken up into its differential form.

Eq. 5 {{V}_{Load}}\cdot d{{V}_{Load}}=\frac{{{P}_{Load}}}{{{C}_{Backup}}}\cdot dt

We know can integrate both sides of Equation 5, as is illustrated in Equation 6.

Eq. 6 \int\limits_{{{V}_{Initial}}}^{{{V}_{Final}}}{{{V}_{Load}}\cdot d{{V}_{Load}}}=\int\limits_{0}^{T}{\frac{{{P}_{Load}}}{{{C}_{Backup}}}\cdot dt}
\left. \frac{V_{Load}^{2}}{2} \right|_{{{V}_{Initial}}}^{{{V}_{Final}}}=\frac{V_{Initial}^{2}-V_{Final}^{2}}{2}=\frac{{{P}_{Load}}}{{{C}_{Backup}}}\cdot T
\therefore {{C}_{Backup}}=\frac{2\cdot P\cdot {{T}_{DyingGasp}}}{V_{Initial}^{2}-V_{Final}^{2}}

As we would expect, both methods yield the same result.

Empirical Example

I threw the calculations into Mathcad and put the screenshot into Figure 1.

Figure 1: Calculation Used for Actual Product.

Figure 1: Calculation Used for Actual Product.

Conclusion

One of the more difficult parts of product development is determining requirements, a process I call "The Treasure Hunt." The analysis shown here describes a very common type of analysis that is done to determine capacitor requirements for backup systems.

Posted in Electronics | Tagged , , | 6 Comments

Computing the Age of the Universe

Posted on 30-January-2011 by mathscinotes

Introduction

While in the lunch room at work, I often look at the paper. The paper one day this week had an article on the farthest object that has yet been observed by astronomers. One of the guys in my group was there and we started to talk about computing the Earth's age and computing the age of the universe. Since I have already covered calculating the age of the Earth in a previous post, I thought it would be worth documenting calculating the age of the universe as well. It is a shorter subject, at least for this level of detail.

Calculation

The only piece of data needed is Hubble's constant. We can see the linear relationship between the recessional velocity and distance from the chart shown in Figure 1 (Source). Note the linearity of the characteristic.

Figure 1: Recessional Velocity Versus Distance.

Figure 1: Recessional Velocity Versus Distance.
I believe the bulge in the middle of the curve is the data from various galaxy clusters (e.g. Virgo), which also have a rotational component to their motion.


I show in Figure 2 that the reciprocal of Hubble's constant equals the age of the universe. Figure 2 is a screenshot of my Mathcad worksheet (I like to use Mathcad's unit checking).
Figure 2: Calculation of the Age of the Universe.

Figure 2: Calculation of the Age of the Universe.

Conclusion

Pretty straightforward. I first got interested in the subject while listening to an audio book called "Horizons of Cosmology" by Silk. It is a great listen and worth your time if you are interested in that sort of thing.

Posted in Astronomy | 2 Comments

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.

Posted in Ballistics | 40 Comments

Ogives Versus Other Shapes

Posted on 23-January-2011 by mathscinotes

The ogive has long been used in projectile design because it simple to manufacture. Over the last few thousand years, people have gotten pretty good at making sections of spheres. However, simple to manufacture does not mean minimum drag. The Wikipedia has a great figure that really does a nice job of summarizing the performance of the ogive relative to other shapes.

Figure 1: Drag Comparison of Different Aerodynamic Shapes.

Figure 1: Drag Comparison of Different Aerodynamic Shapes.
Source:Comparison of drag characteristics of various nose cone shapes in the transonic to low-mach regions. Rankings are: superior (1), good (2), fair (3), inferior (4).

Comparisons are for projectiles of a given length and width. Observe that the ogive is in the inferior category. However, it is still commonly used for projectiles. Ballistics has a long history and change comes slowly.

Posted in Ballistics | 1 Comment

Ogives and Battleships

Posted on 20-January-2011 by mathscinotes

Introduction

The previous two blogs looked at the ogive shape and its use in describing bullet shapes. While cruising around the web, I noticed a rather large ogive shape that I thought was interesting. I am a big fan of anything having to do with battleships, and the 16-inch projectiles fired by the Iowa-class battleships are an excellent example of a large ballistic ogive. They have an ogive radius of 9 calibers (i.e. 144 inches/ 16 inches = 9 caliber). For those interested in a modern discussion of these projectiles and how to improve them, I suggest this forum discussion.

Background

During WWII, the Iowa-class ships fired two types of projectiles:

  • Mk 13 HC (High Capacity)
    A shell designed to carry the maximum amount of explosive. It was used for shore bombardment against "soft" targets. It weighed 1900 lbs. It is shown in Figure 1.
  • Mk 8 AP (Armor Piercing)
    This projectile is designed to destroy structures made of reinforced concrete or armored ships, like other battleships. This projectile weighed 2700 lbs. It is shown in Figure 2.

