Is there an aerodynamic force that would keep this experimental WW2 era prop from flying as fast as an early jet? 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.



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.
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.