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Especially given the relative cost between (say) steel and carbon composites. After all, I assume most fuel is consumed overcoming drag not accelerating mass. Once an aircraft reaches cruising speed it should not matter how heavy/dense the aircraft is as long as the lift/drag remains constant.

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    $\begingroup$ This might be better at Aviation. $\endgroup$
    – ACuriousMind
    Apr 18, 2015 at 12:02
  • $\begingroup$ Don't assume. It makes... you know! How does an airplane fly? By forcing enough air downwards with greater momentum than the desired momentum (vs. gravity) of the airplane. The mass of the plane, not to mention your luggage, is asignificant factor! $\endgroup$ Apr 18, 2015 at 12:04
  • $\begingroup$ Short answer: It's easier to lift lighter things off the ground and hold them in the air. Long answer: read below $\endgroup$
    – Jim
    Apr 18, 2015 at 15:00

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Lift costs drag

Creating lift costs drag. Creating more lift causes more drag. This drag is called induced drag and is the consequence of the wing bending the airflow downwards. To simplify things, let's assume the wing is just acting on the air with the density $\rho$ flowing with the speed $v$ through a circle with a diameter equal to the span $b$ of the wing. If we just look at this stream tube, the mass flow is $$\frac{dm}{dt} = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v$$

Lift $L$ is then the impulse change which is caused by the wing. With the downward air speed $v_z$ imparted by the wing, lift is: $$L = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v\cdot v_z = S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho$$

$S$ is the wing area and $c_L$ the overall lift coefficient. If we now solve for the vertical air speed, we get $$v_z = \frac{S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho}{\frac{b^2}{4}\cdot\pi\cdot\rho\cdot v} = \frac{2\cdot c_L\cdot v}{\pi\cdot AR}$$ with $AR = \frac{b^2}{S}$ the aspect ratio of the wing. Now we can divide the vertical speed by the air speed to calculate the angle by which the air has been deflected by the wing. Let's call it $\alpha_w$: $$\alpha_w = arctan\left(\frac{v_z}{v}\right) = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$

The deflection happens gradually along the wing chord, so the mean local flow angle along the chord is just $\alpha_w / 2$. Lift acts perpendicularly to this local flow, thus is tilted backwards by $\alpha_w / 2$. In coefficients, lift is $c_L$, and the backwards component is $\alpha_w / 2 \cdot c_L$. Let's call this component $c_{Di}$: $$c_{Di} = arctan \left(\frac{c_L}{\pi\cdot AR}\right)\cdot c_L$$

For small $\alpha_w$s the arcus tangens can be neglected, and we get this familiar looking equation for the backwards-pointing component of the reaction force: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR}$$

or in absolute values, where $D_i$ is induced drag and $L$ is lift: $$D_i = \frac{L^2}{\pi\cdot AR}$$

Now you can see that more mass, which requires more lift, will cause a quadratic drag increase at the same speed. In aviation, good designers will give you a dollar value for every pound of mass saved, and this can be a three-digit number. In space applications the number of digits is even higher.

Example for the cost of mass

If you are just interested in a general estimate, the venerable Breguet equation will already give good results. First we need to know how much difference a small change in take-off mass makes: We use an A320-200 with an empty mass (OEW) of 42.6 tons + 36.4 tons of fuel and payload, and use a fuel burn of $b_f$ = 0.000018 kg/Ns and a speed of Mach 0.78, which equates to $v$ = 262 m/s in 11.000 m altitude: $$m_2 = \frac{m_1}{e^{\frac{R\cdot g\cdot b_f}{v\cdot L/D}}}$$ The L/D should be 18 and for the range we use the nominal maximum range of R = 5700.000 m. $m_1$ is take-off mass and $m_2$ landing mass. Now plug in the numbers for $m_1$ = 78 tons, and you will find $m_2$ = 63.0 tons for the landing mass. This means we need to consume and pay for 15 tons of fuel!

To be realistic, you would need to include different fuel flow during climb and descent, but the real mass difference is quite close to what you can get with this simple approximation.

