Tag: envelope

In continuation of our work on envelope fixes, we repeated all the steps from last weekend with the starboard “green” envelope.

Knowing what we are up to, all job went much smoother this time. Inflated envelope seemed to be holding pressure in an acceptable manner.

After making significant progress with our initial milestones, our focus shifted to one last problematic component: the central gondola tube. This particular tube had suffered multiple fractures during previous test runs.

The failures were traced to excessive torsional forces acting on the tube. As the central gondola tube connects directly to the main traverse tubes (each 4 meters long and spaced 2 meters apart through the envelopes), it experienced substantial torsional loading from these points of attachment.

The diagram above illustrates the torsional stress distribution on a cylindrical element. The twisting action creates uneven shear stresses, which the initial central tube design wasn’t adequately prepared to withstand. In hindsight, we’re fortunate that the damage was confined to this tube, preventing a more systemic failure that could have propagated through the entire structure.

After discussing this with Serge, we decided on an alternative solution, which is depicted below.

The new design involves a split aluminum tube at the center, with an additional overlapping section designed to transfer all loads except torsional forces. The aluminum tube is shielded by a rubber hose, which is clamped at both ends using hose clamps. This rubber overlay allows for some controlled torsion flexibility, effectively acting as a damper to reduce stress concentrations. The intent behind this assembly is to isolate and minimize torsional forces on the main central tube, while still allowing the structure to behave elastically and return to its original state post-deformation.

This redesign should improve the resilience of the central tube, reducing the risk of fracture during future test flights and ensuring the gondola’s stability under dynamic conditions.

Obviously replacing super-light CF tube with Aluminum one came with a cost. Weighing revealed that gondola structure is now ~700g heavier. With the excess of 10kg we measured last time it shouldn’t be a big deal and I hope that this investment will pay off.

And that’s it for last weekend! Stay tuned as we are planning all motors test next week!

As nearing to our first motorised flight, there are/were still few things to take care of. Probably the one most pressing was the way how the envelope upper wedge was clamped, where using the Ametalin tape shown to be unreliable (tape was ripping off as envelope inflated).

Anyway, so designed new clamp in OpenSCAD which goes through the envelope and provides means to strap both sides together while distributing the load in some convenient way (Eye foundation is 30 degrees bent). Code is pretty simple:

use <ShortCuts.scad>

$fn=100;

module joint(angleIn) {
rotate_extrude(angle = angleIn)
translate([3,0,0])
circle(3);
}

module clamp() {
Ty(13)
cylinder(10, 3, 3);
T(-3,13,0)
R(90,30,0)
joint(30);
Ry(30)
T(0.4,13,-4.5)
cylinder(3, 2, 3);

Ty(-13)
cylinder(10, 3, 3);
T(-3,-13,0)
R(90,30,0)
joint(30);
Ry(30)
T(0.4,-13,-4.5)
cylinder(3, 2, 3);

T(0, 10, 13)
Rx(90)
cylinder(20, 3, 3);

T(0,10,10)
Ry(-90)
joint(90);

T(0,-10,10)
R(0,-90,180)
joint(90);
}

module pads(upper = true) {
D() {
if (upper) {
minkowski() {
cube([1.5,40,0.01], center = true);
cylinder(1.5,10,10);
}
}else{
Tz(-1.5)
minkowski() {
cube([1.5,40,0.01], center = true);
cylinder(1.5,10,10);
}}
T(0.4,-13,-1.55)
cylinder(3.1, 2.05, 3.05);

T(0.4,13,-1.55)
cylinder(3.1, 2.05, 3.05);
}}

module pegs() {
Tz(-0.1)
cylinder(1.1,1,1);
Tz(1)
sphere(1);

T(0,22,-.1)
cylinder(1.1,1,1);
T(0,22,1)
sphere(1);

T(0,-22,-0.1)
cylinder(1.1,1,1);
T(0,-22,1)
sphere(1);
}

clamp();

*pads(false);
*pegs();

D() {
pads(true);
pegs();
}

Resulting rendering:

Now, printed all parts many times (each side of wedge has those distributed every 50 cm + multiplied twice for each envelope). Then lots of gluing and it worked out very well – see our weekend inflation test which demonstrates that that idea works pretty well.

