# Boosting rocket flight performance without calculus

Fig. 1

The stages of a two-stage rocket have initial masses $m_1$ and $m_2$ respectively and carry a payload of mass $P$. Both stages have equal structure factors $e$ and equal relative exhaust speed $c$. The rocket mass, $m_1+m_2$ is fixed and $\frac{P}{m_1+m_2}=b$.

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two-stage rocket is

$v = -c \log(1-\frac{em_1}{m_1+m_2+P}) - c\log(1-\frac{em_2}{m_2+P})$

Let $a = \frac{m1}{m_2}$, it becomes

$v = -c\log(1-\frac{ea}{a+1+b(a+1)})-c \log(1-\frac{e}{1+b(a+1)})$

where $a>0, b>0, c>0, 0< e < 1$. We will maximize $v$ with an appropriate choice of $a$.

That is, given

$v = c\log(\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1})$

where $a>0, b>0, c>0, 0. Maximize $v$ with an appropriate value of $a$.

The above optimization problem is solved using calculus (see “Viva Rocketry! Part 2“). However, there is an alternative that requires only high school mathematics with the help of a Computer Algebra System (CAS). This non-calculus approach places more emphasis on problem solving through mathematical thinking, as all symbolic calculations are carried out by the CAS (e.g., see Fig. 2). It also makes a range of interesting problems readily tackled with minimum mathematical prerequisites.

The fact that

$\log$ is a monotonic increasing function $\implies v_{max} = c\log(w_{max})$

where

$w = \frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)}$

or

$w(1-e+(a+1)b)((1-e)a+(a+1)b+1) - (a+1)(b+1)((a+1)b+1)=0\quad\quad\quad(1)$

(1) can be written as

$A_1 a^2 + B_1 a +C_1= 0$

where

$A_1 = -bew+b^2w+bw-b^2-b$

$B_1 = e^2w-2bew-2ew+2b^2w+3bw+w-2b^2-3b-1$,

$C_1 = -bew-ew+b^2w+2bw+w-b^2-2b-1$.

Since $A_1 = 0$ means

$-b e w + b^2 w + b w - b^2 - b =0$.

That is

$w = -\frac{b+1}{e-b-1}$.

Solve

$\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = -\frac{b+1}{e-b-1}$

for $a$ gives $a = 0 \implies A_1 \neq 0$ if $a > 0$.

Hence, (1) is a quadratic equation. For it to have solution, its discriminant $B_1^2-4A_1C_1$ must be nonnegative, i.e.,

$(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1) w +(b+1)^2 \geq 0\quad(2)$

Consider

$(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1)w+(b+1)^2 = 0\quad(3)$

If $e^2-2be-2e+b+1 \neq 0$, (3) is a quadratic equation.

Solving (3) yields two solutions

$w_1 = -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2}$,

$w_2=\frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$.

Since $0 < e < 1$,

$w_1 - w_2 = -\frac{4(b+1)\sqrt{b(b+1)}(e-1)e}{(e^2-2be-2e+b+1)^2} > 0\quad(4)$

(4) implies

$w_1 > w_2$

and, the solution to (2) is

$w \leq w_2$ or $w \ge w_1$

i.e.,

$w \leq \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(4)$

or

$w \ge -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(5)$

We prove that (4) is true by showing (5) is false:

Consider $w - w_1= 0$:

$\frac{(b+1)(e-1) e \cdot f(a)}{(1-e+ab+b)(a(1-e)+ab+b+1)(e^2-2be-2e+b+1)^2} = 0\quad\quad\quad(6)$

where

$f(a) = 2a\sqrt{b(b+1)}e^2+a^2be^2+be^2+e^2-2a^2b\sqrt{b(b+1)}$

$-4ab\sqrt{b(b+1)}e - 2b\sqrt{b(b+1)}e - 4a\sqrt{b(b+1)}e$

$-2\sqrt{b(b+1)}e-2a^2b^2e-4ab^2e-2a^2be-4abe-4be$

$-2e+2a^2b^2\sqrt{b(b+1)}+4ab^2\sqrt{b(b+1)} + 2b^2\sqrt{b(b+1)} +2a^2b\sqrt{b(b+1)}$

$+6ab\sqrt{b(b+1)} + 4b\sqrt{b(b+1)} + 2a\sqrt{b(b+1)} + 2\sqrt{b(b+1)}$

$+2a^2b^3 + 4ab^3 + 2b^3 +3a^2b^2 + 8ab^2 +5b^2+a^2b+4ab+4b+1$.

