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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

DR01D
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  • It looks amazing but I think we have the answer – h22 Apr 12 '18 at 06:22
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    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... – Zeus Apr 12 '18 at 09:03
  • @Zeus please post an answer – Federico Apr 12 '18 at 11:19
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    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. – user3528438 Apr 12 '18 at 16:24
  • @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. – AEhere supports Monica Apr 12 '18 at 16:34
  • @DR01D Not disputing that, it seems accurate, and I was mostly nitpicking your previous comments. However, transonic effects start to be relevant below Mach 1, for example, modern airliners fly at around $M = 0.85$ and require careful wing design to avoid excessive drag due to local supersonic areas on the wing. – AEhere supports Monica Apr 12 '18 at 17:01

2 Answers2

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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.

Peter Kämpf
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  • 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. :) – ioctlLR Apr 12 '18 at 19:03
  • @Peter Kämpf Perfect thank so you much! That was the equation I was looking for. In this case I think that yes the engines and cooling system would fit in a smaller aircraft. The XB-42 had a huge bomb bay that held 8000 pounds of bombs and a crew of 3. There is plenty of fat that could be removed from that plane to bring it down to fighter size and still leave room for both engines. – DR01D Apr 12 '18 at 19:09
  • Thanks for the link to the XF-12 Rainbow! https://en.wikipedia.org/wiki/Republic_XF-12_Rainbow I hadn't heard of that aircraft before. Really amazing bomber. – DR01D Apr 12 '18 at 19:12
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    @DR01D: The XF-12 was a reconnaissance aircraft. Not every four-engined airplane is automatically a bomber! – Peter Kämpf Apr 12 '18 at 19:26
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    @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. – Peter Kämpf Apr 12 '18 at 19:31
  • @Peter Kämpf My concept was just a thought experiment. I didn't include induced drag because I wasn't sure how to. Thank you so much for your equation! I'll use those in the future. – DR01D Apr 12 '18 at 21:10
  • @Peter Kämpf I'm using your thrust equation on the XB-42 and could you tell me if I accounted for this correctly? T = (2833.66 * .85)/183.3. And this gives me 13.14 units of Thrust. If I did that correct do you mind if I ask what are the units? Thank you so much! – DR01D Apr 13 '18 at 03:55
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    @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. – Peter Kämpf Apr 13 '18 at 06:15
  • @Peter Kämpf Wow! I didn't realize that early jets only produced about 1/3rd more thrust than a prop. https://en.wikipedia.org/wiki/Junkers_Jumo_004 The Jumo-004B that powered the ME-262 produced 8.83Kn, 004C produced 9.81Kn and the 004D produced 10.3Kn. That's really interesting. – DR01D Apr 13 '18 at 15:59
  • Let us continue this discussion in chat. – DR01D Apr 13 '18 at 16:29
  • @Peter Kämpf Hi Peter, if I used your amazing answer found here, https://aviation.stackexchange.com/questions/420/how-can-i-calculate-maximum-rate-of-climb to determine the XB-42's climb rate would it be possible for you to check my result just to make sure I didn't screw something up? I can put the answer in that thread. – DR01D Apr 14 '18 at 22:56
  • @Peter Kämpf Hi Peter, your equation is amazing but I think you might have made a tiny mathematical error at the end. x^4 = 14144659.448 * 183.3 - 167497545.8 = 221.9155 not 237.76. Maybe I did something wrong? Could you confirm either way? Thanks so much, this is really helpful! – DR01D Apr 25 '18 at 16:18
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    @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. – Peter Kämpf Apr 25 '18 at 16:26
  • @Peter Kämpf Oh.... I get it. v^4 = 14144659.448 * v - 167497545.8. Try different Vs until it works on both sides. Awesome, Thanks! – DR01D Apr 25 '18 at 21:02
  • +1. Insightful answer! – Hans Feb 22 '20 at 16:47
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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.

niels nielsen
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  • 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? – DR01D Apr 12 '18 at 16:51
  • 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? – niels nielsen Apr 12 '18 at 17:17
  • 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. – DR01D Apr 12 '18 at 17:24
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    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. – Gürkan Çetin Apr 12 '18 at 18:14
  • @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. – niels nielsen Apr 12 '18 at 18:21
  • The DO-335 was a ww2 era prop that flew 475mph. How did they solve the prop problem? https://en.wikipedia.org/wiki/Dornier_Do_335 – DR01D Apr 12 '18 at 18:29
  • One factor to remember is that power input dictates the total size and blade-count of the propeller. The XB-42 had two 13ft three-bladed props, which means the propellers' red-line RPM (mach .92) would be roughly 1500rpm, sitting still on the ground. You'd have to drop that RPM quite a bit at 410mph and FL240 to stay under the red-line. – ioctlLR Apr 12 '18 at 18:31
  • @DR01D Voodoo posted 531.53 mph average in four laps last year at Reno, with a fastest pass of 554.69 mph. That's with a really big, really high-performance prop, a "metric ton" of horsepower, and a highly-optimized airframe. It's possible to go that fast, but not practical; Voodoo is effectively purpose-built to go fast and nothing else. – ioctlLR Apr 12 '18 at 18:37
  • @ioctlLR That is interesting, I wasn't aware of the prop limitation. But the DO-335 flew at 475mph back in 1944. That's almost 75 years ago. So clearly there must be a way to solve that problem. – DR01D Apr 12 '18 at 18:41
  • @DR01D The airflow separates from the prop blade at mach .92, causing the blade to vibrate (destructively, if pushed any further) and a bunch of energy to convert to sound instead of thrust. No propeller design can get past that limit (hence the reason everyone called it the sound "barrier"). Even at mach .91 the prop efficiency is so bad it's not really accelerating the aircraft any more. – ioctlLR Apr 12 '18 at 18:49
  • @ioctLR No doubt you are right but I don't mean modern jet speeds. I mean early jet speeds in the 500 mph range. According to this site https://www.fighter-planes.com/jetmach1.htm the speed of sound is 706.9 mph at 20,000 feet. 85% of that is 601mph. I'm referring to planes slower than 600mph. – DR01D Apr 12 '18 at 19:05
  • @DR01D I'll defer to Peter's engineering prowess here, but the prop tip speed problem still applies. Careful engineering would be required, and I'm not certain a scaled-down XB-42 could hold the engines required. You could almost certainly find a better set of props today, though... If you'd like to reply, move us to chat, please. – ioctlLR Apr 12 '18 at 19:12