# What is the correct thrust curve for a solid rocket with a simple circular hole, and why?

Going to wikipedia's article on solid-fuel rockets, I come across some graphs of thrust curves for certain bore-hole geometry. Right now I'm interested in the simplest bore-hole: a circular hole right in the center of the fuel grain: This thrust-curve seems totally wrong. For a circular hole with a linear burn rate, the radius should expand linearly. Since C = pi*d, the circumference should also increase linearly. (Burn rate is proportional to the circumference of the exposed fuel grain.) Therefore, the mass flow should also increase linearly. Therefore, the thrust should also increase linearly.

(That all assumes that all else is equal. I don't know much about solid-rocket calculations. I've tried working with them twice over the years, and I came to the conclusion that solid rocket calculations are even more complex than liquid rocket calculations, which really bugged me cuz solid rockets appealed to my sense of simplicity due to having no moving parts. I'm not sure if combustion pressure goes up linearly, but it seems it would have to...)

Anyway, if all this is true, then the thrust curve should not even be a curve at all. It should just be a straight line going up at some angle...maybe steeply, maybe shallowly, but linearly.

The graph shown is radically different than that. Am I missing something?

What is the thrust curve for a solid rocket with a simple circular hole, and why?

Hill and Peterson "Mechanics and Thermodynamics of Propulsion", third printing, November 1970, page 385, has a diagram that agrees with your intuition.

(sorry for poor scan quality)

You are correct - as the surface area increases, so does the mass flow rate. There must be other factors in play in the graph from Wikipedia.

• Thanks. The scan quality is good enough, but for best results i recommend a cell phone camera to take the photo and crop as needed. – DrZ214 Aug 29 '16 at 17:55

Underneath the diagrams in the Wikipedia article you link, there's a mention of the BATES grain geometry:

Circular bore: if in BATES configuration, produces progressive-regressive thrust curve

https://en.wikipedia.org/wiki/BATES

In which there is combustion occurring on both ends of a cylindrical segment as well as the central bore.

In such a case, there are two factors which decrease the burn surface area over time, counteracting the linear increase expected from a circular bore.

First, the length of the bore decreases as the ends burn away. The inner burn area is the product of the increasing circumference and the decreasing length.

Second, the burning end surfaces are annular, with area decreasing over the burn as the inner diameter of the ring increases. The end-burn area loss is proportional to the square of the burn time.

Depending on the proportions of the initial bore diameter, outer diameter, and length, a variety of thrust curve shapes can be achieved. A quick experiment in a spreadsheet shows that a grain segment with outer diameter of 5 units, initial inner diameter of 2 units, length 14 units, with double-end burning yields a pretty good match to the curve we're trying to understand. • I'm not sure i follow you. What i get from the idea of tubes burning at both ends and also around the inside hole is that it each segment gets shorter and the circumference of the center hole also gets larger with burn time, and those things balance so the total surface area stays fairly constant. Is that the point or am i missing something else? – kim holder wants Monica back Aug 29 '16 at 14:47
• Ah, I wasn't even thinking that through! Three things are happening. The inner circumference is increasing, linearly with time. The length is decreasing, linearly with time. The inner burning surface is the product of those two, so it changes slowly, as you note. In addition, the end surfaces are also burning, and their area is decreasing quadratically. – Russell Borogove Aug 29 '16 at 14:58