0
$\begingroup$

A deep dive into the Cessna 172 POH showed a remarkable linearity between rate of fuel consumption and velocity at various cruise power settings ranging from 55 to 75%. Originally a study of propeller efficiency, it lead to an observation of the drag equation:

Drag = $\rho$ × Area × Cdrag × V$^2$

Power has the units kg V$^2$/time. Fuel as potential energy has the units kg V$^2$, and it's use rate as kg V$^2$/time.

Power is also expressed as Force × distance/time or kg V/s x V.

At steady state thrust force = drag force.

Examination of the drag equation $\rho$×AxCd×V$^2$ seems at first to imply Power is proportional to V$^3$.

Yet, when the units of the drag equation were written out, it was found that, in order to yield force kg velocity/time there could be only one V!

Running through the units revealed that "kg" must come from kg/m$^3$ $\rho$ × m$^2$ Area × m distance, leaving V/s as the remainder of the Drag force kg V/s.

The implications are that parasitic drag increases linearly with velocity, rather than exponentially, as often depicted, and that drag in terms of V$^2$ is actually an expression of Force × Distance or Work.

Consideration of density kg/m$^3$ x Volume of Air m$^3$ (as Area m$^2$ x distance m) has me wondering is the drag force equation being misunderstood as a Work equation?

A possible source of confusion is that, at airspeed "behind the power curve" (slow), induced drag surely makes the drag curve look exponential. But the Cessna data makes the fuel consumption kgV$^2$/s perfectly proportioned between drag kgV/s and airspeed V at cruise speeds. Not V$^3$!

Many people sing songs of praise for cruise climb, and indeed, lowering Angle of Attack and using greater speed seems to help fuel economy. What's going on here?

I will gladly migrate to physics if no one here would like to try to explain this.

Improve this question
$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Jamiec ♦
    Jan 26 at 10:05

3 Answers 3

Reset to default
5
$\begingroup$

You've messed up your units. $\rho$ has units of $\rm kg/m^3$, area has units of $\rm m^2$, and velocity has units of $\rm m/s$. Putting this all together:

$$ \left[\rm Drag\right] = \rm\frac{kg}{m^3}\cdot m^2 \cdot \left(\frac{m}{s}\right)^2 = \frac{kg\,m}{s^2} $$

which has dimensions of force, and the drag equation containing $v^2$ is dimensionally consistent. Side note: $v^2$ scaling is called "quadratic," "exponential" refers $2^v$ scaling.

Regarding cruise climb, higher climb speeds do not improve fuel efficiency on a per hour basis. Either way, the engine is at full power so it doesn't really change at all. But since you are going faster, you go farther in the same amount of time, so they can improve fuel efficiency on a per mile basis.

Improve this answer
$\endgroup$
2
  • $\begingroup$ @RobertDiGiovanni Thrust has units of $\rm kg\,m/s^2$. Power has units of $\rm kg\,m^2/s^3$. You get from thrust to power by multiplying by velocity, not velocity squared. $\endgroup$
    – Chris
    Jan 30 at 18:09
  • $\begingroup$ @RobertDiGiovanni Drag is also a force, and the power required to overcome drag is $v$ times the drag, which scales as $v^3$, not $v^2$. You can't relate these directly to fuel consumption numbers since the propeller efficiency is a function of airspeed and $C_d$ depends on the angle of attack and thus depends indirectly on airspeed, as well as directly to some degree depending on flow conditions. $\endgroup$
    – Chris
    Jan 31 at 1:36
1
$\begingroup$

Seeing as how both lift and drag vary as the square of the airspeed, it would surprise me enormously if parasitic drag, (except for small edge case effects), was any different. An exponential relationship would certainly be out of the question.

Improve this answer
$\endgroup$
0
1
$\begingroup$

Aerodynamic forces can be expressed in dynamic pressure $q$ and a reference area of the aeroplane (usually the wing area $S$). In order to account for the shape of the plane, dimensionless coefficients are used, C$_D$ for drag and C$_L$ for lift. So drag = $C_D \cdot q \cdot S$ and lift = $C_L \cdot q \cdot S$.

Parasitic drag is a component of the total drag, and is therefore according to the above definition a linear function of dynamic pressure $q$. And for subsonic flow, Bernoulli has found that $q$ = ½ $\rho$ V$^2$.

Does parasitic drag increase exponentially with velocity?

No. For incompressible flow, parasitic drag increases quadratically with velocity.

That's. It.

Improve this answer
$\endgroup$
2
  • $\begingroup$ V2/V1 has unit 1. It seems to me that you’re lost in an improper unit analysis. Converting fuel potential energy into net kinetic power has many efficiency losses which can be a function of airspeed, but that has no consequences for the units, which are correct. $\endgroup$
    – Koyovis
    Jan 30 at 22:46
  • $\begingroup$ Drag is a force. Thrust is a force as well. Forces are what moves objects, which will reach a constant velocity if all resulting forces are zero. $\endgroup$
    – Koyovis
    Jan 31 at 2:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .