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.

These and more questions are investigated in this note, as I channel my inner Professor O’Neill as inquisitor in a digital oral defense. My notes are added as <comments>.

Oral exam portion

I asked ChatGPT to explain drag. This isn’t terribly wrong until the induced drag discussion.

Q: "Explain aerodynamic drag from a technical point of view"

Aerodynamic drag is the force that opposes an aircraft's motion through the air. It is caused by the difference in pressure between the front and rear of an aircraft, as well as the friction of the air molecules against the surface of the aircraft. It can be broken down into two main components: parasitic drag and induced drag. <Good>
Parasitic drag is caused by friction of air molecules against the surface of the aircraft, and includes skin friction and form drag. Skin friction is caused by the friction of air molecules against the surface of the aircraft, while form drag is caused by the shape of the aircraft.
Induced drag is caused by the production of lift by the aircraft's wing. As the wing generates lift, it also generates a swirling wake of air behind it, which creates a low pressure area. This low pressure area pulls the tail of the aircraft backwards <Totally wrong>, creating drag.

I would say that this discussion is no worse than most undergraduates who have not taken an aerodynamics course. ChatGPT needs to study Low Speed Aerodynamics, available here. Induced drag is caused by the production of lift and does create a swirling wake of air that convects downstream. The wake is not necessarily a low pressure area, so ChatGPT is not correct. The following statement about pulling the tail backwards is completely nuts; induced drag is not conceptually applied at the aircraft tail.

A previous analysis indicated that the POH fuel-tank recommendations reduced the wing root bending moment. This note attempts to quantify the lifespan impact of the estimated maximum 15% reduction in bending moment.

This video by Airframe Components provides an excellent visual understanding of the PA-28 and PA-32 spar structures. Please review this video before continuing.

The PA-28 aircraft use the NACA 65(2)-415 airfoil (c.f. Piper PA-28 Cherokee Wing Comparison: An Aerospace Engineering Perspective) with wing spar attach points as illustrated in Figure 1. With the spar at the maximum 15% thickness of a 63 inch chord, the spar height is about 9.5 inches. The moment of inertia is about 18.7 [in4] resulting in a stress of 0.253 [psi] per moment [in-lb]. However, the presence of holes significantly increases the stresses at the bolt holes by about 3x.

Aluminum has the unfortunate mechanical property that all loading -regardless of the stress magnitude- contribute to the metallic lifespan.

To be continued.

The Piper Cherokee PA-28-235 recommends the use of the tip tanks last. The purpose of this note is to estimate the reduction in wing root bending moments associated with this strategy. The PA28 model has a known AD history with wing root cracks; reducing the bending reduces the stress and increases the structural life. The POH for a C model says:

To familiarize yourself with the fuel available on the aircraft, the planform view of the aircraft from the POH with fuel tank overlays is illustrated in Figure 1. The aircraft has 4 tanks: two 25 gallon main tanks and two 17 gallon tip tanks for a total of 84 gallons of fuel. The main tanks are approximately 70 inches outboard from the centerline and the tip tanks are approximately 180 inches outboard.

Aerodynamics: Using the process described in https://charles-oneill.com/blog/cherokee-tapered-wing-float/, the wing’s effective spanwise aerodynamic center is located at approximately 43% of the span. To remain conservative, we ignore the structural weight of the wing. The simplified configuration is given in Figure 2.

Summation of moments about the root (left side) gives the following equation. When substituting for the wing panel’s lift (L/2) and the weight of fuel, the overall moment at the wing root is

Plotting this wing root bending moment (per g) over the aircraft’s envelope provides a visual comparison of the differences in loading and operational techniques. All feasible loading and operating conditions lie within the gray shaded region. The worst case (red color) is a light 150 lbf -but perhaps not so bright- pilot with VFR minimum fuel and a maximum payload to achieve gross weight of 2900 lbs; notice that the Cherokee does NOT appear to have a zero fuel weight limitation (ZFW). On the other extreme, a light 150 lb pilot with no payload results in the conditions at lower left, where the POH loading suggestion (main 1st, then tips) is the lower black line. The green line shows the results when tips are fed first until dry and then the mains are fed.

Conclusion 1: Adding fuel reduces wing bending at the root. This immediately shows shows that the tip fuel is almost 5 times more effective at reducing the bending moment per gallon.

Conclusion 2: Each gallon of main fuel is structurally equal to a reduction in payload weight of 4 pounds.

Conclusion 3: Each gallon of tip fuel is structurally equal to a reduction in payload weight of 21 pounds. Adding fuel tip tanks (17 gallons) acts to reduce fatigue stresses similar to reducing the payload weight by 360 pounds. This is not a negligible amount.

Conclusion 4: The maximum benefit of feeding the mains before the tips is approximately a 15% reduction in bending moment.

Statement 1: This analysis does NOT include the effects of maneuvering speed. Please refer to the appropriate POH for guidance and remember that Va reduces at lower weights.

Statement 2: The 15% reduction in bending moment contributes FAR more than 15% to the aircraft lifespan. 15% is in fact a substantial number. Further analysis of this fact will be conducted later (See Cherokee 235 Tip Tanks and Fatigue).

Conclusion: The Piper POH’s suggestion to use the main tanks first does have a substantial structural fatigue benefit.