Jan 29, 2011, 06:51 AM
Registered User
Discussion

# Blimp Thrust Calculations

I would like to know how much thrust I need to make a blimp withstand 40 kmph winds. Unfortunately I do not have the exact dimensions in an AutoCAD format yet, however I can tell you that it will be a tear drop design, with 2 vectoring thrusters (similar to the blimpduino) and 1 thruster on the back (similar to the hyperblimp).

I'm not sure how to work out how powerful I should make the motors to overcome the force of wind against the blimp, I know for a definite answer I'd need to specify the actual geometrics but here I am more interested in a formula (so I can play with the geometries in AutoCAD and come up with an optimal solution).
 Jan 29, 2011, 08:12 AM Graf Zeppelin Fw = A * p/2 * vē * cw Fw - aerodynamic drag A - area of drag p - density of air v - speed cw - indication of air-drag (blimp)
 Jan 29, 2011, 08:55 AM Registered User Thread OP What would CW be in this case? And wouldnt the fact that it's a tear drop shape make a difference to the drag? Or is that what the CW is?
Jan 29, 2011, 09:12 AM
Graf Zeppelin

# Cw

Hi...!

exact... the Cw-value is the individual indicator, which says, how much drag your shape produces. So every shape has a own Cw-value!

Holg-Air
 Jan 31, 2011, 01:29 AM Registered User To get a ballpark numper of drag coefficients Cd (in German Cw), you can calculate it from existing airships or blimps. Actually this number includes propulsive efficiency. If you want to calculate the engine power for your blimp, this is exactly what you need. The equation would be Cd = P/(rho/2*u^3*V^(2/3)) where Cd is the drag coefficient, based on volume P = Engine Power in Watt rho = air density, e.g. 1.225kg/m^3 u = speed in m/s V = Envelope reference volume in m^3 To use the volume to the two third as reference area is helpful since this is the most descriptive parameter of an airship. Ok, let's try the Skyship 600: P1 = 2*300*hp*735*W/hp = 441000W u1 = 105*km/h * 1000*m/km * 1/3600 *h/s = 29.2m/s V1 = 7600*m^3 This would result in a drag coefficient of Cd = 0.0751 Assuming the Cd constant with Reynolds number(*), and assuming an envelope size of, say, V2=2m^3, the power needed for your u2=40km/h=11.1m/s would be approximately P2 = Cd*rho/2*u2^3*V2^(2/3) = 100W (*) Unfortunately, the drag coefficient varies with Reynolds number. To make it simple, one could assume that the increase in friction drag of a turbulent boundary layer is made up by partly laminar flow over the envelope of the small airship. To be on the save side, use a factor of two. To get a more accurate guess, you can scale the drag coefficients by the according friction drag coefficients: Cd1/Cf1 = Cd2/Cf2 solved to Cd2: Cd2 = Cd1 * Cf2/Cf1 where Cd1 is the drag coefficient of the Skyship 600 as computed above Cd2 is the drag coefficient of small airship as computed above Cf1 is the friction drag coefficient of the Skyship 600 Cf2 is the friction drag coefficient of the small airship The friction drag coefficient is the drag coefficient of an equivalent flat plate. The assumption is the ratio of drag to pure friction drag is the same for both the Skyship and the small ship. For the Cf, I'm using the "ITTC57 Friction Line": Cf = 0.075/(log10(Re)-2)^2 Where Re is the Reynolds number computed by Re = u*L/ny Where L = Reference Length ny is kinematic viscosity of air with approximately 14E-6 m^2/s Or you use just a factor of 2 Best, Johannes Last edited by johannes.eissing; Jan 31, 2011 at 01:44 AM. Reason: typo
 Feb 01, 2011, 07:21 AM Registered User Thread OP One thing I am unsure of here is the engine power in watts. Wouldnt using different types of props (for example co-rotational ones or longer ones) give you more thrust per watt?
Feb 02, 2011, 12:52 AM
Registered User

# Propeller design

Quote:
 Originally Posted by Narwhal One thing I am unsure of here is the engine power in watts. Wouldnt using different types of props (for example co-rotational ones or longer ones) give you more thrust per watt?
Narwahl,

sure propeller placement, size and design has an impact on propulsive efficiency.

As a rule of thumb, a stern prop should be as large as possible, but not larger than half the maximum diameter of the envelope. In most cases, this would result in a heavy, slow turning prop, needing a heavy gear. So you would decide for a smaller one, even though efficiency decreases with size, to save weight.

