Does a lift to drag ratio of 100 imply an unattainable performance ratio of 1?

Question

An answer to this question stated that a lift to drag ratio of 100 is "unachievable".

When thinking of ratios, they seem to always be a fraction of one thing to another, with the sum of them being the total amount of what is being compared.

For example a ratio of 1 to 4 means and apportionment of 1/5 and 4/5, 20% and 80%.

From this reference we can see that the DAE 21 airfoil exceeds a lift to drag ratio of 100.

It seems that, as an efficiency, lift to drag ratio (or any ratio) can not be expressed as anything other than:

fraction of a ratio = n/(n+1)

Yes, this can never equal 1.

Is this correct?

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Answer 1

Lift to Drag, (L/D), is just a ratio, so it is perfectly possible to be greater than 1.

Answer 2

When people say that $L/D$ is a measure of aerodynamic efficiency -- they are using the word efficiency very loosely. They are simply using it as a generic term for 'figure of merit'.

When people say that efficiency ($\eta=p_o/p_i$) can not be greater than 1.0, they are referring specifically to a precise definition of efficiency as the ratio of power out to power in. Note that this engineering definition of efficiency does not apply to other quantities (mass, force, speed, time, money, etc.).

Usually this definition of efficiency is applied to power conversion and transmission devices. Things like electric motors that convert electrical power to shaft power, propellers that convert shaft power to thrust power -- or wires and shafts that move power from one place to another. These devices do not store energy. They have an instantaneous state -- if power is going in, then power is coming out.

The fact that efficiency must be less than 1.0 is due to the laws of thermodynamics. 1) You can only break even ($\eta\le1.0$). 2) You can't even break even ($\eta<1.0$).

When we run a power conversion device 'backwards' (say using a motor in a generating mode, or using a propeller as a windmill), we must be careful to re-define what 'input' and 'output' means. Losses like friction and resistance are always there of course.

For all of these devices, the 'inefficiency' ($1-\eta$) is a measure of the power lost to friction, resistance, drag, and other phenomena that eventually wind up as waste heat. The first law tells us that $p_i=p_o+p_l$ -- the input power comes out as useful output plus losses. This lets us write $\eta=p_o/(p_o+p_l)$. Which (with $p_o$ and $p_l$ always positive) shows that $\eta$ must be less than 1.0. T

We can also talk about efficiency of an energy storage device -- say a battery, flywheel, or spring. Here again, we might use the symbol $\eta$, but we have to be extra careful.

First, we must make clear the energy we put in $e_i$, the energy stored $e_s$, and the energy we get out $e_o$. When we put energy into a battery, we might think about a charge efficiency $\eta_c=e_s/e_i$. When we draw energy from a battery, we might think about a discharge efficiency $\eta_d=e_o/e_s$. Or, we might ignore the middle step and think about a round-trip efficiency $\eta_{rt}=e_o/e_i$.

This energy storage efficiency is not an instantaneous measure -- it takes time to charge or discharge a battery. You might only partially charge (or discharge) a battery.

Energy is the integral of power over time $e=\int_{t_i}^{t_f}p\ dt$. So energy storage efficiencies are integral measures of efficiency -- think of them as an averaging process. As we charge a battery, any instantaneous transfer of power has some efficiency. Over the complete charge, there is some average efficiency that reflects the energy (rather than the power).

This is just to say that we must be careful when working with efficiencies that we know exactly what definitions / conventions are being applied and what we can do with them.

Answer 3

There's a bit of confusion here between a generic ratio and the ratio defining efficiency (or performance or figure of merit or whatever you want to the name it).

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The ratio between two quantities is... well, the ratio between two quantities. If the forward chainring of a bike has 52 teeth and the chainring is engaged with a sprocket of 13 teeth, then their ratio is 1:4=0.25 (or 4:1=4). If I have five pens and my roof has 100 tiles, their ratio is 5:100=0.05 (or 100:5=20). If an airplane generates 4000N of lift and 500N of drag, their ratio is 4000:500=8.

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Efficiency is the specific ratio of: (one quantity we are interested in) and (the whole). If 5 roof tiles are broken and 95 are good, the efficiency of my roof tiles is 95/100=0.95 (or the inefficiency is 5/100=0.05). The airplane of before generates a total aerodynamic force of $\sqrt{4000²+500²}=4031N$ and its efficiency in generating lift is therefore 4000/4031=0.992.

Note that being efficiency defined as (a part)/(the whole) then it is a number between 0 and 1. This implies that multiplying it by 100, we get efficiency in percentage.

Answer 4

Terms can get kind of sloppy here. If you are truly talking about the ratio of Lift to Drag, it would be best expressed as a ratio of one number to another. If someone talks about a Lift to Drag ratio of "100", what they really mean is a Lift to Drag ratio of 100:1. This means that Lift/Drag = 100/1 = 100. Nothing unattainable about that.

So could we say that out of 101 mysterious "somethings", 100 of them are "Lifts" and 1 of them is a "Drag"? Sure, maybe, but that doesn't seem like a terribly useful concept, since we are talking about two orthogonal forces.

As opposed to saying that the ratio of Cats to Dogs is 5:2, for example -- here the sum of 5 + 2 = 7 would have some useful relevance, i.e. the actual total number of animals present in a sample size that had 5 cats...

Your question mentions a performance ratio. It's not clear what this would be mean. Are you making the mistake of thinking that a L/D ratio of 100/1 (i.e. 100) somehow has something to do with 100 percent? That would be a misconception.

One can find sources on the internet (example) that imply that when a ratio is converted to a fraction, the denominator should reflect the total number of things present. In other words, in the Cats-and-Dogs example above, we can say that the "Cat fraction" is 5/9 and the "Dog fraction" is 2/9. But that's not what we mean we express an L:D ratio as a fraction. When we talk about "L/D", we simply mean Lift divided by Drag. If L/D is 100, then that simply means that Lift is 100 times greater than Drag, so Lift/Drag = 100/1 = 100.

Are you thinking that a Lift/Drag ratio of 100 somehow implies that (following the phrasing of the previous paragraph) the "Lift Fraction" is 100% or 100 /100, meaning that (Lift / (Lift + Drag)) = 1, and thus implying that Drag is zero? That would be a misconception.

Answer 5

L/D is a ratio of forces, so it can be greater than 1. A reduction gear also produces more output force (torque) than input torque. It does so by trading rotation speed for torque. That's precisely what an aerofoil does - the only difference being that it does so aerodynamically instead of mechanically.

$Power = Force \cdot distance/time$

An aerofoil trades Speed for Force (lift force), thus giving it an L/D ratio greater than 1.

Answer 6

But of course! Perhaps my view of this issue regards coming to terms with understanding the propeller thrust issue, and its relationship to the lift to drag ratio. Somebody asked 'what is performance ratio?' Simply put, this is efficiency. How does one get 100 percent efficiency out of anything? The answer is only with great difficulty. But I digress...

With propeller thrust, we can do a lot of stuff. Going to the primordial equations, we find that if we know the power, velocity, and lift to drag ratio, we can determine the weight of the aircraft we can sustain in flight. What's that, you may ask...

$$W = (L/D)*((lb \cdot ft/sec)/V)$$

Yes, interesting is how that equation resolves. Now, as noted in the previously mentioned answer, we can re-define this equation algebraically to solve for anything else within the equation that we wish to know. The only limitation is that we have to know the stuff on the right side of the equation. Specifically, the term in brackets, $(L/D)$, is the lift/drag ratio, is given properly in the equation by $C_L/C_D$. We note, lb$\cdot$ft/sec is power, V is velocity in ft/sec. The method for determining thrust has already been well sustained, elsewhere. The performance ratio for the best propellers is about 88 to 89 percent. In other words, the best propeller is about 88 to 89 percent efficient. Consequently, propeller thrust is about 88 to 89 percent of rated power available to turn the propeller. Assuming, of course, that the propeller is operating at its optimal speed and pitch relative to V. An aircraft assessment invoking the above equation can be offered if necessary.