Airship Volumes

[AirCaptain2840]

Does anybody know how they found out the volume of an airship? I've looked all over the internet, but I haven't found amy formulas. HELP!!!!!:confused: :cry:

[AirCaptain2840]

and also, how did they know how many tons an airship could carry, after removing the structural weight?

[Charles Adams]

I've been working on a book about this so I should be able to help. :)

Volume is calculated by multiplying the area of the cross section times the depth. This is easiest on an early rigid airship where the hull sides are parallel. Start with the polygon that makes up the cross section. Using standard geometric formulas, calculate the area of this polygon.

Next, multiply that by the length of one "bay." This gives you the total internal volume of that section of the ship. Then, you must subtract for the small volume occupied by the frame and also some volume to allow space or "breathing room" around the gas cell. This breathing room allows the gas to expand when the temperature rises, such as on a sunny day when the hull of the airship gets heated by solar rays.

The cells were designed to contain a round number of cubic feet. These are basically rounded "bags" so the calculations can only be so precise. I'm sure they started with the total available volume of a structural bay, then subtracted enough to allow for expansion, making the patterns from the net desired figure.

Once you know the volume of gas in each cell, you can figure the available lift (called "buoyancy"). Start with the density of the gas. This density yields a specific amount of lift per liter (or cubic foot) of gas at a given temperature and atmospheric pressure. These values are known and readily available for both hydrogen and helium. Most such formulas are expressed in metric or SI units:

Helium: Approx. 1 gram per liter of gas at "standard" temperature and pressure of 20 deg. C and 760 mm of atmospheric pressure. This equates to just over 1 kg per cubic meter at 0 degrees C and 760mm of atmospheric pressure.

Hydrogen: This is slightly higher than helium, about 1.1 kg per cubic meter at 0 degrees C and 760mm of atmospheric pressure. (This was the only figure I had handy.)

Once you know the volume of gas and the lifting capacity in grams, kilograms, or pounds, multiply your volume of gas to determine the total lift of the gas.

Finally, subtract the total weight of the airship frame, engines, and structure to get the net available lift. This will tell you the available payload of the airship. In other words, this is the weight it can carry in passengers, crew, fuel, ballast, and anything else that is not part of the structure of the ship.

[Charles Adams]

I should also note that the lifting capacity of the gas is determined by its density. This density varies with temperature and atmospheric pressure according to the laws of physics. The gas expands when it is heated and this increases the available lift. In the same way, the gas contracts when cooled, reducing available lift.

These variations made it very tricky for airship designers and operators. Something as simple as the sun heating the hull could cause problems. On a hot day, they might need to vent gas to keep the ship in trim (i.e. balance). On the other hand, they could carry a smaller payload on cold days.

Volume also varies with altitude. As the airship ascends, the atmospheric pressure decreases, causing the gas to expand. This varying pressure is why high-altitude balloons have such a strange shape when they are launched. On the ground, these balloons look only partially filled. Then, when they reach altitude, the shape changes and the balloon looks full!

[AirCaptain2840]

Umm, say that in English, please- I may be 14, but i'm not particularly good with translating words to math.

[AirCaptain2840]

so, if I decide to build an airship (granted that I have the money), it will be 800 feet long, by 140 feet tall. The first(top) floor are the passenger quarters, which are 10,000 sq. ft. The bottom floor comprises of the crew's quarters at 6,000 sq. ft, and behind that is the cargo hold at 4800 sq. ft. What would the volume be?

[Charles Adams]

If you want a practical example, let's try to keep it as simple as possible. There is no such thing as a box-shaped airship, but that's easier to calculate. So, for the sake of a simple example, let's say your airship is shaped like a brick.

Let's say the "brick" is 800 feet long X 140 feet tall X 100 feet wide. The cross section is a rectangle that is 140 X 100 feet. The area of this rectangle is 140 * 100 or 14,000 square feet. Take this and multiply by the length to get the total volume of 14,000 * 800 = 11,200,000 cubic feet.

Using an English-to-metric conversion formula available on any number of websites, plug in the volume in cubic feet and convert this to cubic meters. The total volume is about 317,150 cubic meters.

At about 1 kg of lifting force (buoyancy) per cubic meter, this means the total lift of the enormous airship would be about 317,000 kg or 700,000 pounds!

Now, you must subtract the weight of the airship's hull, engines, and structure to figure the available payload. If the airship was non-rigid, the envelope would simply be the weight of the fabric. If it was rigid, you would have a frame to consider. Then there are the engines, etc.

Let's say the airship weighed about 50 tons or 100,000 pounds empty. Subtract this from the available lift of 700,000 pounds to get an approximate payload of 600,000 pounds. That's a lot of lifting capacity!

