Gravity Jet Suit Flight experiences now and Future Racing Events

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Richard Browning spoke at SU Global Summit 2018 and demoed the Gravity Jet Suit twice this week in the SF bay area.

* they have made flying the Jet Suit far easier to fly. There is adjustment of the thrust so that as fuel is used and the flyer is getting lighter the thrust is reduced so the arms can be kept in the same position
* They have made six Jet Suits so far. A couple have been sold for about $440,000 each
* there is a Jet Suit that can fly for four minutes and another that can fly for nine minutes.
* they are developing winged Jet Suit.

Winged Jet Suits will be ten times more fuel efficient and enable far higher speeds. They will also allow gliding and being able to safely move past danger zones in vertical flight for safer travel.

The Gravity winged Jet Suits will be more similar to the Yves Rossi jetpack rather than hang-glider fabric wings.

Right now they offer Jet Suits for sale (about US$440K), Jet Suit flight experiences (about $14K) and Jet Suit training.

They are getting a lot of interest from the military but military procurement is long.

Gravity and Richard Browning are focused on building mass market sustainable business revenues.

This will be from racing events and creating televised shows and mass audience events.

Imagine a racecourse with several people going through and around large hoops and poles.

Richard believes having jetpack racing will drive the rapid improvement of the technology and the capabilities. This is similar to how Formula One and other car racing drive car technology. More powerful engines, lighter materials will be used.

The jetpack flying around a lake is similar to the racing laps that jetpack racing will have.

I believe that Jet Suit flying and Jet Suit adventure sports can become as big or bigger as X-games and motocross and motorcycling.

I will discuss the potential larger societal impact potential of Jet Suiting in follow up articles as well as comparing it to other personal flight technology.

Pictures from Hiller Museum flight demo

Jet Suit Flight Experience

Gravity’s UK location is The Hush House, a repurposed jet testing facility at a former RAF military base.

Clients will get a tour of the Human Propulsion Laboratory, followed by refreshments and a Jet Suit tailoring session.

Following preparation, a fully qualified Gravity Industries Pilot will take the client through:

The Jet Suit power-up sequence
Initial low-power flight experimentation
Higher power testing (dependent on client’s flight capabilities)

36 thoughts on “Gravity Jet Suit Flight experiences now and Future Racing Events”

  1. I recall reading that their test pilot has a very rigorous exercise routine, to be strong enough to control it.

  2. I recall reading that their test pilot has a very rigorous exercise routine to be strong enough to control it.

  5. Hmmm… FUN! Note though that the amount of fuel needed for a 7 to 9 minute buzz-around is significant. Significant enough that the spin is that the jet engines can be throttled (perhaps automatically) to lower their output as fuel is rapidly depleted over the course of the jaunt. Which then kind of begs… how much fuel is involved? Well, let’s “try” from just the very squishy tech specs. 7 minutes, 1,050 horsepower. We know that horsepower converts to watts easily enough: about 740 watts per HP. And minutes goes to seconds easily enough too. 1,050 HP × 740 = 777,000 W 7 min × 60 sec/min = 420 s 777,000 × 420 = 326,000,000 joules Now, since jet engines can’t really break normal Carnot-cycle engine efficiencies, let’s guess 30% efficiency. This would be REALLY good for a small jet engine. But hey… I’m an optimist. 326,000,000 J ÷ 0.30 = 1,087,000,000 J of fuel energy. We know that most petroleum-based jet fuels are a kind of enriched kerosene, having about 44 megajoules per kilogram of distilled Jurassic Swamp juice: 1,087,000,000 ÷ 44,000,000 ≈ 25 kg. 25 kg × 2.21 ≈ 55 lb So there you are, goats. 55 lbs or 25 kg of petrol to keep the bird afloat for 7 minutes: (25 kg, 55 lb) ÷ 7 min = (3.5 kg/min, 7.8 lb/min) Then there is the mass of the rockets. Ahem… jets. The ones on the arms look tame enough, probaby what, 10 lbs or 5 kilos each? But the big honker at the back, the “main thruster” is clearly a bigger puppy. WAY bigger. 25 lbs? I’d not be surprised. 25 lbs — biggie № 1 25 lbs — biggie № 2 10 lbs — LEFT arm № 3 10 lbs — LEFT arm № 4 10 lbs — Right arm № 5 10 lbs — Right arm № 6 4 lbs — a tank for the fuel 3 lbs — hoses for the fuel 15 lbs — control electronics, pumps 25 lbs — welded aircraft aluminum frame 55 lbs — fuel, at TKO ____________ 192+ lbs at TKO 87+ kg at takeoff ON YOUR BACK. Wow. Just wow. While I’m SURE that higher bypass fanjets will enable a lighter output of necessary power, both the Big Boys in the

  6. Fake video ! The noise is no soft music it is an enormous roar, nobody will want to have these contraptions flying near them ! This not the future this is oilpunk SF.

  8. Hmmm… FUN!Note though that the amount of fuel needed for a 7 to 9 minute buzz-around is significant. Significant enough that the spin is that the jet engines can be throttled (perhaps automatically) to lower their output as fuel is rapidly depleted over the course of the jaunt. Which then kind of begs… how much fuel is involved?Well let’s try”” from just the very squishy tech specs. 7 minutes””1050 horsepower. We know that horsepower converts to watts easily enough: about 740 watts per HP. And minutes goes to seconds easily enough too. 1050 HP × 740 = 777000 W7 min × 60 sec/min = 420 s777000 × 420 = 3260000 joulesNow since jet engines can’t really break normal Carnot-cycle engine efficiencies let’s guess 30{22800fc54956079738b58e74e4dcd846757aa319aad70fcf90c97a58f3119a12} efficiency. This would be REALLY good for a small jet engine. But hey… I’m an optimist. 3260000 J ÷ 0.30 = 1870000 J of fuel energy. We know that most petroleum-based jet fuels are a kind of enriched kerosene having about 44 megajoules per kilogram of distilled Jurassic Swamp juice:1870000 ÷ 440000 ≈ 25 kg. 25 kg × 2.21 ≈ 55 lbSo there you are goats. 55 lbs or 25 kg of petrol to keep the bird afloat for 7 minutes:(25 kg 55 lb) ÷ 7 min = (3.5 kg/min 7.8 lb/min)Then there is the mass of the rockets. Ahem… jets. The ones on the arms look tame enough probaby what 10 lbs or 5 kilos each? But the big honker at the back”” the “”””main thruster”””” is clearly a bigger puppy. WAY bigger. 25 lbs? I’d not be surprised. 25 lbs — biggie № 125 lbs — biggie № 210 lbs — LEFT arm № 310 lbs — LEFT arm № 410 lbs — Right arm № 510 lbs — Right arm № 64 lbs — a tank for the fuel3 lbs — hoses for the fuel15 lbs — control electronics”” pumps25 lbs — welded aircraft aluminum frame55 lbs — fuel at TKO____________192+ lbs at TKO87+ kg at takeoff ON YOUR BACK.Wow. Just wow. While I’m SURE that higher byp”

  9. Fake video ! The noise is no soft music it is an enormous roar nobody will want to have these contraptions flying near them !This not the future this is oilpunk SF.

  11. I am thinking that such a push would collide financially with the electric commuter ‘copter styled transport. A collision which could prove disadvantageous given the comparative elements of cost, complexity and ultimately proven reliability of the components.

  12. Why not just develop a small jet that can be crawled into and when closed up launches. All of this can be computer controlled (take off and landings) and the pilot could then do free flight when the computer is done “escorting” the pilot to a safe flight area. Fuel wouldn’t be such an issue if you are flying an actual plane (no matter the size).

  13. I am thinking that such a push would collide financially with the electric commuter ‘copter styled transport. A collision which could prove disadvantageous given the comparative elements of cost complexity and ultimately proven reliability of the components.