Figure 1: Mk 13 HC (High Capacity) 16-inch Shells.

Figure 1: Mk 13 HC (High Capacity) 16-inch Shells.

Figure 2: Mk 8 AP (Armor Piercing), 16 inch Shell (Rotated Horizontal Photo)

Figure 2: Mk 8 AP (Armor Piercing), 16 inch Shell (Rotated Horizontal Photo)
Source Source

Basic Construction

Figure 4 is a good illustration of the differences between the HC and AP projectiles. Notice how the AP shell is basically a big slug of metal that has a windshield on the front of it to make it aerodynamically friendly.

Figure 4: Schematic Diagram of HC and AP 16 Inch Shell Construction.

Figure 4: Schematic Diagram of HC and AP 16 Inch Shell Construction.
Source

Figure 5 shows the actual pieces of an AP shell.

Figure 5: Breakdown of the Mk 8 AP Shell.

Figure 5: Breakdown of the Mk 8 AP Shell.
Source

Figure 6 shows a dimensioned drawing of the HC projectile. Note that the fuze, which attaches to the nose, is not shown. The full length of the projectile with fuze is 64.00 inches.

Figure 6: Dimensioned Drawing of HC Mk 13 Projectile (No Fuze).

Figure 6: Dimensioned Drawing of HC Mk 13 Projectile (No Fuze).
Source

To illustrate that these projectiles have 9 caliber ogives, I fitted a couple of 9 caliber radii circles to one of the photos (see Figure 7).

Figure 7: 9 Caliber Radii Circles Fitted Manually to a Projectile Photo.

Figure 7: 9 Caliber Radii Circles Fitted Manually to a Projectile Photo.

Conclusion

In this post, I showed that even large projectiles use the ogive shape. I collected some useful historical information into a single spot and will use this data in posts to come on ballistics.

Postscript

See Figure 8 for a snippet from a US Navy manual that describes this projectile.

Figure 8: Snippet from 16-inch Gun Range Table. (Source)

Figure 8: Snippet from 16-inch Gun Range Table. (Source)

Posted in Ballistics | Tagged , | 4 Comments

Ballistics, Ogives, and Bullet Shapes (Part 2)

Posted on 12-January-2011 by mathscinotes

Example One: Sierra 308 Caliber, 155 grain, MatchKing.

We will first compute the mass for the Sierra MatchKing projectile (tangent ogive) shown in Figure 10. Observe that this projectile has a flattened nose, called a meplat. Because of the meplat, we do not know the full length of the ogive portion of the projectile. As we show below, however, we can use a numerical solver to find the length very easily.

Figure 10: 308 Caliber, 155 Grain, Sierra Example.

Figure 10: 308 Caliber, 155 Grain, Sierra Example.
Source. All units expressed in inches.

We should review some basic units here.

  • bullet mass is expressed in grains, with 1\text{ grain = }\frac{1\text{ lb mass}}{7000}
  • bullet diameter (at least for North America) is expressed in calibers, with one caliber being one thousandth of an inch.

Determination of Ogive Length

We can use Mathcad to compute the ogive length using its numerical solver. Figure 11 illustrates the calculation. I have defined a length called L' that is the length of the ogive shortened by the meplat. For my equations, I need the length L, which is the length of the full ogive (i.e. not shortened by the meplat). I set up a system of equations that allowed me to solve for L given L' and the meplat diameter.

Figure 11: Variable Definitions for 155 grain Sierra Bullet Example.

Figure 11: Variable Definitions for 155 grain Sierra Bullet Example.
Figure 12: Mathcad Solve Block for 155 Grain Sierra Bullet.

Figure 12: Mathcad Solve Block for 155 Grain Sierra Bullet.


Note that I am estimating the density of the lead alloy by averaging two numbers. I have found various specifications for the density of the alloy and I decided to use the average of the upper and lower values for the range I found. Also, these bullets have jackets made of materials (like copper) with lower density than lead. Some also include steel. The MatchKing actually has a hollow tip. Consider the MatchKing cutaway shown in Figure 13.

Figure 13. MatchKing Cutaway Showing Regions of Lead (Base), Copper (Jacket), and Hollow Tip.

Figure 13. MatchKing Cutaway Showing Regions of Lead (Base), Copper (Jacket), and Hollow Tip.


This will lower the mass of the projectile relative to a solid lead projectile. For my analyses, I will simply use the average of two lead alloy densities.

Mass Calculation

Now that we have the length of the ogive, we can compute the mass of the projectile. Figure 14 illustrates the calculation.

Figure 14: Sierra 155 Grain Mass Calculation.

Figure 14: Sierra 155 Grain Mass Calculation.

