The XB-42 was an experimental bomber developed during WW2. Its twin engines delivered a combined 3,800 HP and top speed of 410 mph at 23,440 ft.

enter image description here

It's an amazing aircraft. But if an enormous aircraft is that fast why not use the same concept for smaller fighter-sized aircraft? I assume there must be a good reason because it didn't happen.

I used a modified version of the drag equation found on NASA's website to estimate the speed increase if the front profile was reduced by 50% and everything else stayed identically the same. The modified version can be found here.

By reducing the front profile by 50% the top speed increased by 25.99%. That brought its top speed up from 410mph to an amazing 517mph. That's almost as fast as a jet. Is that possible or am I missing some other aerodynamic force that would stop the plane from approaching that speed?

I guess at the most basic level is this concept correct?

Smaller frontal area = faster top speed.

Is there some other aerodynamic force that would negate that?

Answer to comment below: Zeus answered that reducing the aircraft's profile wouldn't make a significant difference because most drag was produced by the wing. Yes and no. Wing lift does produce a lot of drag at low speed but that drops as the plane accelerates. At high speed nearly all drag is parasitic (i.e. the shape of the aircraft) as evidenced by the chart below. Moreover my question was theoretical. If 50% of the plane's profile was removed the weight could be removed as well and with it the wing size. Turning the XB-42 airframe into a small, fighter aircraft was the general point of the question.

enter image description here

  • $\begingroup$ It looks amazing but I think we have the answer $\endgroup$
    – h22
    Commented Apr 12, 2018 at 6:22
  • 1
    $\begingroup$ For starters, unlike with cars, most of the drag is created by the wing, and much of it is the by-product of lift. And since you you didn't mention you were going to reduce weight, you can't reduce the wing... $\endgroup$
    – Zeus
    Commented Apr 12, 2018 at 9:03
  • $\begingroup$ @Zeus please post an answer $\endgroup$
    – Federico
    Commented Apr 12, 2018 at 11:19
  • 1
    $\begingroup$ This question is really unclear because the speed of "a" jet could be anything. There are fast jets and slow jets, and in fact quite a few of them fly slower than 500mph. $\endgroup$ Commented Apr 12, 2018 at 16:24
  • $\begingroup$ @DR01D The speed of sound is not a constant, but a function of temperature and the thermodynamic properties of the gas (air in this case). The speed of sound decreases noticeably as altitude increases. $\endgroup$ Commented Apr 12, 2018 at 16:34

2 Answers 2


First of all, there were several propeller aircraft which were much faster than the XB-42. But let's stick with it and scale it to see where that brings us.

For that we need two things: Equations for drag $D$: $$D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\left(c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right) = \frac{\rho\cdot v^2}{2}\cdot S\cdot c_{D0}+\frac{(m\cdot g)^2}{\pi\cdot AR\cdot\epsilon\cdot\frac{\rho\cdot v^2}{2}\cdot S}$$ and for thrust $T$: $$T = \frac{P}{v}\cdot\eta_p$$

The symbols are (values are for the XB-42 at 7,140 m):
$\kern{5mm} \rho\:\:\:\:\:$ air density = 0.581 kg/m³
$\kern{5mm} v\:\:\:\:\:$ velocity = 410 mph = 183.3 m/s
$\kern{5mm} S\:\:\:\:\:$ wing surface area = 51.6 m²
$\kern{5mm} c_{D0} \:$ zero-lift drag coefficient
$\kern{5mm} c_L \:\:\:$ lift coefficient
$\kern{5mm} \pi \:\:\:\:\:$ 3.14159$\dots$
$\kern{5mm} AR \:\:$ aspect ratio of the wing = 8.95
$\kern{5mm} \epsilon \:\:\:\:\:\:$ the wing's Oswald factor = 0.9
$\kern{5mm} m \:\:\:\;$ aircraft mass = 15,000 kg
$\kern{5mm} g \:\:\:\:\;$ gravitational acceleration
$\kern{5mm} P \:\:\:\:\:$ engine shaft power = 3800 hp = 2833.66 kW
$\kern{5mm} \eta_P \:\:\;$ propeller efficiency = 0.85

In order to determine the unknown zero-lift drag, drag is set equal to thrust, the values above are inserted and the equation is solved for $c_{D0}$: $$c_{D0} = \frac{\frac{P}{v}\cdot\eta_p - \frac{(m\cdot g)^2}{\pi\cdot AR\cdot\epsilon\cdot\frac{\rho\cdot v^2}{2}\cdot S}}{\frac{\rho\cdot v^2}{2}\cdot S} = 0.02272$$ which looks about right.

