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I did some research on Minimum Sink Rate, and saw much information defining it, (such as it is the speed at which the aircraft will remain in the air for the longest time, etc.) and how it is generally lower than Best Glide speed, and is generally a few knots above stall speed. There was also some information about what is used for (when you are trying to remain airborne for as long as possible after engine failure in order to deal with an emergency, or for gliders to maximize the climb effect from a thermal),

…. But I could not find, anywhere, an analysis/explanation for how this works, from a Physics perspective, as you can easily find to explain Vx, (best angle of Climb airspeed), and Vy (best rate of climb airspeed), that explain/analyze them from a Physics perspective (Vx is speed at which you have the highest Excess Thrust, whereas Vy is the speed at which you have Maximum Excess Power).

Also, from my limited understanding, what I did read seems to violate my basic understanding of what Min Sink Rate means, from a Physics Perspective. As I would understand it, Min Sink Rate Speed is the speed at which your rate of descent is the lowest, (you will remain airborne for the greatest amount of time). This is equivalent to the airspeed at which the aircraft is losing altitude, (Potential Energy), at the slowest possible rate. With no power on the aircraft, loss of energy (altitude) is directly related to total drag, (energy must be conserved!), and this occurs at L/Dmax. All the complexities involved with determining Maximum Excess Power, or Maximum Excess Thrust, as are required to analyze Vx and Vy, become moot. So although it appears to be accepted wisdom that Minimum sink is different from Best Glide, this seems to violate basic physics. They should both occur at the airspeed (AOA actually) where total Drag is minimized, i.e., at L/Dmax. Where am I going wrong?

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From first principle, assuming the airplane is a point mass, no wind, small angle of attack, and thrust acts in-line with drag:

$$T-D-W\sin \gamma=m\dot{V}$$

where $T$ is thrust, $D$ is drag, $\gamma$ is angle of climb, $m$ is mass of the airplane, $W=mg$ is weight, $V$ is airspeed/forward speed (assuming no wind).

In a total power off scenario, $T=0$, and assuming steady-state, $\dot{V}=0$:

$$\sin \gamma = -\frac{D}{W}$$

The rate of climb ($\dot{z}$), which is the negative of sink rate, is related to angle of climb by: $\dot{z}=V\sin \gamma$. Therefore, we have:

$$\dot{z}=-\frac{DV}{W}=-\frac{P_R}{W}$$

Therefore, for minimum rate of descent, we would like the minimum power required ($P_R$).

Note: for minimum glide slope, it corresponds to $L/D_{max}$, but not for minimum rate of descent.

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    $\begingroup$ So power required is product of total drag times velocity. And min rate of descent occurs when this product is at minimum. So to see where this occurs you need to express D as function of V and then determine value of V where this product expression is at minimum. $\endgroup$ Feb 20 '20 at 23:09
  • $\begingroup$ i.e., set dP/dV = 0 and solve for V. $\endgroup$ Feb 21 '20 at 21:09
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You are right that in order to achieve minimum sink rate, you need to minimize energy losses. And yes, energy losses are related to the drag. But here you need to stop and think again what 'related' exactly means.

The drag is a force, not an energy. But when you multiply the force acting on moving body by its speed, you get power (that is energy per unit of time) inflicted by this force. (Actually, you have to multiply only the fraction of force, which is parallel with the velocity vector, but drag is, by definition, acting directly against movement, so this is fulfilled automatically).

Therefore energy losses are drag multiplied by speed. At L/Dmax you are flying with minimal drag force acting on the airplane (for steady flight), but energy "consumption" is not at the minimum.

If you decrease airspeed a bit, drag force increases by some small amount, but the product of airspeed and drag decreases thanks to decreased speed. So you keep slowing down until these two effects cancels out and you end up at the point with minimal energy loss rate. That is the least "Watts" of drag, not "Newtons".

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  • $\begingroup$ Thanks. This is exactly in line with JZYLs answer. I gave him the check because he included mathematical derivation. $\endgroup$ Feb 20 '20 at 23:15
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    $\begingroup$ @CharlesBretana Of course, accept the answer which works best for you. I have seen JZYLs answer in middle of my writing, but wanted to post my one too, so there can be another explanation in different words. $\endgroup$
    – Martin
    Feb 20 '20 at 23:32
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In addition to some excellent content in other answers, it's worth noting that V-bestglide occurs at the angle-of-attack where L/D (and therefore also Cl / Cd) is maximized, while for shallow glide angles, it's a good approximation to say that V-minsink occurs at the angle-of-attack where (Cl^3 / Cd^2) is maximized1. The difference between the two formulae means that V-minsink will always occur at a lower airspeed than V-bestglide. In fact the reduction in sink rate obtained by slowing down is the reason that the Cl term is cubed, while the Cd term is only squared, in the expression for V-bestglide. This is related to the concept that required power is minimized at V-minsink.

loss of energy (altitude) is directly related to total drag

No, we can show that total Drag is minimized at V-bestglide, but we if fly more slowly down a slightly steeper glide path, we can end up with a lower vertical speed. That's the reason for the difference between the airspeeds for min sink and best glide.

Footnotes--

  1. Derivation: the glide ratio can be geometrically be shown to be equal to L/D, which is arithmetically equal to Cl/Cd. The sink rate is proportional to airspeed * (1/ glide ratio), which is the same as airspeed * D/L or airspeed * Cd/Cl. At shallow glide angles, nearly all the Weight is supported by the Lift vector, so it's a good approximation to say that Lift is constant and airspeed is therefore inversely proportional to the square root of Cl. Therefore at shallow glide angles, it's a good approximation to say that the sink rate is proportional to Cd / CL * 1/(Cl^1.5), which works out to Cd / (Cl^1.5), which is the same as (Cd^2 / Cl^3). A derivation of this appears in "Model Aircraft Aerodynamics" by Martin Simons (3rd edition, 1994) on pp. 40-41, or in more detail, on pp. 238-239. (link to PDF).

Related content appears in this ASE answer--

Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?

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  • $\begingroup$ This answer would benefit from adding some content that is not limited to the case of shallow glide angles. For example, when the glide angle is very steep, the sink rate is proportional to the square root of Cd (think of a parachute). There is a formula that gives the sink rate at all Cl/Cd ratios, but it has not yet been included in this answer. Stay tuned $\endgroup$ Nov 11 '20 at 14:34

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