  14. Why not just develop a small jet that can be crawled into and when closed up launches. All of this can be computer controlled (take off and landings) and the pilot could then do free flight when the computer is done escorting”” the pilot to a safe flight area. Fuel wouldn’t be such an issue if you are flying an actual plane (no matter the size).”””

  15. Duh, that’s why you use an Arc reactor. Supposedly that produces about 4m HP which should be more than sufficient for the task. How’s the research on that coming?

  16. I guess it gets a 10 for fun factor potential and proprietary uniqueness … but like most cutting edge tech, falls short on practicality and general value. IMHO, an electric prop driven platform would or could be significantly less expensive, carry far more and potentially safer … not to mention likely quieter as well … granted, not as fun perhaps, but like a high end two seat sportster, the province of the rich.

  17. Was thinking the same thing … along with the suspicion that someone was/is a fan of Ironman. IIRC, maneuvering in the movie is partially accomplished by the glove repulsor units? I am banking more on a winged based unit with some kind of vectored thrust that takes the arms out of the equation.

  18. Duh that’s why you use an Arc reactor. Supposedly that produces about 4m HP which should be more than sufficient for the task. How’s the research on that coming?

  19. I guess it gets a 10 for fun factor potential and proprietary uniqueness … but like most cutting edge tech falls short on practicality and general value. IMHO an electric prop driven platform would or could be significantly less expensive carry far more and potentially safer … not to mention likely quieter as well … granted not as fun perhaps but like a high end two seat sportster the province of the rich.

  20. Was thinking the same thing … along with the suspicion that someone was/is a fan of Ironman. IIRC maneuvering in the movie is partially accomplished by the glove repulsor units? I am banking more on a winged based unit with some kind of vectored thrust that takes the arms out of the equation.

  23. Actually, the physics for that doesn’t work out any better until the diameter of the electric-prop becomes the size of a similarly loaded helicopter. Really… Although admittedly it takes an especially creative formula monkey to push around the fairly complicated physics-of-lift-and-power-to-do-it equations, it is also fairly easy to find useful reductions. In this case: P = k⋅√( (MG)³ / (πρd²) ) … the power (absolute) to lift an object with a fanblade, moving air. Where… P → absolute power invested in the exhaust thrust stream k → a scaling factor — empirically determined for an airframe and propulsion system M → mass of whole contraption & people riding along with it G → 9.81 N/kg π → 3.14159265… ρ → 1.28 kg/m³ density of normal sea coast air d → diameter of rotor, in meters Armed with this, and just “looking at it”, you can see that the power scales per √( 1/d² ) or 1 / d, the inverse of the diameter of the lift fan. You can surmise that if it takes 100 units of power to lift a fat body with a 1 meter total fan diameter (back converted from all the areas of individual fans), that if you quadruple that, then power will drop to about 25%, ¼ of the 100 units. Bonus! You also can see that TOTAL MASS has a scaling of √(M³) or the ³/₂ root of M. Power needed (without adjusting the design of propellers) rises by the 1.5 power of mass. That’s pretty heavy. (Bad pun.) So, for instance, if we compose a platform able to lift 2 people, or 300 lbs worth, AND that has a quadracoptor setup having 4 props each 1 meter across for an effective diameter of 2 m, and given an airframe, motors, fuel, electronics and so on overhead of some 200 lbs or a TKO mass of 500 lbs → 225 kg, then: P = √((225 × 9.81)³ / (π⋅1.28⋅2²)) P = 26,000 watts → 35 horsepower Well… that’s refreshingly modest. Haven’t figured in slipstream turbulence, prop wash inefficiencies, efficiency of the jet-turbine motors, and so on, but hey… the theoretical minimal power is reasonable. Probably nee