It can be written as

$A_2a^2 + B_2a + C_2\quad\quad\quad(7)$

where

$A_2 = be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}$

$+2b^3+3b^2+b$,

$B_2 = 2\sqrt{b(b+1)}e^2-4b\sqrt{b(b+1)}e -4\sqrt{b(b+1)}e$

$-4b^2e-4be+4b^2\sqrt{b(b+1)}+6b\sqrt{b(b+1)}+2\sqrt{b(b+1)}+4b^3+8b^2+4b$,

$C_2=be^2+e^2-2b\sqrt{b(b+1)}e-2\sqrt{b(b+1)}e-2b^2e-4be$

$-2e+2b^2\sqrt{b(b+1)}+4b\sqrt{b(b+1)}+2\sqrt{b(b+1)}+2b^3+5b^2+4b+1$.

Since $A_2 > 0$ (see Exercise 1) and,

solve (7) for $a$ yields

$a = -\sqrt{1+\frac{1}{b}}$.

It follows that for $a > 0, f(a) > 0$.

Consequently, $w-w_1$ is a negative quantity. i.e.,

$w-w1 < 0$

which tells that (5) is false.

Hence, when $e^2-2be-2e+b+1 \neq 0$, the global maximum $w_{max}$ is $w_2$.

Solving $w = w_2$ for $a$:

$\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$,

we have

$a = \sqrt{1+ \frac{1}{b}}$.

Therefore,

$e^2-2be-2e+b+1 \neq 0 \implies w$ attains maximum at $a = \sqrt{1+ \frac{1}{b}}$.

In fact, $w$ attains maxima at $a = \sqrt{1+\frac{1}{b}}$ even when $e^2-2be-2e+b+1 = 0$, as shown below:

Solving $e^2-2be-2e+b+1 = 0$ for $e$, we have

$e_1= -\sqrt{b(b+1)}+b+1$ or $e_2 = \sqrt{b(b+1)} + b + 1$.

Only $e_1$ is valid (see Exercise-2),

When $e = e_1$,

$w(\sqrt{1+\frac{1}{b}} )- w(a) = - \frac{(b+1)g(a)}{4 \sqrt{b(b+1)} (\sqrt{b(b+1)}+ab) (a\sqrt{b(b+1)}+b+1)}\quad(8)$

where

$g(a) = (2a^2b+4ab+2b+2a+2)\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1$

Solve quadratic equation $g(a) = 0$ for $a$ yields

$a = \sqrt{1+\frac{1}{b}}$.

The coefficient of $a^2$ in $g(a)$ is $2b\sqrt{b(b+1)}-2b^2-b$, a negative quantity (see Exercise-3).

The implication is that $g(a)$ is a negative quantity when $a \neq \sqrt{1 + \frac{1}{b}}$.

Hence, (8) is a positive quantity, i.e.,

$e^2-2be-2e+b+1 = 0, a \neq \sqrt{1+\frac{1}{b}} \implies w(\sqrt{1+\frac{1}{b}})-w(a) > 0$

We therefore conclude

$\forall 0 < e < 1, b > 0, w$ attains its maximum at $a = \sqrt{1+\frac{1}{b}}$.

Fig. 2

Exercise-1 Prove:$00 \implies$

$be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}+2b^3+3b^2+b > 0$

Exercise-2 Prove: $b > 0 \implies 0 <-\sqrt{b(b+1)} + b +1 <1$

Exercise-3 Prove: $b > 0 \implies 2b\sqrt{b(b+1)}-2b^2-b < 0$