Long lasting kid fun

Firstly, we must find a relationship between the energy used by the helicopter, the time it can fly and the mass of the helicopter WITH the batteries.

Firstly how much energy is used to keep the helicopter flying?

Let the mass of the helicopter (with batteries be M). Notice that work is done in pushing the air down.

Thus, over a period of time $dt$ a mass of air $dm$ gains velocity $v$

Since $dP/dT = F$

$dP = F dt$

$dP = Mg dt$

By conservation of momentum, $dP = dm v = Mg dt$

and thus,

$dm = \frac{Mg}{v} dt$

Consider the energy gained by the air, $dE_{gained by air} = \frac 12 dm v^2$

and therefore, the work done by the helicopter is: $dE_{heli} = \frac12 dm v^2$

Sub in the value of $dm$

$dE_{heli} = \frac 12 Mg v dt$

By integrating (and seeing that no variables are functions of time,

$E_{heli} = \frac12 Mg v t$

Now, we have a function that relates the mass with time and energy. the energy is proportional to $l^{2.7}$ and the mass is related to l (in meters) by $M = 4000l^3 +0.05$

we now sub in the values and combine all the coefficients in to a single variable $k$ .

Thus, we get: $l^2.7 = k (4000l^3+0.05) t$

and thus the time of flight is given by:

$t = \frac {l^{2.7}}{k (4000 l^3+0.05)}$

And now we differentiate with respect to l, and to find the max value, we let $dt/dl = 0$

We get a max value of l, when

$l = \frac {3^{2/3}}{20 \cdot 10^{1/3}}$

$l = 0.0482745$

we then must convert this to mm, (as the question requests) and we get $48.27 mm$ .

The question is asking what size of the battery will maximize the time the helicopter can hover in space, but since the energy is not directly related to time, it is hard to create an equation at the first time.

However, since the question did not specify how the helicopter will move in the air, we can create one so that the problem will get easier to solve.

Let's think that the helicopter will move at a certain constant speed vertically until the battery lasts. Then, since the speed is constant, the helicopter that will move highest will have the most time hovering in space.

Then, we can create an equation,

$(m_{0} + m_{b})gh = l^{2.7}$ ,

where the left side represents the maximum gravitational potential energy the helicopter can have and the right side the energy the battery contains ( $m_{b}$ represents the mass of the battery.).

Then, since $m = V\rho$ , we can change $m_{b}$ to $l^{3}\rho$ . Then, the equation is now

$(m_{0} + l^{3}\rho)gh = l^{2.7}$ .

Then, we can express $h$ as

$h = \frac {l^{2.7}} {m_{0} + l^{3}\rho} = \frac {l^{2.7}} {50 *10^{-3} + l^{3} * 4000}$ .

Then, we can differentiate this equation in respect to $l$ and find a maximum point.

Then, the maximum point in that equation is when $l = 4.8274469 * 10^ {-2}$ . However, since that $l$ is in meters, we have to multiply it by $10^ {3}$ to find a value in millimeter. Therefore, the answer is $l = 48.27mm$ .

The total mass of the system in kg is $m=0.05+4000\cdot l^3$ .

The air flows at constant velocity throw the blades of the helicopter and the total force $F$ of the blades in the air molecules is equal to the height of the helicopter, i.e., $F=mg$ .

As the force and the velocity of the air molecules are constant when they are passing throw the blades, one can write, being $P$ the power of the battery and $v$ the air molecules velocity:

$P=F\cdot v=mgv$

Moreover, we know that $E=P\cdot t$ and $E\propto l^{2.7}$ , so $E=k\cdot l^{2.7}$ for some constant $k$ . Then:

$t=\dfrac{E}{P}=\dfrac{k}{gv}\cdot\dfrac{l^{2.7}}{0.05+4000\cdot l^3}$ $\dfrac{d t}{d l}=\dfrac{k}{gv}\cdot\dfrac{8.4375\cdot 10^{-9}\cdot l^{1.7}-7.5\cdot 10^{-5}\cdot l^{4.7}}{(l^3+1.25\cdot 10^{-5})^2}$