Hope this helps. . . :)

[AirCaptain2840]

so, that's it? no subtractions for the rounded ends?

[Charles Adams]

No, I was trying to illustrate the process in the simplest way possible. In general, volume is length X width X depth, but calculating the volume of a shape like a rounded airship hull is much more challenging. In fact, I'd much rather build a 3D model of it and let the computer calculate the volume. That would be much easier than trying to do all the math yourself!

Hope that helps. . . :)

[Charles Adams]

To see how much the components of an airship might weigh, check this out:

http://www.aht.ndirect.co.uk/airships/r37/R37Parts.htm

This was the R-37, a British airship built to 90% completion but then cancelled before it could be flown. The grand total weight if you add all that up is 69,720 pounds. It seems my estimate of 50 tons was pretty good! :)

As another math example, it was stated on that website that she had a disposable lift of 50%. Assuming this means that her lift was twice her weight, that would be a lift of about 140,000 pounds. Converting to kilograms, that is 63,500 kg.

Hydrogen has about 1.1 kg of lift per cubic meter. We know how many kg of lift this airship had, but we need to figure the resulting volume that provided this lift. A formula we can use to do this is:

(?) cubic meters of volume * 1.1 kg/cubic meter = 63,500 kg

Using basic Algebra, "x" will be the unknown quantity:

x * 1.1 = 63,500

x = 63,500 / 1.1

x = 57,727 cubic meters

Converting this to cubic feet, we have 2,038,609 cubic feet.

According to the website, her stated volume was actually 2,101,000 cubic feet. So, you can see we are pretty close! :)

[AirCaptain2840]

Thank you very much. Also, do you know what programs I can buy for 3-D designing? (that are on the cheap side, and that a 14-year-old can use)

[Charles Adams]

There are several good programs you can use to get started for free. I don't think these programs will calculate volumes, but will allow you to build shapes in 3D. To calculate volumes, you may need a CAD program that has 3D features.

Read these threads for more info on free and inexpensive 3D software:

http://www.starshipmodeler.net/cgi-bin/phpBB2/viewtopic.php?t=20715

http://www.hobbytalk.com/bbs1/showthread.php?t=183107

http://www.scifi-meshes.com/forums/3d-questions-answers/

[Danthekiwi]

I worked out roughly based on the wartime zeppelins that if you worked out a box based on the dimensions then mutiplied that by 0.6 then that was about the capacity.

I like to dream about flying ships (both models and real things) and i always work it out to 0.52 just to be safe. that way if theres any extra lift its a bonus.

[AirCaptain2840]

so, you just make a 'box', insert dimensions, then do the math; then with the total answer, you multiply that answer by 0.52?

For example-

V= L x W x H
5 x 2 x 2
V= 20 cu. ft.

20 x 0.52=
10.4 cu. ft? :confused:

[Danthekiwi]

take the maximum width
so say a flying model thats say 5 metres thats around 60 cm

divide that by 2 and thats the radius of your circle

area of a circle pi time radius squared:

(3.14 x 30)2
then times that by length say x 500

then take that total and multiply by 0.62, and that is a VERY ROUGH aproximation of volume, its easier to keep the calculations in cubic meters as 1 cubic meter of helium can lift 1 kg (as a general guide)

but as I say this is only a general guide based on volumes and dimensions of war time zeppelins and doesnt take into account how much space the frame and bomb bays took up.as well as the fact the war time zeppelins were the same width most of the way down as opposed to later desgins whcih had no two rings the same diameter for a more aerodynamic efficiency.

[Danthekiwi]

oh another thing, an airships width is not a pluckable figure, there are alot of things that are effected by the length to width ratio. For example american designs such as the MACON and AKRON had a ratio of around a width = 1/5 of length. War time zeppelins had a ratio of around 1/7. 1/5 gave greater structural strength, where 1/7 gives a better aerodynamic and inertia qualities ( dont know how that second one works but apparantly it does)

[joec]

I'm the worst at reading all the threads so if I repeat forgive me, I'm sure this is in there somewhere.

If your ship is patterned from a known airship with a known volume and you are scaling it down (for a model) you can calculate the volume by dividing the actual volume by the scale cubed. Example; if the real airship is 5,000,000 cubic feet and your model is the same but 1:80 scale, the volume is 5,000,000/80^3=9.76 cubic feet or (x1728 ci/cf) = 16875 cubic inches. I use 1 oz lift per cubic foot for the helium, but have little real experience with it, but from searching around it seems a reasonable number (oddly enough it's almost exactly one gram per liter).

