Here are some videos that I posted to Linkedin on this DARPA Drone Topic
Update of 01-21-26. Happy New Year to all! I started a discussion on Linkedin about my design for this DARPA Heavy Lift drone. The DARPA specifications include a set of flying activities that DARPA wants the drone to do. These include take off, flying a specified course, landing, off loading the payload, taking off again, flying another different specified course, and then landing. I plan some videos to discuss a math discussion that I prepared that shows my opinion that my DARPA Drone design can perform the flying activities that DARPA specifies. I made some videos about my DARPA Drone design and uploaded them to Linkedin, the links below will let you watch those videos.
Math Analysis for the Little DARPA Heavy Lift Drone
The plan was this drone will weigh 55lbs. I describe its aerodynamics elsewhere. The airfoil for the wing is the AH 80-140 chord 2.5 ft. Length of each wing 2ft
The fuselage is 3.18ft chord, width 19 inches. The fuselage airfoil is a modification of the AH 80-140 airfoil.
The fuselage is in place as an upside down version of the AH 80-140.
This upside down airfoil for the fuselage creates (with not much drag) a positive moment on the drone as it is flying in fixed wing mode.
This positive moment from the fuselage counteracts the negative moment created by the wings, thus the aircraft in its cruising mode of airspeed 90mph and drone wt of 55lbs and payload wt of 148lbs (total drone wt = 203lbs), it flies with zero moment.
The advantage of zero moment is that it does not need a tail and if it needs any
pitch control, this can be done by adjustments of the various propeller rpm’s, this allows the drone to be on the ground for take off in the vertical attitude of tail-down, nose-up.
The prop is from the Mejzlic website, 42 in diam, 201 grams, it develops 25.35 kg thrust at 2658 rpm needing 11.70 Nm torque. The fuel engine is two of the Jakadofsky turboshaft jet engines, each engine weighs 3.3kg, and develops 10 kw power output of its output shaft, this shaft running at 19,000 rpm, this gives a torque for this output shaft of 5.06 Nm.
Each engine drives a Parker Hannifin PGP- 505-0070 pump.
These pumps are also of the Parker Hannifin aluminum gear pump/motor line. These pumps weigh 2.4 kg each. Each pump is driven at 3108 rpm.
Each pump’s output is 19.58 L/min at 248.6 Bar pressure. The torque needed to drive each pump is 30.69 Nm. The 19,000 rpm output shaft of each jet is geared down by (19,000/3108) = a gear down ratio of 6.11. This gives a Nm of the geared down shaft running at 3108 rpm of
30.69 Nm (5.026, the 19,000rpm shaft torque) x (6.11, gear down ration) = 30.69 Nm resulting (5.026 Nm) x (6.11) = 30.69 Nm
There are 4 Parker Hannifin motors on the drone, each drives one propeller. The motor is the Parker Hannifin PGM 502-0033 motor. An aluminum motor weighing 2.6 lbs per motor. The motors need 9.75 L/min to create an rpm for the propeller of 2658 rpm. Each jet engine provides pump flow to two propeller motors, thus each jet engine’s pump puts out 19.5 L/min.
If these pumps are driven at 3108 rpm, their output flow will be 19.58 L/min
The torque needed by the propellers is 11.70 Nm. If the pressure given to these motors is
248.6 Bar, they will develop 11.71 Nm of torque.
30.69 Nm of torque is given to each pump. If the pump receives 30.69 Nm of torque on its input shaft, it can give an output pressure of 248.6 Bar.
Enjoy! Dr. Gray
Here are the links to the Linkedin videos I sent to Linkedin about this drone.
Update from 12-29-25 Well… I decided to try a drone design for this DARPA Drone Challenge. I decided to just use equipment that really does exist and see how much weight I could put up into the air. I used the Jakadovsky turboshaft jet engines. There is a version of this engine with an output of 10 kW. I chose to use two of these engines and four propellers, the propellers are 42 inch diameter. I chose to move the power to the propellers via hydraulic motors turning the propellers, and I let the turboshaft jet engine output shafts turn hydraulic pumps to supply the needed flow and pressure of the hydraulic fluid flowing to these hydraulic motors. My calculations demonstrate that this approach can allow a 55lb carbon fiber drone with wing chord 2.5 ft and wing width for each wing of about 2 ft, attached to a fuselage of chord 3.1 feet and width of 18 inches, this drone can lift a 148 lb payload up and then fly it via fixed wing flying (Thus the take off weight of the drone is: 55lbs drone wt + 148lbs payload weight = total take off wt = 203lbs). The drone takes off and lands via VTOL. It does VTOL-Fixed Wing and Fixed Wing-VTOL transition as needed. I created a pdf (48mb free for download) that has my math discussion about the take off sequence of this drone, and this pdf has some images of various topics I think apply. I plan a series of videos on this topic. I will send them to Linkedin as a series of 15 minute videos (Linkedin only allows videos that are 15 min long or shorter). I think I may also create a YouTube which would be a combination of all these shorter videos. I added below a link to that 48mb pdf.
