Economics of Fire Protection

There is a set of interesting articles in the Louisville Courier-Journal regarding suburban fire protection districts. There is great debate in Jefferson County Kentucky regarding merging fire districts together forming one big department or a few small ones. The debate goes on with regard to this merger.

I tend to look at the fiscal part of the fire services on a macro level or a per capita basis of providing protection. While this is not an exact science and there should be a greater matrix on looking at the quality of a particular department as well as the costs that it takes to run it, the per capita cost can give one a guide to the cost of fire protection. Services vary for example, some departments provided EMS to its residents, while others are combination departments having both career and volunteers provide services. Then there are all volunteer departments as well as all paid departments having all career firefighters. When looking at the per capita spending it is important to understand exactly the services that are being offered for the money.

Per capita spending on public safety and specifically in fire protection varies widely across the country states and counties. For example in my community we are providing fire protection services for $143 per capita per year. In the next community near my house (1/4 mile) they are spending $25 per capita on fire protection, and in the city nearby they spend $250 per capita on fire protection.  In Jefferson County Kentucky the proposal is that for $125 million, fire protection can be provided to the entire community. That would be 17 departments merging into one. The per capita cost of that in Jefferson County would be $167.

The services provided in each of the communities mentioned above do vary.  My community has a part career and volunteer force.  The community next to me is an all volunteer force.  As you can see the funding of an organization by examining per capita spending at a macro level is one metric that should not be ignored in public safety.  It will be difficult to balance the services in the mergers between an organization spending $25 per capita and another spending $250 per capita as there would be some extensive service balancing.

In general if efficiencies can be increased and levels of service increased while maintaining steady per capita spending over time merger should happen. The fire service is here to help people and render care. We should all do this to the highest level that we can. If we analyze the financial impact of providing services, increasing resources where they are needed and decreasing in other areas that are over funded, we can provide a better net product.

It seems that the fire service, police and EMS are not immune to budget constraints in every community across the country. I do feel in Indiana we are in better shape than in other states around us. Fire departments and fire administrators must be cognizant of the financial implications of providing public safety. We must remain diligent in looking at data and providing the highest level of service that we can afford for the taxpaying public. If merger is in the cards, and it provides a better service to the public, then it should happen and as public officials we should not be in the way.

Public safety is near and dear to my heart. I serve on the Board of Trustees for the Georgetown Township Fire Protection District  I'm an alumnus of the National Fire Academy and served as a volunteer firefighter for 13 years.

Rocket Equations

This summer I decided that I would spend some time with my 13-year-old daughters building model rockets. The process of doing this would teach them how to measure and build as well as engineer something that actually did something cool such as fly.  Building rockets is a good way to study a wide range of physics, engineering, math and science. I built rockets when I was a kid and it was such great fun I thought I would do it again.

I brushed up on my rocket skills and read the official NAR handbook. In the book they described rocket equations to estimate the height and velocity that a rocket will go theoretically ignoring aerodynamic drag. As I started looking at the equations I was slightly confused as to how they were derived. I looked around on the Internet and did not see a clear  representation of the math in one place that my kids would understand. My daughters have had basic algebra but most of the derivations on the net leave a lot of the steps out. I decided to derive these equations by brushing off my college physics book and going through it in a very verbose way. I’m writing about this because along with my daughters I thought it would help others who are teaching younger students on the basics of the equations of motion. 


Between reading the NAR handbook and this you will have a good understanding of the forces on a rocket.  The equations of motion are well documented in general. I’m going to glue together the basics of finding out how fast a model rocket will go and how high theoretically it will go. We will calculate the velocity of the rocket after its initial burn and then after the coasting of the rocket in the air. Remember this only works ignoring aerodynamic drag which is a very big piece of how a rocket flies.  Later I may address the more complex math but for now we’ll start with the easy stuff.

By definition these equations work with constant acceleration. We use Newton’s laws of motion.  All of Newton’s laws work on a rocket and affect how it flies. Remember this is ideal.  For much better real world models use RockSim or OpenRocket.  

Given the graph below with a line in a coordinate plane where Y is equal to velocity and X is equal to time. In this basic graph of velocity versus time I have defined acceleration as the change in velocity over the change in time. If we add up the area under the line this would be the area of a rectangle plus the area of the triangle. This area is called the displacement. For our equations this will be the altitude of a rocket.

Again looking at the graph of the line of acceleration as:

Remember that acceleration is the slope of the graph.  Rearranging gives us:

This final equation looks a lot like the basic equation for a straight line that everyone is used to.

We want to find the area under the line which is our displacement. In order to do this accurately we will use calculus and integrate velocity with respect to time.

Plugging in equation 1 to the integration and solving gives us:

By integrating we add up all the area under the line, that’s all integration really is. “s” is the disposition.  Physicist use s for some reason (note s = d in our graph).  This equation, we will call equation 2, has acceleration in it. In order for us to use it for our rocket equation we need to get acceleration out by substituting our definition of acceleration. We’re going to do some substitution and simplify.

The next equation we will take equation 3 and eliminate time.  The algebra on this is a little messy but I have all the steps and I’m using the binomial theorem.  Substitute t in equation 3 and simplify.

After all that work equation 3 is our altitude at burnout and equation 4 is our coasting altitude.

In order to get the altitude at burn out we will need the velocity at the time our motor runs out of burn. In order to get this we will use Newton’s Second Law F=ma:

T = thrust or force to put the rocket in the air 

t = motor burn time

g = acceleration due to gravity. 9.8m/s^2

wavg = average weight of rocket

vm =   maximum velocity during motor burn

In order to use this law we need to include gravity with mass to figure the force on our rocket and that is the gravity for acceleration. Given our equation above for velocity we will substitute.

Finally we can calculate some ideal values for our rocket. Given the picture of our rocket below lets figure out our altitude at burnout and our maximum altitude theoretically.

Using my kids Alpha rocket with an A8-3 motor in it weighs 38.8 grams at lift off. The weight of the propellant is 3.12 grams giving an average weight of 37.2 grams during the thrust of the flight. The motor thrusts for .32 seconds and has an impulse of 2.50 Newton seconds. In our equations you will want to convert grams to kilograms as the standard unit of measurement.

The maximum velocity of our rocket ideally will be 64.03 m/s or 143 mph.  Now let’s figure our altitude after the rocket motor burns.

Using equation 3:

Our rocket altitude at motor burnout is 10.24 meters or 32 feet above the ground theoretically.

Using equation 4 we can find out the total altitude theoretically that our rocket will go.

Our rocket theoretically will go 219.22 m or 719 feet in the air ignoring aerodynamic drag with constant acceleration.

A few notes on this last equation to solve for the coasting altitude. Many of the units I’ve used here are called vectors. They have a magnitude and direction. Gravity has a  downward direction. This is why the 9.8 is in the negative direction. It works out because negative divided by a negative is a positive and altitude needs to be positive. I didn’t mention vectors throughout because it may have been more confusing. Be aware however that a rocket doesn’t go just straight up it has an X, Y and Z coordinate.  We know when we shoot off a rocket it tends to point into the wind which would decrease its overall altitude.