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Total Airship Lift Capacity

November 13, 2010 1 comment
All Four Gas Bags

All Four Gas Bags

I have now calculated the airship’s lifting capacity for all 4 of it’s gas bags in previous posts. There will be one in the nose, two in the middle, and one in the tail. So here is what I have calculated just to recap.

Note: There will be almost no space between the gas bags within the finished airship. I’ve added a couple inches between them in the diagram (right) for illustration only.

Nose

Volume: 2.4469 ft3
Lift Capacity: 69 grams

Middle (Fore)

Volume: 5.4002 ft3
Lift Capacity: 152.2856 grams

Middle (Aft)

Volume: 5.4002 ft3
Lift Capacity: 152.2856 grams

Tail

Volume: 4.3412 ft3
Lift Capacity: 122.42 grams

So it looks like the airship will have a total gas volume of 17.5885 ft3 and a lift capacity of 495.9912 grams. Google says there are around 453.5924 grams in a pound. So the airship’s helium volume should, hopefully, be able to lift around 1.09 pounds. At this point it seems to be more than enough, but we’ll see as I work through the rest of the frame and equipment weight calculations..

Categories: Design, Lift Capacity

Calculating Gas Bag Volume / Tail

November 11, 2010 2 comments

I decided it would be a good idea to calculate the lift capacity for the whole ship before continuing the build work, and have now moved on to the tail gas bag. Once again, there will only be a single gas bag in the tail, but I have drawn it in ring segments (Fig. 1) to make calculating the volume easier. We will calculate the volume of each ring segment (Fig. 2), then add them all together, to get the total volume of the aft (tail) gas bag.

Tail gas bag (In segments)

Fig. 1 — Tail gas bag (In segments)

Since Sketchup will not draw smaller increments than 1/16″, so once again I have again rounded up 1/32″ (radius) in a couple of places. The calculations should still be fairly accurate though.

We will be using the same equation we used to calculate the volume of the nose gas bag to calculate the  volume of the tail segments. There is also an online calculator that can also make this easier, but for now we will work the equations manually to illustrate the process.

V = (pi * h / 12)(d2 + db + b2)

Gas bag segment (6 inches)

Fig. 2

Each segment of the aft (tail) gas bag, with the exception of the tail cone at the end, is a “frustum of a cone” shape (a cone with the top cut off). And each segment is separated by the frame ring where the bevel/angle happens.

We will work the calculations from left (largest) to right (smallest) as pictured above, denoting each as section A, B, C, D, E and F (cone). I am again using only 4 decimal points for everything, but if you notice an error in my math (it happens) don’t be shy and let me know!

Section A

V = (3.1459 * 6 / 12)(21.6252 + 21.625 * 22.25 + 22.252)

V = 1.570795 * 1,443.8594 = 2,268.0071 in3 (or 1.3125 ft3)

Section B

V = (3.1459 * 6 / 12)(19.8752 + 19.875 * 21.625 + 21.6252)

V = 1.570795 * 1,292.4531 = 2,030.1789 in3 (or 1.1749 ft3)

Section C

V = (3.1459 * 6 / 12)(17.02 + 17.0 * 19.875 + 19.8752)

V = 1.570795 * 1,021.8906 = 1,605.1806 in3 (or 0.9289 ft3)

Section D

V = (3.1459 * 6 / 12)(12.8752 + 12.875 * 17.0 + 17.02)

V = 1.570795 * 673.6406 = 1,058.1513 in3 (or 0.6124 ft3)

Section E

V = (3.1459 * 6 / 12)(6.752 + 6.75 * 12.875 + 12.8752)

V = 1.570795 * 298.2344 = 468.4651 in3 (or 0.2711 ft3)

Section F (Tail cone)

V = ((pi * r2) * h) / 3

V = ((3.14159 * 3.3752) * 6.0) / 3 = 71.5693 in3 (or 0.0414 ft3)

Tail gas bag (Whole)

Fig. 3 — Tail gas bag (Whole)

So it looks like the aft (tail) gas bag should have a total volume of approximately 4.3412 ft3. Since a cubic foot of helium can lift around 28.2 grams, this should give the aft gas bag a total lift capacity of around 122.42 grams.

I have not yet figured out what the frame, and tail fins, will weigh though. This is probably enough to lift them though, since a good portion of the tail will be built from 1/16″ balsa sticks. But if not, there is some excess lift capacity in the two middle sections to compensate for it. I will finish calculating the whole frame weight in a future update just to be sure.

Categories: Design, Lift Capacity

Calculating Gas Bag Volume / Nose

November 3, 2010 1 comment

I decided it would be a good idea to calculate the lift capacity for the whole ship before continuing the build work, starting with the forward gas bag in the nose. There will only be one single gas bag in the nose, but I have drawn it in ring segments (Fig. 1) to make calculating the volume easier. We will first calculate the volume of each ring section (Fig. 2), then add them all together, to get the volume of the forward (nose) gas bag.

Nose gas bag (In segments)

Fig. 1 — Nose gas bag (In segments)

Since Sketchup will not draw smaller increments than 1/16″ I have rounded up 1/32″ (radius) in a couple of places. There is enough space in between the rings for the gas bag to expand a little that the calculated volume should still be okay.