Setting $\dfrac{dt}{dl}=0$ :

$8.4375\cdot 10^{-9}\cdot l^{1.7}=7.5\cdot 10^{-5}\cdot l^{4.7}$ $l^3=\dfrac{8.4375\cdot 10^{-9}}{7.5\cdot 10^{-5}}$ $l\approx 0.04827\ m=48.27\ mm$

The total energy stored in the battery is directly proportional to $l^{2.7}$ . Meanwhile, the total energy required to keep the helicopter in the air is directly proportional to the total mass of the helicopter, which is $4000l^3 + 0.05$ . Therefore, the amount of time the helicopter is in the air is directly proportional to $\dfrac{4000l^3 + 0.05}{l^{2.7}} = 4000l^{0.3} + 0.05l^{-2.7}$ . To find the maximum of this, we find the derivative and equate it to zero. Thus, $1200l^{-0.7}-0.135l^{-3.7}=0$ . Solving this gives $l = 0.04827$ . This is indeed the maximum, as can be checked from the second derivative test. Therefore, the linear size of the battery that will maximize the time the helicopter can hover in place is $0.04827 \textrm{ m}$ , or $48.27 \textrm{ mm}$ .

The force the rotors must provide to the helicopter to hover in place must counteract the force of gravity, i.e. $F=(m_0+m_b)g$ , where $m_b$ is the mass of the battery. The total momentum the rotors provide to the air per unit time is therefore

$\frac{dp}{dt}=(m_0+m_b)g$ .

If we think about a collection of air molecules with some mass $m_a$ that get momentum over some short time frame from the rotors, the energy of the air is simply $E=p^2/2m_a$ . We therefore find that $dE=vdp$ , where $v$ is the velocity acquired by the air. Substituting this in above, we have that

$\frac{dE}{dt}=v(m_b+m_0)g$ .

Since the voltage provided by the battery is constant so is $v$ and therefore $dE/dt=E/T$ where $E$ is the total energy in the battery and $T$ is the hover time. Solving for $T$ and substituting how $E$ and $m_b$ change with $l$ gives

$T=\frac{E_0 l^{2.7}}{v(\rho l^3 + m_0)g}$ .

We can minimize this with respect to $l$ , which yields

$l^3=\frac{2.7 m_0}{0.3 \rho}$

and $l=0.04827~\mbox{m}=48.27~\mbox{mm}$ .

From the assumptions, we know that, battery lifespan $T$ is directly proportional to $l^{2.7}$ and inversely proportional to the total mass of the helicopter including the battery which is $4000l^{3}+0.05$ . That is: $T\quad \alpha \quad \frac { { l }^{ 2.7 } }{ 4000{ l }^{ 3 }+0.05 }$ $T$ is maximum when $\frac { { l }^{ 2.7 } }{ 4000{ l }^{ 3 }+0.05 }$ is maximum or $X=\frac { 4000{ l }^{ 3 }+0.05 }{ { l }^{ 2.7 } }$ is minimum. That is when: $\frac {dX}{dl} = \frac {12000l^{2}}{l^{2.7}} - \frac {10800l^{3}+0.135}{l^{3.7}} = \frac {1200l^{3}-0.135}{l^{3.7}}$ $X$ is minimum when $\frac {dX}{dl} = 0$ or $1200l^{3}=0.135$

$\Rightarrow l = \sqrt [ 3 ]{ 0.0001125 } = 0.04827 m = \boxed{48.27} mm$

As the energy is proportional to the volume of the battery, the power P is also proportional to that of the battery. We can express it as $P=kl^{2.7}$ , where k is a constant.

The total mass is the mass of the battery and the mass of the base. $M_{total}=\rho l^3+m_0$

Since $P=Fv, v=\frac{P}{F}$ , where P denotes power, F denotes force and v stands for velocity.

From these we can obtain an equation,

$v=\frac{P}{F}=\frac{kl^{2.7}}{(\rho l^3+m_0)g}$ ,

Substitute the values in. Upon differentiating, we get,

$\frac{dv}{dl}=\frac{8.60969\times 10^{-10} l^{1.7}-7.65306\times 10^{-6} l^{4.7}}{(0.0000125+l^3)^2}$

Setting the derivative to be 0 to achieve the maximum time the helicopter can hover in air and solve for $l$ ,

Since $l$ is a positive real number, the only acceptable solution is $l=0.0482745m\approx48.27mm$

We're told that the behavior of the battery is all-or-nothing, i.e. the battery outputs some constant voltage $V_0$ as long as there is any energy left and then has a voltage of zero after the energy runs out.

The downward thrust delivered by the engine must be given by some function of the battery's output voltage $f\left(V_0\right)$ and since $V_0$ is constant, the delivered thrust must be constant at well. In general, this implies that the engine delivers a constant amount of power.

In a hover, the thrust of the blades is equal in magnitude to the downward pull of gravity, given by $\vec{F_g} = \vec{g}\left(m+\rho l^3\right)$ . The blades do work by accelerate air molecules downwards, i.e. the lifting thrust is equal to the momentum imparted on the molecule of air. This is expressed by $\vec{F_L} = \vec{F_g} = \frac{d\vec{p_{air}}}{dt}$ .

The rotors deliver a constant velocity to the air molecules, so the time derivative of momentum reflects the flow of massive air molecule through the blades $\frac{d\vec{p_{air}}}{dt} = \frac{dm_{air}}{dt}\vec{v_{air}}$ .

The problem tells us to assume that the flow of mass is a constant of the helicopter, independent of the rotor speed. In reality, the velocity imparted on the air molecules is a function of the thrust. Let's carry through with the calculations in parallel, noting one as the physical case and one as the constant case. In the constant case, the mass of air moving through the blades per unit time is given by a constant $J_m$ . In the physical case, the mass of the air moving through the blades per unit time $dm_{air}$ is given by $\rho_{air}A_{blades}v_{air}dt$ . This reflects the idea that there is a column of air above the blades, roughly cylindrical, that gets sucked through the blades. $A_{blades}$ is the area of the circle swept out by the blades and the heigh of the cylindrical air column pulled through in a time $t$ is $v_{air}t$ , therefore, the volume of air pulled through per unit time is $A_{blades}v_{air}$ . We therefore have:

$F_T= \begin{cases} J_mv_g & \mbox{constant}\\ \rho_{air}A_{blades}v_{air}^2 & \mbox{physical} \end{cases}$

The work done by a force acting through a distance is given by $W = \int\vec{F}\cdot d\vec{x}$ . Similarly, the power $P = \frac{dW}{dt}$ is given by $P = \frac{d}{dt} \int\vec{F}\cdot d\vec{x} = \frac{d}{dt}\int\vec{F}\cdot \frac{d\vec{x}}{dt}dt = \vec{F}\cdot \vec{v}$

Carrying on with the two cases, the power required for the hover scales as

$P_{hover}= \begin{cases} v_g^2 & \mbox{constant}\\ v_{air}^3 & \mbox{physical} \end{cases}$

or in terms of the thrust

$\displaystyle P_T= \begin{cases} F_T & \mbox{constant}\\ F_T^{3/2} & \mbox{physical} \end{cases}$

The amount of time that the power required for flight $P_T$ can be sustained by a battery of total energy $E$ is given by $t_{flight} = E/P_T$ .

$E$ scales as $l^{2.7}$ , so $t_{flight}$ scales as

$\displaystyle t_{flight}\sim \begin{cases} \frac{l^{2.7}}{m_o+\rho l^3} & \mbox{constant}\\ \frac{l^{2.7}}{\sqrt[\frac{3}{2}]{m_o+\rho l^3}} & \mbox{physical} \end{cases}$

We aim to set $l$ in such a way the time of flight is maximized, so we set the derivative $\displaystyle \frac{d}{dl}t_{flight}$ equal to zero and solve for $l$ :

In the constant case, we find

$l^3 = \frac{2.7}{3-2.7}\frac{m_0}{\rho}$

or

$\displaystyle l = 3^{2/3}\left(\frac{m_0}{\rho}\right)^{1/3}$

In the physical case, we find

$l = \frac{3m_o}{2\rho}^{1/3}$