As we can see in Figure 14, the computed mass of 160 grains is within 5 grains of the manufacturer's mass specification. This is well within the range possible due to variations in the density of the bullet alloy and the fact that I am ignoring the presence of a jacket.

Example Two: JLK, 7 mm, 180 Grain Bullet.

I found the following drawing for the JLK bullet and thought it would be a good example to use to test my algorithm. This bullet has a secant ogive, which is different than the tangent ogive of Example 1 (i.e. it appears to be more tapered). Figure 15 is a dimensioned drawing of this projectile.

Figure 15: JLK 7 mm, 180 Grain Example.

Figure 15: JLK, 7 mm, 180 Grain Example.
Source. All units expressed in inches.

Determination of Ogive Length

Figures 16 and 17 show the calculation of the ogive length.

Figure 16: Definitions for JLK 180 Grain Bullet.

Figure 16: Definitions for JLK 180 Grain Bullet.
Figure 17: Solve for L in180 Grain JLK Bullet.

Figure 17: Solve for L in180 Grain JLK Bullet.


Mass Calculation

Figure 18 illustrates the calculation of the JLK bullet's mass.

Figure 18: Mass Calculation for JLK Bullet.

Figure 18: Mass Calculation for JLK Bullet.

The calculated mass is within 4 grains of the manufacturer's specification of 180 grains. This is well within the range possible due to variations in the density of the bullet alloy and the fact that I am ignoring the presence of a jacket.

Conclusion

This blog post went through the basic geometry of the ogive and derived a set of equations that allow basic physical parameters such as volume and mass to be computed. Two examples were worked that illustrated how the equations could be applied to real problems. This was a good exercise in the use of basic geometry and applying a computer algebra system to solving a real engineering problem.

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Ballistics, Ogives, and Bullet Shapes (Part 1)

Posted on 11-January-2011 by mathscinotes

Introduction

I have always been interested in the shooting sports, but I have not pursued any of them while I was raising my kids. I suddenly find myself with my kids gone and my interest in shooting has reappeared. As part of my interest in shooting, I have been reading the book "Modern Practical Ballistics" by Pejsa. While reading this book, I quickly learned that not all bullet shapes are created equal. So I started to look at how bullet shapes are defined. To give you an idea of the diversity of bullet shapes, I have included Figure 1, which shows a small number of bullet shape examples.

Figure 1: Example of a Few Bullet Shapes.

Figure 1: Example of a Few Bullet Shapes.
Source

To make this learning exercise more concrete, I decided to focus on developing a general algorithm for computing the mass of a spitzer bullet. I made this choice after reading an interesting article that computed bullet mass using a BASIC program. I used to do a lot of BASIC programming myself, but in recent years I have found computer algebra systems to be more convenient for the working engineer to use on a daily basis. I thought the mass calculation would be a good example to use to demonstrate the power of computer algebra system to solve real problems. I will develop an algorithm for computing the mass of bullet and will test this algorithm on a real bullet design.

The overall exercise was a good exercise in applied mathematics. It also demonstrates the power of a computer algebra system. In this case, I am using Mathcad.

Descriptive Geometry of a Spitzer Bullet

The Pejsa book is focused on the spitzer (German for pointed) bullet shape, and I will concentrate on the spitzer shape as well. Today, it is the most commonly used bullet shape for hunting and target shooting applications. Figure 2 shows how this shape is defined.

Figure 2: Descriptive Geometry of a Spitzer Bullet (Tangent Ogive with Radius of 6 Diameters).

Figure 2: Descriptive Geometry of a Spitzer Bullet (Tangent Ogive with Radius of 6 Diameters).

As shown in Figure 2, the spitzer bullet can be modeled geometrically in three pieces:

  • Ogive (pronounced "Ojive")
    The ogive shape forms the front of the bullet. The ogive shape is formed from the arcs of two circles. The ogive may or may not be tangent at the point of intersection to the cylindrical portion of the bullet. When the circles are tangent to the cylinder portion, we call say this is a tangent ogive. When the circles are not tangent to the cylinder portion, we say we have a secant ogive. The rationale behind the use of the term "secant" can be seen in Figure 6, where there are two horizontal reference lines (brown color) that are both secant lines.
  • Cylinder
    The cylindrical portion of the bullet is what engages the rifling of the barrel.
  • Frustum of a Cone
    The back of the bullet (aka "boattail") geometrically is in the shape of the frustum of the cone Tapering the back of the bullet reduces drag, particularly at speeds less than supersonic.

I have included three examples of ogive-shaped projectiles. Figure 3 shows a bullet with a tangent ogive nose. Figure 4 shows a bullet with a secant ogive nose. Figure 5 shows one of the most famous secant ogive noses, which is on the Honest John missile. I cannot imagine the Honest John ogive being used for a bullet, but it does give a feeling for the range of shapes possible using an ogive model.


Figure 3: Tangent Ogive Example.