Same is with fancy propeller design. Contra rotating props might be more efficient, but they add weight and complexity. Weight at the very stern of the hull is quite an issue concerning static trim and "hogging" of the envelope.

You gain more efficiency from focussing on the propper size, slope and rpm of the prop, than on unconventional design. Unfortunately there aren't many "performance maps" around for model propellers off the shelve.

But I agree, contra rotating props would add an unbeatable coolness factor, though ;)

Johannes
 Feb 02, 2011, 05:14 AM Registered User Thread OP So take for example this contra-rotating BL motor: http://www.hobbycity.com/hobbyking/s...dProduct=10592 Down the bottom it says the Power (W) is 254, I take it then this should be more than enough to propel the blimp at 40 kmph +? Or am I missing something when it comes to this?
 Feb 02, 2011, 02:43 PM Registered User Yummi. 250W would be about the right delivered SHAFT- power for a Skyship 600 scaled to 2cbm. Those stated 254W is consumed electrical power. Engine efficiency might be in the range of maybe 80% at the most favourable rpm (rough guess). Using a stern prop instead of two mounted underneath the belly would improve performance ...if the prop disk area would be same as the sum of the latter two, and if the stern is not too blunt. For a 2cbm ship, 7cm prop diameter is not much. The smaller the prop, the higher the disk loading, and the worse the propeller efficiency. As I said, I'd go for a larger, stern mounted prop if speed and efficiency is an issue.
 Feb 02, 2011, 09:05 PM Registered User Thread OP I'll probably opt for a larger prop (I assume I can just replace them but keep the contra-rotating motor) on the stern. But what engines would you recommend for the vectoring thrusters, their use being ascending, descending, turning and perhaps providing extra power if necessary to the forward movement.
 Jun 19, 2011, 03:47 AM Registered User Thread OP I was going over these equations again, and I decided to use the LZ 129 Hindenburg as the example because it's fineness ratio of 6 is close to mine. So for the equation Cd = P/(rho/2*u^3*V^(2/3)) I used these values: P = 1,200 (Horsepower per engine) * 4 (Number of engines) * 735 (W/Hp) = 3,528,000 W rho = 1.225 kg/m^3 u = 135 km/h * 1000 m/km * 1/3600 h/s = 37.5 m/s V = 200,000 m^3 And I obtain a Cd of 0.0319, is this a valid assumption given my airship has a 6:1 length to diameter ratio?
Jun 22, 2011, 12:07 PM
Master of Flying Wings......
Quote:
 Originally Posted by Holg-Air Fw = A * p/2 * vē * cw Fw - aerodynamic drag A - area of drag p - density of air v - speed cw - indication of air-drag (blimp)
seriously man just do trial and error like the old guys did in WWI.....

damn mathematics. I can understand measuring something to be precise but trying to figure something completely random like that is just silly.....
Jun 22, 2011, 05:07 PM
Graf Zeppelin
Quote:
 Originally Posted by flyingwing12 damn mathematics
But the mathematicians have one advantage:

Every morning they wake up with a "unknown" !!!
 Jun 24, 2011, 08:43 PM Master of Flying Wings...... i mean really. Do you think for one second that the Germans figured the drag and set a speed for the zeppelins BULL MESS! Mathematics just means your scared to experiment..... hmmmmm...maybe a new inspirational quote?? A Mathematician can do many things, but when placed next to an expert mechanic, he is a fool. Everything we see around us was made by the wonder of the human mind. The pinheads in the early days just manipulated letters and numbers to make the same measurements able to scale. There is no creativity with mathematics. Everything is EXACT or PRECISE. When just simple trial and error gives you a wonderful result, overall satisfaction. You do not need math to be happy with your design. I admit it makes you feel smart, but it is all for enjoyment and not for a bloated ego. Go have fun with it dammit, experiment and you may learn something no one else knows. Last edited by flyingwing12; Jun 24, 2011 at 08:48 PM.
 Jun 25, 2011, 07:58 PM Registered User Thread OP I came across this research paper the other day: "TECHNICAL FEASIBILITY OF LOITERING LIGHTER-THAN-AIR NEAR- SPACE MANEUVERING VEHICLES" On page 12 they give the drag coefficient vs fineness ratio, and it turns out that 4.62 is the best fineness ratio, so that is what I will use.