Then we need some clever formula how mass will change with a scaled airframe. Since both the engine and the pilot will stay constant, to assume that mass will scale with the square of the length is quite generous. So we have a second aircraft of 70.7% size which will have half the frontal area, half the mass but the same power. Now we need to solve for the unknown speed with the now known zero-lift drag: $$v^4 = \frac{2\cdot P\cdot v\cdot\eta_p}{\rho\cdot ½\cdot S\cdot c_{D0}} - \frac{(2\cdot ½\cdot m\cdot g)^2}{\pi\cdot AR\cdot\epsilon\cdot\left(\rho\cdot ½\cdot S\right)^2\cdot c_{D0}}$$ $$v^4 = 14.144.659.448\cdot v - 167,497,545.8$$ $$v = 237.76\:\text{m/s} = 531.85\:\text{mph}$$

Now all what is left to be solved is how to squeeze the existing engines and their cooling system into half the fuselage cross section.

I guess at the most basic level is this concept correct?

To answer your question directly: I used a slightly different approach and got an even higher speed. If I leave mass constant, the higher induced drag will limit the speed to 223.37 m/s, which is just shy of 500 mph. I wonder how you got a speed of 517 mph. The car example on the linked site does not include induced drag, so I guess your concept needs a little improvement.

  • $\begingroup$ What would you say about the prop(s) required to drive such a hypothetical aircraft? Rough calc's say the XB-42's props would have to support a pitch of at least 237.8m (i.e., blades nearly edge-on to the direction of flight) just to generate any thrust at all at 531.85mph. Feel free to correct my math. :) $\endgroup$
    – ioctlLR
    Commented Apr 12, 2018 at 19:03
  • 1
    $\begingroup$ @DR01D: The XF-12 was a reconnaissance aircraft. Not every four-engined airplane is automatically a bomber! $\endgroup$ Commented Apr 12, 2018 at 19:26
  • 1
    $\begingroup$ @ioctlLR: Yes, you would need more prop solidity to absorb the power and a lower prop speed in order to avoid supersonic tips. The efficiency will most likely drop - after all, this is already Mach 0.767. But 520 mph should be feasible with some adjustments. $\endgroup$ Commented Apr 12, 2018 at 19:31
  • 1
    $\begingroup$ @DR01D: If you use kW for the power, the result is in kN force. 13.14 kN at 15 tons mass gives you a thrust/weight ratio of 0.089, so the L/D at max speed is 11.2. This looks about right. $\endgroup$ Commented Apr 13, 2018 at 6:15
  • 1
    $\begingroup$ @DR01D The second $v$ is the same as the $v$ in $v^4$ on the left side (and not 183.3). I simply put the equation into Numbers and adjusted $v$ until the result was equal on both sides. $\endgroup$ Commented Apr 25, 2018 at 16:26

all propeller-driven aircraft are subject to the same velocity limit. to make the plane go fast, the propeller has to spin faster, and the propeller blades achieve mach 1 before the airframe does. as they go supersonic, their thrust output goes down and they waste work in making noise instead of propulsion. the propulsive efficiency loss is sufficient to prohibit the airframe from achieving mach 1.

  • $\begingroup$ those forces are sufficient to stop a prop driven aircraft from reaching mach 1. But would they in theory be sufficient to stop a plane traveling at 410mph from reaching 517mph? $\endgroup$
    – DR01D
    Commented Apr 12, 2018 at 16:51
  • $\begingroup$ well, have a look at the relationship between top speed and horsepower first. HP requirement goes up as the cube of airspeed. can you fit that HP into the existing airframe? $\endgroup$ Commented Apr 12, 2018 at 17:17
  • $\begingroup$ In this case I think yes. The bottom portion of the XB-42 was a big, empty bomb bay that held up to 8,000 pounds. With that gone you could reduce the wing area and shorten the tail. Plus a WW2 fighter only needs 1 pilot and the XB-42 had a crew of 3. So with all of that gone you could maybe reduce the size by half and still keep the same engines and HP. $\endgroup$
    – DR01D
    Commented Apr 12, 2018 at 17:24
  • 1
    $\begingroup$ You can increase the pitch of the prop and by doing so you can keep its RPM low (subsonic). So the propeller tip speed problem can be a trade off and can be potentially solved. $\endgroup$ Commented Apr 12, 2018 at 18:14
  • $\begingroup$ @gurkancetin, i think not, for the following reasons. on an airplane that is moving through the air at, say, 1000fps, the relative speed of the rotating propeller moving through that air will always be greater than the speed of the plane. it is inevitable that the prop blades will hit sonic velocity before the airframe will. $\endgroup$ Commented Apr 12, 2018 at 18:21

You must log in to answer this question.

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