[take a cube that is 4' x 4' x 4', the volume is 4x4x4=64 cubic feet. At 1:80 scale it would be 0.05'x0.05'x0.05' (.6" on ea side) or .000125 cf, same as 64/80^3=.000125]

I use a spread sheet to calculate the volume of my airship as well as to find the surface area of the outside and the surface area of the gas bags. It calculates everything based on truncated cones, if I could post a .xlr I'd put it on here, but if you want I can email it to anyone interested. It's not rocket science, you just need the formula. For each line of the spread sheet you put in the radius of the ring (zero for first and last) and the distance to the next ring. Like Charles said if you have 3D cad use that, much easier, I use 2D cad, but getting the numbers off it is easy, and I basically had to so I could calculate the weight of the bags and the get the center of bouyancy and center of mass etc. If you only calculate it at a few locations it's not very accurate but once you've gotten up to say 20, my guess, the answer won't change a whole lot the more sections you add.
JoeC

[JoeVanCleave]

... where 1/7 gives a better aerodynamic and inertia qualities ( dont know how that second one works but apparantly it does)

As I understand it, an airframe that is better streamlined, such as the 1/7 aspect ratio of the hypothetical WWI Zeppelin hull listed above, will have less aerodynamic drag. The inertial quality referred to comes into play when landing, for instance, there's a point where the captain orders the throttles cut back; knowing how much further the ship will glide under its own inertia (here I'm assuming zero ground winds) depends on how much drag the hull presents to the air. An airship hull of greater drag - such as a 1:5 aspect ratio - may slow down faster (have less inertia) than the hypothetical 1:7 hull, all else being equal.

This is a bit off-thread, but somewhat related...

I'm also interested in the thermodynamics of these massive airships, in terms of them being modelled as massive gas systems. Charles touched on this earlier in his discussion of how the net lift is affected by gas temperature and pressure, as well as that of the surrounding atmosphere.

Some of the behaviors of the rigids were non-intuitive. For instance, in reading literature on the 'preflight' weighing-off process of the giant rigids, it seems that if flying early in the morning, and after an overnight stay in the shed, the airship's lifting gas can actually be colder than the surrounding outside air, caused by the enthalpy of the closed system of the gas cells; this resulted in reduced lift at weigh-off. However, once in the air and underway as a dirigible, the properties of convection came into play, as the outside air moved across the hull; this resulted in rapid heating of the interior volume by the warmer outside air, resulting in a rapidly changing situation where the static condition of the airship changes from heavy to light. Rookie pilots were cautioned not to react too early to this situation by weighing off the airship neutral, but under these conditions to weigh off a bit heavy, and use a takeoff with dynamic lift, using a short takeoff run; once underweigh and the air temperatures begin to equalize, the ship's static condition becomes closer to neutral, rather than becoming too light.

There's also some evidence I've gleaned from books that these massive airships can have peculiar properties when flying at a nose up or nose down attitude, for instance when compensating for a static inequilibrium by using positive or negative dynamic lift, or when flying in turbulent weather; it seems that differences in air temperature of only a hundred feet or so of altitude between one end of the ship and the other can cause the gas cells at one end of the airship to heat or cool at a faster rate than those of the other, due to their altitude difference, and also due to rapid changes in altitude.

For instance, a rapid change in pitch attitude (let's say caused by turbulent air), say nose down, can cause the cells in the rear of the ship, which have now changed altitude rapidly, to rapidly expand in volume; the rapid expansion causes them to lose heat rapidly, because in the short term they can be considered a closed thermodynamic system, and the absolute gas law says that if a fixed volume of gas increases volume then the temperature decreases; this causes them to decrease lift, which would function as a self-correcting system to limit the magnitude of such pitch excursions. However, there are other instances where the large-scale behavior of such airships is counter-intuitive, and an inexperienced pilot or elevator man can get the ship into serious trouble due to an incorrect (but logical) response.

It's interesting to me how the large-scale behavior of these flight systems is little understood today from a practical point of view; sure, computer modelling of a flight system can arrive at some of these same behavioral conclusions, but I'm under the assumption that, at least in the case of heavier-than-air, the modelled system's behavior is constantly corrected and updated by real-world flight experience; this is not the case with large rigid airship systems, as the only imperical evidence we can use to correct the theoretical model is hearsay and secondary evidence from 67+ years ago.

As I see it, were a large rigid airship be developed today, the biggest success challenge would be in understanding the system's flight behavior in the atmosphere in a way that is non-intuitive from merely understanding the behavior of smaller airships. Most of that knowledge was imperical, learned through trial and error, and most, if not all, of those folks have passed.

