2
$\begingroup$

Suppose that we have a quadcopter weighing 3 kg (without a battery) and a battery (let it be Li-Ion type) weighing 2 kg. With this battery our quad can hover for 20 minutes until the battery runs flat. How much time can it hover with a 4 kg battery (supposedly our quad has enough power to lift it)?

I know it depends on a tons of factors, so I've made some assumptions for the sake of easier calculations. Here's the math:

I've picked a specific energy definition: $S=\frac{Pt}{m_b}$, where P - power drain, mb - battery mass, t - time until full discharge,

and combined it with a static thrust equation from here:

static thrust equation

Here I assume dry aircraft mass (without the battery) to depend linearly on the battery mass (as we need larger, more powerful motors and a larger hull to fit in and lift a larger battery) with a certain multiplier $m_c$:

$m_b=m*m_c$

After the calculation I've got:

$t=Sm_cD\sqrt\frac{\pi\rho m_c}{2m_bg^3}$,

which means we add more battery and have less flight time! Which can't be true, as otherwise we could use a tiny capacitor to fly for hours. I've done this calculations 3 times with 3 different paths and kept getting the same result. Where did I make a mistake? Or what is the real relation between $m_b$ and t?

Added: Can't reply to the comments yet, but I see some confusion and have to explain - it's a pretty general question, I didn't intend it to be of a "check my work" type, but as I did some work actually, I decided to share it, maybe it will help someone and provide a clue of what am I asking about. The main question is in the title, you may use any means to answer it.

$\endgroup$
  • 4
    $\begingroup$ I strongly disagree with the close vote. This is quite on topic. We want to encourage people to think about physics, which is what is being done here. $\endgroup$ – Ross Millikan Mar 26 '16 at 2:19
  • 1
    $\begingroup$ I didn't analysed your approximations; with roughter naive ones I get 28.6 mn. Still, your conclusion is not valid: this is the famous rocket fueling recursive problem, with more fuel requiring even more fuel to carry the rocket + fuel. This is very non linear, so it could happend that more battery doesn't help if the weight grow faster than effective power, and it's untrue that it implies that an epsilon-battery would allow hours (since then the dominant weight is the dead mass of the copter). The curve possibly has a maximum, then decrease with more weight. $\endgroup$ – Fabrice NEYRET Mar 26 '16 at 2:33
  • $\begingroup$ You didn't make a mistake if you get that "larger" isn't necessarily better for flying. There are insects and certainly birds that can fly as long (and as far) as a 747 (just not as fast)... and they do it on a lot less fuel. A rocket, no matter how large you make it, can only hover its initial mass for the time of the specific impulse of its fuel/engine combination, which rarely exceeds 400s. How long something can fly depends mostly on the efficiency of its lifting mechanism and quad-copters are about as inefficient as things can be. Having said that, you certainly got your equation wrong. $\endgroup$ – CuriousOne Mar 26 '16 at 7:40
  • $\begingroup$ @RossMillikan: This is a check my work question which is considered off-topic. $\endgroup$ – Kyle Kanos Mar 26 '16 at 11:32
  • 1
    $\begingroup$ This isn't "check my work." This is "where did my well-reasoned concepts go wrong?" $\endgroup$ – user10851 Mar 26 '16 at 21:46
0
$\begingroup$

If we eliminate the constants from your equation $6$, we get $$m\propto (DP)^{2/3}$$ which says you get flight time for free by increasing $D$. That is silly because increasing $D$ increases the dry mass of the copter. You assumed that dry mass increases linearly with energy storage, which is a reasonable first cut. Under that assumption, the flight time is proportional to $\frac mP$ because the battery mass is proportional to $m$. That gives $$t \propto \frac mP \propto D^{2/3}P^{-1/3}$$ This says, not surprisingly, that we want bigger rotors to lift the bigger battery. We can afford that because we have more dry mass to play with. At this scale the thickness of the blades is probably driven more by manufacturing than by strength, so it doesn't need to increase (yet) with diameter. It is hard to say how fast you can increase $D$ with mass.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.