  24. Actually the physics for that doesn’t work out any better until the diameter of the electric-prop becomes the size of a similarly loaded helicopter. Really… Although admittedly it takes an especially creative formula monkey to push around the fairly complicated physics-of-lift-and-power-to-do-it equations it is also fairly easy to find useful reductions. In this case:P = k⋅√( (MG)³ / (πρd²) ) … the power (absolute) to lift an object with a fanblade moving air. Where…P → absolute power invested in the exhaust thrust streamk → a scaling factor — empirically determined for an airframe and propulsion systemM → mass of whole contraption & people riding along with itG → 9.81 N/kgπ → 3.14159265…ρ → 1.28 kg/m³ density of normal sea coast aird → diameter of rotor in metersArmed with this and just looking at it””” you can see that the power scales per √( 1/d² ) or 1 / d the inverse of the diameter of the lift fan. You can surmise that if it takes 100 units of power to lift a fat body with a 1 meter total fan diameter (back converted from all the areas of individual fans) that if you quadruple that then power will drop to about 25{22800fc54956079738b58e74e4dcd846757aa319aad70fcf90c97a58f3119a12} ¼ of the 100 units. Bonus!You also can see that TOTAL MASS has a scaling of √(M³) or the ³/₂ root of M. Power needed (without adjusting the design of propellers) rises by the 1.5 power of mass. That’s pretty heavy. (Bad pun.)So for instance if we compose a platform able to lift 2 people or 300 lbs worth AND that has a quadracoptor setup having 4 props each 1 meter across for an effective diameter of 2 m and given an airframe motors fuel electronics and so on overhead of some 200 lbs or a TKO mass of 500 lbs → 225 kg then:P = √((225 × 9.81)³ / (π⋅1.28⋅2²))P = 26000 watts → 35 horsepowerWell… that’s refreshingly modest. Haven’t figured in slipstream turbulence prop wash inefficiencies efficiency of the jet-turbine motors and”

  26. Actually, the physics for that doesn’t work out any better until the diameter of the electric-prop becomes the size of a similarly loaded helicopter. Really…

    Although admittedly it takes an especially creative formula monkey to push around the fairly complicated physics-of-lift-and-power-to-do-it equations, it is also fairly easy to find useful reductions. In this case:

    P = k⋅√( (MG)³ / (πρd²) ) … the power (absolute) to lift an object with a fanblade, moving air.

    Where…
    P → absolute power invested in the exhaust thrust stream
    k → a scaling factor — empirically determined for an airframe and propulsion system
    M → mass of whole contraption & people riding along with it
    G → 9.81 N/kg
    π → 3.14159265…
    ρ → 1.28 kg/m³ density of normal sea coast air
    d → diameter of rotor, in meters

    Armed with this, and just “looking at it”, you can see that the power scales per √( 1/d² ) or 1 / d, the inverse of the diameter of the lift fan. You can surmise that if it takes 100 units of power to lift a fat body with a 1 meter total fan diameter (back converted from all the areas of individual fans), that if you quadruple that, then power will drop to about 25%, ¼ of the 100 units. Bonus!

    You also can see that TOTAL MASS has a scaling of √(M³) or the ³/₂ root of M. Power needed (without adjusting the design of propellers) rises by the 1.5 power of mass. That’s pretty heavy. (Bad pun.)

    So, for instance, if we compose a platform able to lift 2 people, or 300 lbs worth, AND that has a quadracoptor setup having 4 props each 1 meter across for an effective diameter of 2 m, and given an airframe, motors, fuel, electronics and so on overhead of some 200 lbs or a TKO mass of 500 lbs → 225 kg, then:

    P = √((225 × 9.81)³ / (π⋅1.28⋅2²))
    P = 26,000 watts → 35 horsepower

    Well… that’s refreshingly modest. Haven’t figured in slipstream turbulence, prop wash inefficiencies, efficiency of the jet-turbine motors, and so on, but hey… the theoretical minimal power is reasonable. Probably need 2× or more over that just to “get ‘er done”.

    At 50,000 watts, at 15 loiter and 5 TKO/landing minutes, such an aircraft needs to carry 5 gallons of fuel (the legal limit for “Ultralight” exemptions), and weigh in empty less than 200 lbs. Fueled. Big order. Using pixies and garden gnomes as passengers would help.