EDIT: I would also mention that much of the success of the German rigid airship in transoceanic travel was in the understanding and practical application of 'pressure pattern flying', where the captain took advantage of tail winds from high or low pressure systems, choosing not the most direct route in distance, but the most efficient route in terms of tail winds. This would be hard if not impossible to apply today in commercial air travel, given that commercial air traffic is usually routed through precisely defined air corridors; the requirements of large rigid airships to fly pressure patterns rather than predefined navigational corridors makes them completely incompatible with global air travel. Unless much political pressure could be applied, at an international scale, to give rigid airships a chance to compete, their future is merely theoretical, I'm afraid to say.

Anyway, thanks for such an interesting thread.

~Joe

[Jake]

Does anybody know how they found out the volume of an airship? I've looked all over the internet, but I haven't found amy formulas. HELP!!!!!:confused: :cry:

It really depends on the airship.

If you don't want to do much math, a close approximation would be to use the formula for the volume of a cylinder:
V= (pi*radius^2)*length

Of course, this would be an overestimate, since the airship is curved. But to find an exact formula for the volume of the airship, you need to know exactly how it's curved... It's not the same for every airship.

A somewhat better estimate would be to use a cylinder with a cone on each end. Still, that's far from precise.

Do you have a particular airship in mind that you want to find the volume of? Or a shape for that matter?

Because if you know the equations that describe the shape of the airship, then we can pretty much find the exact volume.

As an example, here's an excerpt from the book Airship Technology:
The National Physical Laboratory in England suggested a low drag shape for an airship hull. A cross section along the axis of symmetry was defined mathematically.

I've redrawn the figure, with the equations of the shape included.
http://img.photobucket.com/albums/v696/ghaleon88/PerfectAirship.jpg
*Edited to save bandwidth*

If you don't like math, skip to the end of this post about now. :P

If you do like math: For of an airship of this shape, you can find the volume quite easily. Basically, here 'y' becomes the radius of a cylinder, but it continually changing along the length of the airship (x). If you find the volume of that funny-shaped cylinder (you'll probably have to use calculus), that's the volume of the airship. For this example,

Radius:
r = y = b*sqrt(1-x^2/a^2) [front half]
r = y = b*sqrt(1-x^2/2a^2) [rear half]

The area of a circle is pi*r^2. So in this case, pi*r*2 = pi*(b^2)(1-x^2/a^2) [front half]
This is the same as: pi*b^2 - pi*(b*2/a^2)(x^2) [front half]
For the rear half, it's similar: pi*b^2 - pi*(b*2/2a^2)(x^2) [rear half]

If you integrate this along the length of the airship:

The front half gives you 2/3(pi)(b^2)(a)
The rear half gives you 0.9427(pi)(b^2)(a)

Adding these up gives you:

1.609(pi)(b^2)(a)

Since 2b = width, and 2.404a = length, you can get the following equation:

(You can pay attention again even if you don't like math):

Volume = 0.1674 (pi)(width^2)(length)

This describes the volume of the above airship. :D

Not all airships look exactly like that, but most look pretty similar... so it should be close enough.

[Charles Adams]

Calculus -- UGH! I've tried hard for years to forget calculus. . . ;) ;) :LOL:

[Jake]

Sorry!

(I've done all the hard work for you though!) :p

BTW - I might have made mistakes, so, don't bet anybody's life on that formula. But I think it's right.

*EDIT* Holy moly! I can't believe I did all that work just to get this equation. If you multiply 0.1674 by pi, you get 0.5259.

So that formula becomes V = 0.5259(width^2)(length)

Which is pretty much exactly what was posted before by DantheKiwi as an approximation derived from historical figures.

Interesting!

[toddk]

[Pendant Mode On]

Actually, the formula stated in the book is wrong, something which I was able to confirm via correspondence with the author of that particular chapter (Cheeseman if I recall) a number of years ago.

Technobabble follows (for those disinclined to follow, skip to the end of the post). (With further apologies to Charles.)

While the prolate ellipsoid is correct for the bow, the stern actually should be a parabola rotated about the x axis rather than a prolate ellipsoid. (Note that the formula given in the book describes a prolate ellipsoid with a semimajor axis of sqrt(2)*a and a semiminor axis of b, which clearly is at odds with the stern tapering to a point as in the diagram. It is also at odds with the description in the text that the radius of curvature strictly increases from bow to stern.)

The correct formula for the stern is:

(x^2)/(2a^2) + y/b = 1

or,

r = y = b(1 - (x^2)/(2a^2)

with the area for the circle bing

pi*b^2(1 - (x^2)/(a^2) + (x^4)/(4A^4).

Which upon integration gives the volume for the stern as:

(8*sqrt(2)/15)*pi(a)(b^2)

and a total volume of

((10+8*sqrt(2))/15)*pi(a)(b^2) (exact)

or approximately

1.4209pi(a)(b^2)

or in terms of length and width:
(Laypersons' Results)

Volume = 0.1471pi(length)(width)^2

Also, other results from that chapter suggest that for a streamlined hull, the lowest air resistance actually comes from an aspect ration (width/length) between 1/4.5 and 1/6.