Figure 3: Tangent Ogive Example.
Figure 4: Secant Ogive Example.

Figure 4: Secant Ogive Example.
Figure 5: Secant Ogive Example.

Figure 5: Secant Ogive Example.
Source Source Source

Bullet Mass Calculation

Approach

We will compute the volume of the spitzer bullet as follows.

  • Compute the volume of the frustum portion (VFrustum)
  • Compute the volume of the cylinder portion (VCylinder)
  • Compute the volume of the ogive portion (VOgive)
  • Compute the total volume by summing all the volumes of the pieces
    ({{V}_{Total}}={{V}_{Frustum}}+{{V}_{Cylinder}}+{{V}_{Ogive}})
  • Compute the mass using the density of lead (M=\rho_{LeadAlloy} \cdot {{V}_{Total}})

Frustum Volume

The volume of the backside of a bullet (known as the boattail) has the shape of the frustum of a cone. The formula for the frustum is well known and is given in Equation 1.

Eq. 1 {{V}_{Frustum}}=\frac{\pi \cdot L_F}{3}\cdot \left( {{R}^{2}}+R\cdot r+{{r}^{2}} \right)

Cylinder Volume

Equation 2 is used to calculate the volume of the cylindrical portion of the bullet.

Eq. 2 {{V}_{Cylinder}}=\pi \cdot {{R}^{2}}\cdot L_C

Ogive Volume

Types of Ogives

The real work is in computing the volume of the ogive. Let's begin with the variables defined in Figure 6.

Figure 6: Basic Ogive Variable Definitions.

Figure 6: Basic Ogive Variable Definitions.

Using Figure 6, we can see that the different types of ogives are defined by the angle γ. There are three cases (for each case, two criteria are listed– their equivalence is shown at the bottom of this post) :

  • γ < π/2 \left( \rho >\frac{{{R}^{2}}+{{L_O}^{2}}}{2\cdot R} \right)
    The ogive's circular arc is not tangent to the cylinder at the point of intersection. This case results in a rather pointy bullet.
  • γ = π/2 \left( \rho =\frac{{{R}^{2}}+{{L_O}^{2}}}{2\cdot R} \right)
    The ogive's circular arc is tangent to the cylinder at the point of intersection. This results in a rather curved bullet.
  • γ > π/2 \left( \rho <\frac{{{R}^{2}}+{{L_O}^{2}}}{2\cdot R} \right)
    The ogive's circular arc is not tangent to the cylinder at the point of intersection. This case results in a bulbous shape like that of the Honest John missile (see Figure 5).

Analysis

The most important cases for bullet design are when γ ≤ π/2 and the following drawings will focus on these cases. However, the equations are general and apply to all cases. We begin our ogive analysis with Figure 7, which contains the definitions of the critical angles and lengths.

Figure 7: Definitions of Ogive Angles.

Figure 7: Definitions of Ogive Angles.


We need to calculate angles α and γ. To accomplish this task, we will be working with two triangles from Figure 7 (see Figures 8 and 9).

Figure 8: Angles Alpha and Gamma Derivation.

Figure 8: Angles Alpha and Gamma Derivation.
Figure 9: Triangle for Deriving Angle Beta Equation.

Figure 9: Triangle for Deriving Angle Beta Equation.

Using Figures 7, 8, and 9, we can derive the key equations (Equations 3-6). I have used basic trigonometry and will the let the figures stand for themselves. These equations will be used in part 2 to compute the mass of two projectile examples.

Eq. 3 \alpha ={{\cos }^{-1}}\left( \frac{\sqrt{{{R}^{2}}+{{L_O}^{2}}}}{2\cdot \rho } \right)-{{\tan }^{-1}}\left( \frac{R}{L_O} \right)
Eq. 4 \gamma =\pi -{{\tan }^{-1}}\left( \frac{R}{L_O} \right)-{{\cos }^{-1}}\left( \frac{\sqrt{{{R}^{2}}+{{L_O}^{2}}}}{2\cdot \rho } \right)
Eq. 5 y(x)=\sqrt{{{\rho }^{2}}-{{\left( \rho \cdot \cos \left( \alpha \right)-x \right)}^{2}}}-\rho \cdot \sin \left( \alpha \right)
Eq. 6 V_{Ogive}=\int_{0}^{L_O}{\pi \cdot y{{(x)}^{2}}\cdot dx}

Note that two equivalent criteria were listed for identifying the type of ogive. Equation 2 can be used to demonstrate this equivalence. Included below is a short derivation showing that \gamma < \frac{\pi }{2} => \rho > \frac{{{R}^{2}}+{{L_O}^{2}}}{2\cdot R} . The equivalence between the other criteria can be demonstrated similarly.

Derivation Demonstration Using Equation 2

Derivation Demonstration Using Equation 2


Continued on Part 2

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