    GoatGuy

  28. Duh, that’s why you use an Arc reactor. Supposedly that produces about 4m HP which should be more than sufficient for the task. How’s the research on that coming?

  29. I guess it gets a 10 for fun factor potential and proprietary uniqueness … but like most cutting edge tech, falls short on practicality and general value. IMHO, an electric prop driven platform would or could be significantly less expensive, carry far more and potentially safer … not to mention likely quieter as well … granted, not as fun perhaps, but like a high end two seat sportster, the province of the rich.

  30. Was thinking the same thing … along with the suspicion that someone was/is a fan of Ironman. IIRC, maneuvering in the movie is partially accomplished by the glove repulsor units? I am banking more on a winged based unit with some kind of vectored thrust that takes the arms out of the equation.

  31. I am thinking that such a push would collide financially with the electric commuter ‘copter styled transport. A collision which could prove disadvantageous given the comparative elements of cost, complexity and ultimately proven reliability of the components.

  32. Why not just develop a small jet that can be crawled into and when closed up launches. All of this can be computer controlled (take off and landings) and the pilot could then do free flight when the computer is done “escorting” the pilot to a safe flight area. Fuel wouldn’t be such an issue if you are flying an actual plane (no matter the size).

  33. Hmmm… FUN!

    Note though that the amount of fuel needed for a 7 to 9 minute buzz-around is significant. Significant enough that the spin is that the jet engines can be throttled (perhaps automatically) to lower their output as fuel is rapidly depleted over the course of the jaunt.

    Which then kind of begs… how much fuel is involved?

    Well, let’s “try” from just the very squishy tech specs.

    7 minutes, 1,050 horsepower.

    We know that horsepower converts to watts easily enough: about 740 watts per HP. And minutes goes to seconds easily enough too.

    1,050 HP × 740 = 777,000 W
    7 min × 60 sec/min = 420 s
    777,000 × 420 = 326,000,000 joules

    Now, since jet engines can’t really break normal Carnot-cycle engine efficiencies, let’s guess 30% efficiency. This would be REALLY good for a small jet engine. But hey… I’m an optimist.

    326,000,000 J ÷ 0.30 = 1,087,000,000 J of fuel energy.

    We know that most petroleum-based jet fuels are a kind of enriched kerosene, having about 44 megajoules per kilogram of distilled Jurassic Swamp juice:

    1,087,000,000 ÷ 44,000,000 ≈ 25 kg.
    25 kg × 2.21 ≈ 55 lb

    So there you are, goats. 55 lbs or 25 kg of petrol to keep the bird afloat for 7 minutes:

    (25 kg, 55 lb) ÷ 7 min = (3.5 kg/min, 7.8 lb/min)

    Then there is the mass of the rockets. Ahem… jets. The ones on the arms look tame enough, probaby what, 10 lbs or 5 kilos each? But the big honker at the back, the “main thruster” is clearly a bigger puppy. WAY bigger. 25 lbs? I’d not be surprised.

    25 lbs — biggie № 1
    25 lbs — biggie № 2
    10 lbs — LEFT arm № 3
    10 lbs — LEFT arm № 4
    10 lbs — Right arm № 5
    10 lbs — Right arm № 6
    4 lbs — a tank for the fuel
    3 lbs — hoses for the fuel
    15 lbs — control electronics, pumps
    25 lbs — welded aircraft aluminum frame
    55 lbs — fuel, at TKO
    ____________

    192+ lbs at TKO
    87+ kg at takeoff ON YOUR BACK.

    Wow. Just wow. While I’m SURE that higher bypass fanjets will enable a lighter output of necessary power, both the Big Boys in the back and the pipsqueaks for guidance, you’ve STILL got a lot of mass in frame, fuel, electronics, main fans an tankage. And person. Don’t forget the person. 150+ lbs ÷ 70+ kg.

    Just saying
    [b]Goat[/b]Guy

  34. Fake video ! The noise is no soft music it is an enormous roar, nobody will want to have these contraptions flying near them !

    This not the future this is oilpunk SF.

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