[/Pendant Mode Off]

Any errors in the results are purely my one.

[Jake]

There's an error in the book? Wow! Thanks the correction. That makes a lot more sense, and you're right, it fits much better with the diagram.

*Posts a sticky note in his copy of Airship Technology*

The new constant is 0.46, then. Somewhat lower than 0.52.

[toddk]

No problem, it really isn't obvious unless you're paying very close attention.

When I first noticed it, I brought it up on the Airship List and Arnold Nayler was able to put me in touch with Professor Cheeseman (retired). As previously stated, he confirmed that the correct curve was indeed a parabola. At the time, he claimed (and I have no reason to doubt him) that he was unaware of the error, so I am forced to assume that it is very easy to overlook. (Especially seeing as I was reading it purely for my own edification.)

[Ray E.]

Holy Cow people , can any of you do the math with out a computer ? When I was 13 or 14 I just worked it out with a pencil and paper . While your formula's for the volumes of ellipses and parabolas " shapes " would be accurate for a non rigid ship ; to get an accurate volume for a rigid airship figuring the the volume for each cell would be more accurate . It's then a matter of figuring the volume of cylinders , cones , and frustums of cones and adding them together . If Aircaptain still needs help ,the various formulas can be found at places like Wikipedia . When I was his age I found them in a machinists reference book . They have a million formulas in them and they're still out there . If you work with your hands and you have one it will get used . Don't get me wrong I think calculators and computers are great it's that many people think that you HAVE TO HAVE THEM to figure anything out .:confused: And certain technology requires them .But really anyone posting to this thread could figure it out on paper if they wanted to , there's just no reason to . Thinking back makes me realize that I was really focused at the time because I was figuring everything out to 8 decimal places . Have fun with it :cheers: .

[Charles Adams]

I agree it's most helpful to know the formula, how it was derived, and exactly how to use it before making calculations. But, after spending years in the study of math (my least favorite aspect of engineering), I know from experience that doing the work by hand (pencil to paper) allows for the possibility of making mistakes at many different points in the process. Many of us are not math wizards by nature (including me) and we can sometimes make silly mistakes.

Not only is it simply quicker and easier to use a computer or calculator, but this approach is also less prone to error. Maybe it's an "old school" vs "new school" difference in mentality, but many folks prefer the quickest and easiest solution to the problem. Why do more work than you absolutely have to do? :)

Still, with that said, if you just find some formula and start plugging in numbers without knowing exactly what it is you're doing, you could be in for all sorts of problems. So, at the very least, know what's going on in the formula and what it is you're doing!

[toddk]

Ray, no it doesn't give the actual gas volume for a rigid but it does give an upper limit. Further for a rigid one needs to know various factors (depth of longeron girders and ring spacing for example) before beginning to calculate the volume of the truncated cones. And considering that this thread started with approximating the volume with a box and progressed through hand waving, it seems slightly odd that you feel the need to criticize slightly more rigorous methods. As to mathematics by hand, that is precisely what I did to obtain my final formula. I suppose that you actually mean arithmetic by hand, in which case not often, I'm not even sure that they teach it any longer.

Charles, I think that your most effective argument is actually the knowledge one. Also, I generally use a computer for specific results as there is little software (and fewer calculators) which are capable of producing general results (and those which are capable of doing so assume that the user is knowledgeable to correctly set up the problem). Silly mistakes are possible with either method.

In point of fact, it was a formula followed blindly which drew me into this discussion. I then provided my own solution to the correct formula which I had worked out quite some time ago for my own private use. I wouldn't have added it to this thread otherwise because, it is one of many different streamlined designs each of which will yield slightly different results and, as has already been pointed out, it only provides an upper limit on the volume of a rigid.

Hopefully, I haven't misunderstood either of you too badly. I've delved too deeply into mathematics to think calculations are anything more than arithmetic. To make my point, I'll trot out an old story which I'm sure that Charles has heard...


A man wants to know what three times three is, so he decides that he will as an Engineer, since they work with numbers. The Engineer whips out his calculator and pronounces it to be, "9.000000000001."

The man is suitably impressed but wonders if there might be a simpler answer so he decides to as a Physicist. The Physicist immediately answers that, "It is about ten."

The man is now very confused, he wants to know once and for all what three times three is. Is it 9.0000000000001 as the Engineer said or about ten as the Physicist said. He decides the only way to settle this once and for all is to as a Mathematician. When he does, the Mathematician gets a glazed look on his face and wanders off...