Now repeat the calculation with a heavier structure, maybe by exchanging expensive graphite-epoxy and titanium for simple steel, and assume an OEW of 50 tons. Here it is helpful to turn the equation around and solve for the take-off mass with a known landing mass: $$m_1 = m_2 \cdot e^{\frac{R\cdot g\cdot b_f}{v\cdot L/D}}$$

With a landing mass increased by our structural mass increase of 7.4 tons, our $m_2$ becomes now 70.4 tons, and $m_1$ will be 87.14 tons. Increasing structural mass by 7.4 tons will translate into a fuel consumption of 16.74 tons for the same trip with the same payload, 1.74 tons more than with the lighter airframe. I leave it as an exercise to the reader to figure out how this will add up over an airframe life of 30 years with three trips per day!

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  • $\begingroup$ I think a simple summary goes like this: A heavier aircraft (i.e. a higher density) needs greater lift than a ligher one. It can either get this lift by traveling faster (more drag) or in denser air (more drag) $\endgroup$
    – user56903
    Apr 20, 2015 at 12:02
  • $\begingroup$ @DirkBruere: Yes, and more drag makes flying more expensive. Besides flying faster or in denser air, it can also fly at a higher angle of attack. But this again increases drag. $\endgroup$ Apr 20, 2015 at 14:32
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It seems like you are incorrectly thinking that the lift force opposes the drag force.

There are essentially four forces on an airplane.

enter image description here

There is a lift force from the wings - this force points upwards. There is a force of Gravity, which points downwards. There is a force of thrust from the engines pointing forwards, and a force of drag pointing backwards.

If you're visualizing these four forces, then you should see that the drag force opposes the thrust force, and the lift force opposes the force of gravity.

So, if you have a more massive plane, then the force of gravity is greater. That means to maintain a constant altitude, you need a larger lift force. The amount of lift you get is positively related to your speed, and the faster you go, the greater the drag force, which means you need more thrust, and that takes more fuel.

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  • $\begingroup$ Thanks for the diagram @ACuriousJim! Exactly what I was envisioning. $\endgroup$
    – Brionius
    Apr 18, 2015 at 18:37
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Check out this explanation.

Basically, when a wing is flying, it feels a force vector which is not vertical.

It points up and back.

We call the upward component lift, and the backward component drag.

Suppose a plane is loaded to twice its normal weight.

Then it needs twice its normal lift to support that weight. It pays a price for this in twice the normal drag, therefore twice the normal power needed.

I'm simplifying this of course, to an embarrassing degree. There are many variables to consider - speed, wing area, angle of attack, etc. etc., but the site I linked to gives great explanation of all of this.

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Building on Carl's comment:

The way any heavier than air aircraft works is that the wings exert a downward force on the air - increasing the downward momentum of the air results in a net lift force in the wing.

Now if you are heavier, you need to either move more air down per unit time, or move it down faster, in order for you to generate sufficient force.

You can move more air by flying faster - increasing drag; or you move the air down further by changing the attack angle of the wing - increasing the projected area and increasing the drag.

A simplistic calculation suggests that if you have a wing span $\ell$ and velocity $v$, you sweep an area $\ell v$ per unit time. If air density is $\rho$ and the attack angle is $\theta$ you move the air down with a velocity of $v \tan\theta\approx v\theta$ for small \theta$.

Now the projected area of the wing scales with the angle $\theta$ - it follows that if you want to generate more lift, whether by flying faster or increasing the angle of attack, you need more power.

The penalty is surprisingly large. Over the operational life of a plane, one kilo of additional mass can result in hundreds (thousands) of dollars of additional fuel costs. I did the math once - I will see if I can find it.

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    $\begingroup$ Why is you have always done the math for every problem and have it written somewhere? I'm actually impressed. I never keep my math doodles. +1 $\endgroup$
    – Jim
    Apr 18, 2015 at 15:01
  • $\begingroup$ The lift is not entirely from the downward movement of air. Most of the lift is generated from the Bernoulli principle of air flowing over the wing, creating a reduced pressure over the wing. en.wikipedia.org/wiki/Bernoulli%27s_principle $\endgroup$
    – LDC3
    Apr 18, 2015 at 15:10
  • $\begingroup$ @LDC3 No, that is flat out incorrect - albeit a very cpmmon "Bad Science" view of flight. $\endgroup$ Apr 18, 2015 at 15:12
  • $\begingroup$ @LDC3 - please see this excellent question and the answers given. I agree with Carl on this one. $\endgroup$
    – Floris
    Apr 18, 2015 at 15:13
  • $\begingroup$ @CarlWitthoft Do you have a reference that states lift is only from the downward movement of air? $\endgroup$
    – LDC3
    Apr 18, 2015 at 15:13

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