The Rocket Equation is a fundamental solution for rocket conceptual understanding and development. But did you know that there is an analytical solution to the distance, the integral of the rocket equation?

ΔV Solution

The ideal rocket equation links the change in velocity ΔV to the propulsion system’s effective exit velocity and the mass fraction. We often see these written in terms of the specific impulse Isp.

The critical conceptual points are that the velocity change depends linearly on Isp but non-linearly with the mass fraction. The trivial zero fuel case (i.e. final = initial) has a natural log of 1 which computes to a velocity of zero. As Gus Grissom says, “No bucks, no Buck Rogers.” Adding fuel increases the rocket’s velocity change, but only in a log proportion to the fuel added. The concept boils down to one concept: Accelerating fuel requires fuel.

The rocket equation is valid across the propulsion burn, such that the instantaneous ΔV is computed from the burn parameters. If we compute the final mass as:

then ΔV(t) as a function of time is

The second form consolidates the terms into two canonical parameters: (1) a propulsion term with Isp, and (2) a mass flow ratio to the initial mass. Implied in this form is the need to calculate the total burn time. The total burn time multiplied by the mass flow rate is the total propellant available, such that the final delta V is only a ln function of the ratio of propellant to initial mass and a linear function of the Isp. The following figure shows the delta V versus time for two different propellant mass flows.

Integral Distance Solution

Now for the rest of the story. The integral of ΔV(t) is the rocket distance, which has an exact analytical solution. We will start with the canonical form from above.

Now, the rate of change of distance is velocity, such that the differential distance versus differential time relationship is:

An analytical integral is available for this form, such that the distance traveled as a function of time when starting from rest in empty space is:

This can be particularly useful for verification of rocket solutions.

More to follow

This is a talk for the Texas Flying Club.

This note demonstrates how to quickly find the volume of a rotationally symmetrical enclosure. I find this approach useful for estimating and designing tank volumes. As an example, you can quickly find the volume of a toroidal tank with a 1 inch circle revolved about a line 2 inch from the circle’s center is slightly under 10 cubic inches (precisely π squared).

The equation is simple: Volume = 2π·A·L, where A is the area of the revolved cross section and L is the distance from the rotational axis to the center-of-gravity of the cross sectional area. For the torus, the area is 2π·πD²/4·L which simplifies to π²D²L/2. Neat and fast.

Here’s a design application: develop an equation for elliptical ended cylindrical tanks to hold precisely G gallons. Using the revolution equation in the figure below gives a total volume of V=πR²(h+4/3·b).

Now conversion to gallons is a linear transformation with 231 in³ per gallon.

This approach is also useful for crazy, bizarre, strange, and otherwise normal volumes of revolution. For instance, you can now find the volume of a specialty triangular o-ring seal.

The triangular cross sectional area is half the base times height, so 50 mm². The center of gravity of a triangle is 2/3 the way from the tip to the base, so that’s 6.6 mm. The total distance from the axis to the CG is 20+6.6 = 26.6 mm. Multiply and you obtain: Volume = 2π 50 * 26.6 = 8377 mm³

Why does this work?

The key trick here is recognizing that in cylindrical coordinates, the differential volume is dV=r·dr·dθ·dz, which appears to contain the moment equation dM=x·dx in a 2D Cartesian frame. A derivation is presented below.