Several days later, the man gets a phone call from the Mathematician who tells him, "About your question, I don't have a solution yet but, it does exist and it is unique."

[Ray E.]

First of all it seems I both failed to make my point and offend Todd and Charles at the same time by what was intended as humor . By the by they still teach kids how to do math without calculators in school , I just asked my 15 year old who is in the advanced math program . I was merely trying to make a point for a 14 year old ,not come up with a solution for string theory . In fact I was enjoying the progression of complexity of math in the thread , but it seemed that a step was skipped that a 14 year old could actually use . I suspect that not all people interested in airships are either mathematicians or engineers . Now I know a lot of people will know what Ockham's Razor is " start with the simplest possibility and work to more complex ones " it's longer and more complicated in the latin he wrote it in . The Military calls it KISS "keep it simple stupid " less things to go wrong .My point is if you know the sizes of the gas bags you know the potential lift without needing the exact size or shape of the airship . Of course you need the model's weight for static lift but that's about it . Lift from vectored thrust , a lifting body shape or from planning are not going to be factored in .I imagine more then one MODEL airship has been modified while under construction and adding a cylinder section would be a quick and dirty way of adding lift to an airship . There is most likely an infinite number of factors that could be taken in to account covering lift in an airship from air temp. to the sun shining on the envelope to using chaos theory to figure the the paths of helium molecules in the gas bags or even the wrinkles in the fabric covering , but I thought this was about MODELS . Also I didn't say I didn't like to use computers and I did mean math because it's fun to do.What do you think would be the ideal size/volume for a model rigid be ?:confused:

[joec]

Interesting stuff, I didn't know the hull shapes were fitted to mathmatical formulae. I will say in reading about airships the calcualtions they did were astounding, especially since they used only slide rules. When the airship weighs overall somewhere in the hundreds of tons, yet they could tell if there was a 150 lbs stowaway on board they obviously kept their pencils sharp, ie lots of significant digits. Of course when filled with gas the majority of the weight is offset but regardless, they still worked with small numbers relative to the whole. JoeC

[toddk]

First off, no offense was truly taken and, I apologize for having seemingly done so.

Secondly, as for your question, there are several possible answers. The first one is as large as you have room to. While that may sound flip, it isn't intended that way. The simple truth is that a rigid, whether a model or not, requires a lot of lift. Surplus lift means that one can use larger batteries which can either be used to give longer flight times or power larger and/or more motors.

There is a thread under R/C airships where several of us are discussing our own rigid projects with balsa frames. The design elements which seem to be generating the most concern are the weight of the gasbags and the weight of the covering. My project, which has yet to progress beyond the design phase, is for a .9m by 4.5m streamlined hull based on the NPL formula already mentioned. The total gas capacity has been calculated at slightly over 1.5 cubic meters if the cells are at 100% inflation. If it were physically possible, I think that I would increase the diameter to between 1.25 and 1.5m and increase the the length to width ratio from 1:5 to 1:6 which would result in a 7.5 to 9m long hull. Merely increasing the radius without changing the length ratio would yield something roughly between 4 and 7 cubic meters of total gas capacity... That provides for a lot of lift. :)

I suppose that my ultimate answer to your question would be a 1.5m radius 9m long model would be close to ideal. It might even provide enough lift to add a camera. :D

[Ray E.]

If I were to move some other projects and were willing to deal with the wife's wraith I could go as large as 2.5 meters in dia. and as long as 10 meters . A more reasonable dia. would be .8 meters by around 4 meters. Oh snap ! I just don't think in metric , not on purpose anyway . Let's try 30 in. by 12 ft. But this go round I haven't made my mind up on the airship project yet . I'm going to try and check out the RC threads you mentioned.

[AirCaptain2840]

Wow- I never knew that this thread would be so pouplar....... Anyways, some of you are saying that airships are designed by a ratio depending on structural stability, and some of you are saying that they are designed by a mathematical formula..... WHICH IS IT???????

Well, now I think I know another reason why they stopped making airships- it's REALLY complicated! I wonder if there are any agencies in the US or Germany where they have a whole store of this information.... watch the formula be something really easy! :LOL:

*everybody groans in frustration of the math above*

Anyways, I must say thank you to everyone for the info so far.... I hope it helps with my airship blueprints, which I can reccomend that people do NOT use WORD for making it....... it seems easy, but after awhile, it gets complicated. Anybody have any reccomendations for 2D blueprint programs? Or should I keep tearing out my hair using WORD?

:p

[toddk]

If you are referring to the ratio of the length to the width, that was chosen for many different reasons. In the end, it comes down to a design choice.