Enjoy ! Dr. Gray
Download 48mb pdf This is a pdf file. It contains the math discussion and some images. It is about the Take-Off Discussion about the DARPA Heavy Lift Drone Challenge topic (37 downloads )Update from 11-25-25:
Hello everyone, when I first was thinking about this DARPA Lift Challenge, it did not seem to me there would be any method to get a drone that weighs only 55lbs to have a power source that could generate enough kW to run a group of propellers where this group of propellers together created enough static thrust to lift off the ground a take off weight of (55lb drone +220lb payload) 275lbs. It is still not clear to me that this is possible.
Of interest moving a weight up in the air, if a time interval and distance are specified, this becomes a definition of energy. DARPA states they want the device to fly along at 350 feet off the ground. If 275lbs went straight up at 1 ft/sec, then this would be 275 ft-lbs/sec. And 275 ft-lbs/sec is 0.372 kW. To get up to an altitude of 350 feet at 1 ft/sec would require 350 seconds, and at 0.372 kW x 350 seconds, this becomes an energy expenditure of (0.372 kW x 350 sec) = 130.2 kW-sec which is 130,200 Joules. Looking at JP-9 fuel, it has an energy density of 39.6 million Joules per liter. So, (130,200 Joules) / (39,600,000 Joules/Liter) is .00328 liters (3.28 ml). Google Ai states normal energy conversion efficiency of small turbine jet engines is in the range 15-30%, I will choose 20%, meaning one would need (3.28ml) x 5 = 16.4 ml of JP-9 fuel just to satisfy the energy equation for getting 275 lbs up in the air by 350 feet in 350 seconds.
For propellers the issue will be essentially static thrust since the device is moving up so slowly. Google Ai gives this example for calculating propeller thrust vs energy expenditure for a propeller moving through the air at 1 ft/sec. Google Ai does make the point, that for the aerodynamics community, the propeller efficiency for static (zero airspeed) thrust is defined as zero, and Google Ai uses a work-around to get the thrust efficiency.
Here is my actual Google Ai question, “energy conversion efficiency of modern propellers at 1 ft/sec velocity thrust output in lbs vs energy input in Joules”
I think you could type this question into Google Ai and ask Google Aito go into the “more depth discussion” option and you should get Google Ai’s discussion of the work-around that Google Ai used.
So, for a 275lb drone, we would need (13.56 Joules/sec) x (275lbs, for a 275lb drone) = 3729 Joules per sec for the propeller related thrust. This needs to last for 350 seconds to get the drone to an altitude of 350 feet, thus (3729 Joules) x (350 seconds) = 1,305,150 Joules to get the drone to 350 feet. If the energy conversion efficiency of the jet turbine engine is 20%, then we will need the jet engine to be using (1,305,150 Joules needed) x (5, energy efficiency penalty) = 6,525,750 Joules to get the drone to altitude.
Looking at JP-9 fuel, it has an energy density of 39.6 million Joules per liter. This would mean, in order to get the 6,525,750 Joules of energy input the engine would need to consume (6,525,750 Joules)/(39,600,000 Joules/Liter) = 0.164 Liters or 164 ml of JP-9 to get the drone up to 350 feet altitude.
DARPA also wants the drone to fly 5 nautical miles (30,380 ft) mostly at this 350 ft altitude. Thus the propellers will need to keep creating the thrust and the energy input rate will be 164 ml JP-9 per second. If the drone flies along at 30mph (a fast pace for a loaded drone) which is 44 feet/sec, the drone will be flying for (30,380 ft)/(44 ft/sec) = 690 seconds. If it is using 164 ml JP-9/sec, then it will need (690 seconds) x (164 ml/sec fuel input flow rate) = 113,160 ml of JP-9 fuel. The density of JP-9 fuel is 0.8gm/ml, so it will need in order to fly the 5 nautical miles (113,160 ml needed) x (0.8 gm/ml density) = 90,528 grams of JP-9 fuel, which is 90.5 kg of fuel, and at 2.2 lbs/kg, this is 199.1 lbs of JP-9.
So…. because of these calculations, I decided to just walk away from a 275lb drone and instead do calculations on a 55lb drone using JP-9 fuel turboshaft jet engines to power a distributed hydraulic drone propeller system and see how much (drone wt + payload wt) I could get to.
I plan some videos about this. More to come 🙂 pg