Like many, I’m no expert with geometry. And some formulas I found are a little too math nerd for me. But I found a very useful online calculator that makes it easy to calculate the volume and surface area of the frustum of a cone. The volume calculation is based on the following formula:

V = (pi * h / 12)(d2 + db + b2)

Gas bag segment (6 inches)

Fig. 2

Each segment of the forward gas bag, with the exception of the cone at the end, is a “frustum of a cone” shape (a cone with the top cut off). And each segment is separated by the frame ring where the bevel/angle happens.

We will work the calculations from left (largest) to right (smallest) as pictured above, denoting each as section A, B, C and D (cone). I am only using up to 4 decimal points for everything, but if you notice an error in my math (it happens) don’t be shy and let me know!

Section A

V = (3.14159 * 6 / 12)(20.752 + 20.75 * 22.25 + 22.252)

V = 1.570795 * 1,387.3125 = 2,179.1835 in3 (or 1.2611 ft3)

Section B

V = (3.14159 * 6 / 12)(18.02 + 18.0 * 20.75 + 20.752)

V = 1.570795 * 793.3125 = 1,246.1313 in3 (or 0.7211 ft3)

Section C

V = (3.14159 * 6 / 12)(6.8752 + 6.875 * 18.0 + 18.02)

V = 1.570795 * 495.0156 = 777.568 in3 (or 0.445 ft3)

Section D (Nose cone)

V = ((pi * r2) * h) / 3

V = ((3.14159 * 3.43752) * 2.75) / 3 = 34.0288 in3 (or 0.0197 ft3)

Nose gas bag (Whole)

Fig. 3 — Nose gas bag (Whole)

So from what I gather so far it looks like the forward (nose) gas bag should have a total volume of approximately 2.4469 ft3. Since a cubic foot of helium can lift around 28.2 grams, this should give the forward gas bag a total lift capacity of around 69 grams.

This may not be enough to lift the frame of the nose (I have not calculated that yet), but there will be some excess lifting capacity left over in the middle sections that can compensate for this, if needed. I will most likely also build most of the nose section from smaller 1/16″ balsa sticks too, so that should help keep the frame weight down.

Categories: Design, Lift Capacity

Gas Bag Volume/Lift and Weight Calculations

September 18, 2010 2 comments
Small Rigid Airship Gas Bag

Gas bag dimensions, area and volume

I was initially trying to decide on a diameter for my airship. At first I was working some math on both a 3′ diameter and a 2′ diameter. Then I remembered that I live in an apartment, and I will need to be able to fit this thing through a normal single door.. So 2 foot diameter it is!

To start I created a 2′ diameter ring, with 16 sides, using Sketchup. The program makes it easy to then draw a circle (arc) from the center out until it meets up with the inside of the rings, and then I can get the area of the circle too with a right-click menu choice. But this ended up wrong since it’s not measuring a circle, but a polygon.. So I used 1′ 10 1/4″ for the diameter (rounded up a bit).

Most folks that graduated high school learned how to calculate the volume of a cylinder. But if high school was a couple of decades ago there are online tools to do it for us! Basically, the area of the base x the height of the cylinder = volume. So..

Area = πr2

3.14159 x  11.1252 (11 1/8″ radius) = 388.8208 in3

..and we know that there are 144 (12 x 12) inches in a square foot, so..

388.8208 / 144 = 2.7001 in2

Model Airship Frame

2' test section frame dimensions

So we know that each 1 foot length of this cylinder has a volume of 2.7001 ft3, and our test section is 2 feet long, so the gas bag should have a volume of 5.4002 ft3.

Anyway.. Now we need to find out if this gas volume can lift the frame as we’ve designed it. So we will add up the weight of the construction materials (minus glue) to see if this design might actually work.

I may post some data on the calculations with carbon fiber rods later, but at this point I’m pretty sure I’ll be building the test section with balsa wood sticks. So for now we’ll focus on the weight of a balsa wood frame.

The outer part of the rings, and the longitudinal stringers, will use 1/8″ balsa sticks. And the inner bracing of the rings will use 1/16″ balsa sticks. I weighed a 36″ section of the 1/8 balsa and it weighed 2 grams. So a 36″ section of the 1/16″ should weigh 0.5 grams.

This 2′ test section of the frame will use 63′ 3″ of the 1/8″ balsa and 124′ 6″ of the 1/16″ balsa (probably a bit less, since I’m not accounting for bevel cuts). The weight of the 1/8″ sticks should be about 42.1667 grams, and the weight of the 1/16″ sticks should be about 10.2083 grams. So the total weight of the balsa wood frame should be around 52.395 grams.

We know that 1 cubic foot of helium can lift about 28.2 grams. So the 5.4002 ft3 our gas bag holds should be able to lift 152.2856 grams (about 5.4 ounces). So we’re doing well so far since the frame only weighs 52.396 grams! Stay tuned and we’ll factor in the weight of the exterior covering and the gas bag materials next.

Categories: Design, Frame, Lift Capacity
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