If you are referring to the ratio between the volume and the product of the length and the width, that will vary according to the design. Most designs before the end of WWI were essentially cylinders which were rounded at the bow and tapered at the stern. After WWI, streamlined designs were produced whose hulls were roughly teardrop shaped. This shape can be produced by a number of curves, one of which was given in this thread.

Dan gave an empirical answer based upon WWI airship designs, while I gave an answer for one particular streamlined design.

Airplanes are no simpler to design and contain many more exotic curves than an airship ever did.

I wasn't even aware that MS Word had the facilities to make drawings. Look amongst the endless CDs of old programs for a CAD program. Unless you are either willing to pay some serious money or try your luck with open source, that is your best bet. But be warned, CAD programs often have a pretty steep learning curve.

[Richard_2001]

the 2d Cad program I use is called CadStd, it is pretty powerful for its type can use either English or Metric dimensioning and best of you can try it out for free.

Http://www.cadstd.com

it works really good at making line drawings and bluprints :cheers:

Richard

[AirCaptain2840]

Would it work to get a rough estimate of the volume if you imagine that the gas cells are cylinders, and then find the volume of each cell?

Ex:

10 feet long
1.4 feet diameter (widest point)
oh, say 10 gas cells (1 foot long each)

Widest point:

Volume= Pi * R^2 * H
V= 3,14 * 1,4^2 * 1
V= 3,14 * 2,8 * 1
V= 8,792 cubic feet

then you do this for each gas cell, then add all of them together. Does this look right?

[joec]

Yes Aircaptain you can do that, it's just a matter of how close you need to estimate the volume.

Your first estimate could be the max diameter * 1/2 the length or 30.8 cubic feet.

Cut it into one foot cylinders and you'll have a better estimate, cut it into 6" cylinders and you'll have a better estimate, etc etc (Calculus is a form of mathematics where you make the sections infinitely thin, you can then calculate the exact volume (no longer an estimate), however you need a mathematical formula to work with.

The formulas for the area of a circle, volume of a sphere, etc are derived from the mathematical formula that describes a circle x^2 + y^2 = r^2 using calculus. Despite what you might hear it a lot of it isn't very complicated, if you can learn algebra, you will be able to learn calculus, geometry, trigonometry pretty much needed too.

If you use excel I can post or email the spread sheet I used for calcalting the volume of my airship, it uses cones, and its only slightly more accurate that using cylinders. However you can split the volume into as many sections as you have the patience to measure and input into the spreadsheet.

Good luck, are you building an airship? Cap

[johannes.eissing]

Hi all,
sorry for rolling up an old thread (01-24-2008), not even heaving read all the postings completely.

There is a simple rule of thumb, relating to the so called prismatic coefficient or block coefficient, which are similar in use. The idea is to relate the actual volume of a ship or an airship to an enveloping prism. In case of a body of revolution, like a blimp envelope or a simple submarine, this prism simplifies to a cylinder. A prismatic coefficient then expresses, how good the body fills a surrounding 'can'.

As an example, any ellipsoid, regardles if being a prolate spheroid, an oblate spheroid or a general ellipsoid, shows a prismatic coefficient of 2/3 or approximately 0,67. Many airships built are showing a prismatic coefficient similar to this number. Those with a more pointed bow and/or stern do not fill an enveloping 'can' well, and the prismatic coefficient is smaller, see the R100/R101. Those who look more bulky, like e.g. the Hindenburg, are showing a slightly higher prismatic coefficient. As a rule of thumb, an airship shows a volume of
V=L*pi*D^2/4*Cp
where V is the actual volume, L is the length, pi is about 3.14, D is the diameter and Cp is the prismatic coefficient, as a fist assumption being 0.67.

To play around with the prismatic coefficient, you might try the "Gertler Series 58 Generator (http://www.airshipworld.info/software/2008/08/generator-version-05-download/)" coded by Andreas Grunwald and my humble self (where I did the math and Andreas made it accessible by coding it in java and putting it to the blog)

A block coefficient is very similar, just the reference volume is not a cylinder ('can') but a box, measuring the same length, width and height as the body. The idea behind this is a dockyard or a lock that fits a ship under consideration. Any ellipsoid shows a block coefficient of pi/6 or approximately 0,52. So a good approximation for an airship would be
V=L*D^2*Cb
wher V is again the actual volume, L the length, D the diameter and Cb the block coefficient, as a first approach 0.52.

Best, Johannes

[WaikatoTKM]

Hi All,

Thanks to Johannes for the useful Gertler Series generator and for the volume calcs, but can anyone tell me how I can use the numbers to figure out the surface area of an airship's envelope?

