Lift, drag and power requirments

[Bergee]

I have been trying to calculate the lift drag and power requirements of this hypothetical flying saucer which uses an external maglev fan. Problem is I am an electrical engineer and not sure if I got my numbers right. Is there anyone who can check my numbers and give me an idea if I am on the right track power wis?Basically it is circular with a 40 foot diameter engine payload area. Around the edge are attached 2' x 10' fan blades spaced two feet apart that ride on a maglev linear induction motor joined at the fan root. I have attached my calculations. Thank you.

 

[rb1957]

1) do not post your email, at least not without disguising it. (ATgmailDOTcom) you may be able to edit your post.

2) I think your lift is optimistic. 35* a single blade ... doesn't sound right to me.

3) I think the power needs to include the vehicle inertia ... overcoming drag is one thing, accelerating the vehicle another.

another day in paradise, or is paradise one day closer ?

 

[rb1957]

oh, I forgot ... induced drag (you've got profile drag only)

another day in paradise, or is paradise one day closer ?

 

[rb1957]

part 3 ...

\why is speed different for lift and drag ?

are the 35 blades turning, ie spinning around the body ? (then blade speed is much higher (and lower) than aircraft speed ?

is your Cd based on frontal area ? (normally Wing Cd is based of S, plan area)

another day in paradise, or is paradise one day closer ?

 

[Bergee]

There are 35 blades turning on the periphery of the saucer on a maglev track. Each blade is 2 feet wide and 10 feet long with a 2 foot gap between each one.
I am simply trying to determine how much lift and drag they will produce at a blade velocity of 140 m/s and how much power it will take to produce that lift and overcome the drag.

I have the Cd based on the frontal area, is that correct?

 

[rb1957]

1) ok, rotating blades. so lift should use the vehicle forward velocity and the blade rotating velocity. but 35* one blade is "wrong", maybe be ok for hover. Have a look into helicopter calcs

2) i'd've thought the body of the vehicle (sorry, I can't bring myself to write saucer) would contribute the lift, but maybe you've being conservative (neglecting it)

3) is there a "drive motor" somewhere ? ie does the lift (from the rotating blades) overcome weight and move the vehicle ? if the baldes are the motive force, then the vehicle will need to pitch (like a helicopter) to create a forward acting component of lift.

4) "I have the Cd based on the frontal area, is that correct?" it depends on the area used to derive the Cd ... if they used frontal area then ok. But i'd also consider drag from the vehicle body, and blade drag is not 35*one blade, and blade drag needs to include the blade rotational speed (like lift) and needs to include "induced drag" (drag due to lift).

another day in paradise, or is paradise one day closer ?

 

[Bergee]

1. You are right, this is like a helicopter with a lot of blades but just using the outer 10 feel of blades for lift. or you might consider it like a jet engine that is sitting vertical and has all those blades turning to move it upwards.
2. So far I am just trying to find out how much this arrangement of blades and their speed can lift.
3. Once I get the vehicle vertical off the ground I will start working on forward movement and the power and vehicle generated lift associated with that.
4. I'm stumped on how to calculate the lift and drag for the 35 blades if it is not 35 * the one blade lift. I wonder how they calculate the thrust with the multiple blades on a fan jet engine. That is just the fan bypass thrust contribution. Basically I am making a huge bypass fan.

 

[GregLocock]

Why wouldn't you use the helicopter calculations? Big clue they don't use n times a single blade.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[rb1957]

I think i get your line of thought ... in hover the blade is "just" advancing through the air, what does it matter what side of the disc it is on ? That somehow doesn't sit well with me. It may be ok (because of hover) but ...

research helicopters.

another day in paradise, or is paradise one day closer ?

 

[moon161]

Back of envelope comparison of power/weight:


R22

https://en.wikipedia.org/wiki/Robinson_R22

Max power / max TOW. = 93 kW/635 kg = .15 kW/kg
Disc loading @ max TOW: 14 kg/m^2


Robinson R66

https://en.wikipedia.org/wiki/Robinson_R66

Max power / max TOW = 201 kW/1225 = .16 kW/kg
Disc loading @ max TOW: 15.7 kg/m^2

K-Max (meshed lift rotors)

https://en.wikipedia.org/wiki/Kaman_K-MAX

Max power / max TOW. = 1118 kW/5400 kg = .20 kW/kg
Disc loading @ max TOW: 16 kg/m^2

Bell 206

https://en.wikipedia.org/wiki/Bell_206

Max power / max TOW. = 234 kW/1451 kg = .16 kW/kg
Disc loading @ max TOW: 18 kg/m^2


CH46 (counterrotating lift rotors)

https://en.wikipedia.org/wiki/Boeing_Vertol_CH-46_Sea_Knight

Max power / max TOW = 2800 kW/11,000 = .25 kW/kg
Disc loading @ max TOW: 22kg/m^2

Ka-27 (coaxial counter rotating lift rotors)

https://en.wikipedia.org/wiki/Kamov_Ka-27

Max power / max TOW = 3320 kW/12,000 = .28 kW/kg
Disc loading @ max TOW: 30kg/m^2

Ka-50 (coaxial counter rotating lift rotors)

https://en.wikipedia.org/wiki/Kamov_Ka-50

Max power / max TOW = 3600 kW/10,400 = .34 kW/kg
Disc loading @ max TOW: 33kg/m^2

CH-53

https://en.wikipedia.org/wiki/Sikorsky_CH-53_Sea_Stallion

Max power / max TOW = 5854 kW/19100 = .30 kW/kg
Disc loading @ max TOW: 50kg/m^2

Flying saucer

Max
Max power / max TOW = 9243 kW/82,855 kg = .11 kW/kg
Disc loading @ max TOW: 60kg/m^2

Power / MTOW seems to climb as the disc loading does.
The flying saucer comes in much lower than the lightly loaded R22, R66 and Bell 206.

 

[moon161]

There's a well defined group in this plot, a few russian helicopers (Mil Mi 1, Mi 10, Mil V12) and the UFO lie outside of it.