[rick c faust]

What if the cells were heated from inside, and the pressure also was controled inside??? Tie it altogether with computors and watch how it all inter acts with temp , pressures, and weather conditions.:yeow: or I'm i just dreaming rick

[johannes.eissing]

Hi All,

Thanks to Johannes for the useful Gertler Series generator and for the volume calcs, but can anyone tell me how I can use the numbers to figure out the surface area of an airship's envelope?
Hi WaikatoTKM,
thanks for the flowers.
Yes there are several approaches for surface coefficients. The surface coefficient used by Gertler for the Series58 bodies is defined as
Cs=S/(L*pi*D)
The wetted surface S is related to, say, the amount of paper you would need to wrap it. I dont have my literature handy, but a typical surface coefficient would be 0,78. It is depending on the prismatic coefficient, less on the length to diameter ratio.
The surface area is then computed by

S=Cs*L*pi*D

There is a great NACA report on surface coefficients of ariship shapes, try and google it. The next version of the Series58 generator (comming soon!) will calculate the surface coefficient for you.
Best, Johannes

[WaikatoTKM]

Thanks Johannes. That helps to get some idea of what the spare lifting capacity might be once I've figured out what to cover the envelope with. Cheers.

[johannes.eissing]

...I dont have my literature handy, but ...
Home again.
The report I cited is NACA TN 86 (http://naca.central.cranfield.ac.uk/report.php?NID=257) from Feb. 1922:
"Surface area coefficients for airship envelopes"
52 airship shapes were investigated showing various slenderness ratios and prismatic coefficients. The data can be cooked down to an empirical formula*:

S=2.87*V/D+0.967*D*L

where S is wetted surface area, V is volume, D is diameter and L is length of the body of revolution under consideration.

Ok, here comes some math, just stop reading if you dislike it. Let's check this emperical formula against the Series58 data [1].

Example one, Body 4154:
L = 9.000 ft; Length
D = 2.250 ft; Diameter
V = 23.26 ft^3; Volume
S = 50.18 ft^2; Surface Area
abovementioned formula:
S = 2.87*23.26 /2.250 +0.967*2.250 *9.000 = 49.25
Relative difference is (49.25-50.18)/50.18 = -1.9%

Example two, Body 4175:
L = 9.000 ft; Length
D = 1.800 ft; Diameter
V = 13.74 ft^3; Volume
S = 37.70 ft^2; Surface Area
abovementioned formula:
S = 2.87*13.74 /1.800 +0.967*1.800 *9.000 = 37.57
Relative difference is (37.57-37.70)/37.70 = -0.3%

For those two arbitrary examples it works pretty fine I think.

Best, Johannes

[1] ADA800144 (http://oai.dtic.mil/oai/oai?verb=getRecord&metadataPrefix=html&identifier=ADA800144)"Resistance Experiments on a Systematic Series of Streamlined Bodies of Revolution - For Application to the Design of High-Speed Submarines"

* For those who want to dig deeper:
1.) Approximate the surface coefficient
Cs=S/(L*pi*D)
as a function of prismatic coefficient
Cp=V/(L*pi*D^2/4)
to the linear function
Cs=m*Cp+b
2.) Find the gradient m and the axis intercept b by linear regression.
3.) Insert for Cs: S/(L*pi*D) and for Cp: V/(L*pi*D^2/4)
4.) Solve for S and you gain the abovementioned empirical formula.

[johannes.eissing]

Hi all,

Andreas Grunewald just uploaded the new version of our Series 58 profile generator, see:
http://www.airshipworld.info/software/2009/08/generator-v06-available/
http://www.airshipworld.info/software/wp-content/uploads/2009/08/profilegenerator06-300x170.jpg
It features computation of the centre of buoyancy, surface coefficient,
polynomial coefficients plus several further improvements. You can
download the java jar file here
http://www.airshipworld.info/software/wp-content/uploads/2009/08/airshipworldprofilegenerator-v06.jar
My computer prefered to rename the jar file to a zip file, so I had to
rename it manually, but as default it should work without problems.
Please dont hesitate to post questions.

Best regards and enjoy,

Johannes

[Tim]

I have been calculating the volume of the 4 gas bags in my own model airship project recently. In the middle of the ship I am calculating it like a cylinder, then for the nose and tail I have been breaking it into individual ring segments, where I can calculate the volume of each (each is the shape of the “frustum of a cone”), and then add them up to get the whole.

I've attached a couple of images of the bag broken up into its individual ring segments, as well as whole in another post (http://www.airshipmodeler.com/forums/showpost.php?p=7524&postcount=5) and the project blog (http://modelairship.wordpress.com), to help illustrate how I am doing this. Then I use the following equation on each section (except for the “cone” shape at the end).

V = (pi * h / 12)(d^2 + db + b^2)

More information, as well as more detail in working the calculations, is on my airship project blog at http://modelairship.wordpress.com - Hope that helps.