Another part installed was envelope gas relieve vent as deflating envelope through the maintenance opening didn’t work out well.

Installed valves actually brought a nice opportunity to see our envelope from inside for a first time showing some nice details.

Finally we did another inflation test which worked out in envelope bursting out in the front and around one meter long rip through it. Luckily it was anticipated that something like this can happen and we were able to fix quickly with a new weld. Inflated envelope then seemed to be keeping its shape for ~10 minutes without any signs of deflation and we took some nice pictures.

Having the port-side envelope done, the next step is to repeat all this for the starboard one, but that’s for another day. 😉

Having a good go over the Hydrogen Summit 2024, I decided to do another complete assembly test back at the base, mainly to get some nice pictures for our next step.

Apparently I’ve picked hottest day in Queensland’s winter (33C) and not just that, it’s been pretty windy as well. Huge thanks to Kristian that he made himself available and came to help! Thanks to him we did some more work on envelope, mainly as those envelope openings / sleeves were not being satisfactory and also cut main tubes to their final lengths.

Then, when inflated, we took it out to our cul-de-sac to take few pictures.

I also did a 360 video myself:

I think it looks awesome! Still few things missing:

• Tape closing the upper wedges is not sticking properly and needs some different idea on how to make it working
• The envelope openings are not sealed properly and I don’t like that system in general. That clearly needs replacing with something clever
• Intakes needs to be redesigned and printed again.

Finally, Seb did an awesome video with his DJI Mini 4 drone. If you’ll watch it long enough you’ll see how we’ve been transporting whole airship back to our carport.

As usual, huge thanks to a whole team, mainly to Kristian who spent practically whole Saturday putting this together and then back in pieces, Seb for his assistance & video, then Oli, Xavier and “a guy in red shorts” who made sure our airship haven’t flown away! 😀

It took a while, but both envelopes are ready to go. Gallery below shows that they fit nicely into 128l box, weighing ~1100g each. Can’t wait to see them both inflated! 🙂

We had a great & busy weekend. Let’s let images to do the talking first.

As you can see, our project is getting from being decent to a massive fast. This is just a left envelope, while second is in the development. It took about an hour to inflate and with prevailing winds it felt like being alive (watch video below).

It was overall very satisfying feeling to see envelope’s shape for a first time. There are few deficiencies (found one insufficient welding in a rear part and slenderness of the front part can be better), but hey, its just prototype. 🙂

There are still envelope beans openings missing, main duct needs to be adjusted on its length and potentially some filler material needs to be put in to set it in properly a wedge’s centre, filling valve and pressure monitoring opening needs to be mounted … and all that once more on the other side (25% envelope welding done by now). The plan is to have it all ready to go in next two weeks, while planning for another test next weekend!

Needing to find where to put holes in an envelope we needed to find its centre of gravity first. Surprisingly there is a term for that already – a centroid! Well I’ve been lazy and explained all that to ChatGPT and it actually came back with a very sensible set of answers which I compiled into following.

Finding the Centroid of a Composite Shape: Half-Ellipses and a Rectangle

In this post, we’ll explore how to find the centroid of a composite shape made up of two half-ellipses and a rectangle. This process involves both analytical and graphical methods to ensure accuracy and clarity.

Step 1: Centroid of a Half-Ellipse

For a half-ellipse with the semi-major axis aaa and semi-minor axis bbb, the centroid (xˉ,yˉ)(\bar{x}, \bar{y})(xˉ,yˉ​) is given by:

• xˉ=0\bar{x} = 0xˉ=0
• yˉ=4b3π\bar{y} = \frac{4b}{3\pi}yˉ​=3π4b​

Step 2: Centroid Calculation for Given Shapes

Given two half-ellipses:

• First half-ellipse with a=−3a = -3a=−3 and b=1b = 1b=1
• Second half-ellipse with a=1.5a = 1.5a=1.5 and b=1b = 1b=1, centered at (2,0)(2,0)(2,0)

For a horizontally sliced half-ellipse:

• First half-ellipse centroid: (0,43π)(0, \frac{4}{3\pi})(0,3π4​)
• Second half-ellipse centroid: (2,4×13π)(2, \frac{4 \times 1}{3\pi})(2,3π4×1​)

Step 3: Centroid of a Rectangle

For a rectangle with coordinates starting at (0,−1)(0, -1)(0,−1) and ending at (2,1)(2, 1)(2,1):

• Centroid: (xˉ3,yˉ3)=(0+22,−1+12)=(1,0)(\bar{x}_3, \bar{y}_3) = \left(\frac{0 + 2}{2}, \frac{-1 + 1}{2}\right) = (1, 0)(xˉ3​,yˉ​3​)=(20+2​,2−1+1​)=(1,0)

Step 4: Composite Shape and Its Centroid

To find the centroid of the composite shape consisting of the two half-ellipses and the rectangle:

1. Calculate the areas (weights) of each shape:
   • Half-Ellipse 1: A1=12π×3×1=3π2A_1 = \frac{1}{2} \pi \times 3 \times 1 = \frac{3\pi}{2}A1​=21​π×3×1=23π​
   • Half-Ellipse 2: A2=12π×1.5×1=1.5π2=0.75πA_2 = \frac{1}{2} \pi \times 1.5 \times 1 = \frac{1.5\pi}{2} = 0.75\piA2​=21​π×1.5×1=21.5π​=0.75π
   • Rectangle: A3=2×2=4A_3 = 2 \times 2 = 4A3​=2×2=4
2. Find the centroids of each shape:
   • Half-Ellipse 1: (xˉ1,yˉ1)=(−4π,0)(\bar{x}_1, \bar{y}_1) = \left(\frac{-4}{\pi}, 0\right)(xˉ1​,yˉ​1​)=(π−4​,0)
   • Half-Ellipse 2: (xˉ2,yˉ2)=(2+2π,0)(\bar{x}_2, \bar{y}_2) = \left(2 + \frac{2}{\pi}, 0\right)(xˉ2​,yˉ​2​)=(2+π2​,0)
   • Rectangle: (xˉ3,yˉ3)=(1,0)(\bar{x}_3, \bar{y}_3) = (1, 0)(xˉ3​,yˉ​3​)=(1,0)
3. Compute the weighted average of the centroids:
   • Total area: A1+A2+A3=3π2+0.75π+4=9π4+4A_1 + A_2 + A_3 = \frac{3\pi}{2} + 0.75\pi + 4 = \frac{9\pi}{4} + 4A1​+A2​+A3​=23π​+0.75π+4=49π​+4
   • Centroid of the composite shape: xˉ=A1⋅xˉ1+A2⋅xˉ2+A3⋅xˉ3A1+A2+A3=3π2⋅−4π+0.75π⋅(2+2π)+4⋅19π4+4≈0.3806\bar{x} = \frac{A_1 \cdot \bar{x}_1 + A_2 \cdot \bar{x}_2 + A_3 \cdot \bar{x}_3}{A_1 + A_2 + A_3} = \frac{\frac{3\pi}{2} \cdot \frac{-4}{\pi} + 0.75\pi \cdot \left(2 + \frac{2}{\pi}\right) + 4 \cdot 1}{\frac{9\pi}{4} + 4} \approx 0.3806xˉ=A1​+A2​+A3​A1​⋅xˉ1​+A2​⋅xˉ2​+A3​⋅xˉ3​​=49π​+423π​⋅π−4​+0.75π⋅(2+π2​)+4⋅1​≈0.3806
   • yˉ=0\bar{y} = 0yˉ​=0

Visualization

Below is the graphical representation of the composite shape along with the calculated centroid and cut points.

import matplotlib.pyplot as plt
import numpy as np

# Weights and centroids
W1 = 12.57
W2 = 6.0
d = 6 / np.pi + 2

# Coordinates of centroids
x1, y1 = 0, 0
x2, y2 = d, 0

# Centroid of composite shape
bar_x = 0.3806
bar_y = 0

# Points A and B
point_A = (-0.6194, 0)
point_B = (1.3806, 0)

# Define rectangle
rect_x = [0, 2, 2, 0, 0]
rect_y = [-1, -1, 1, 1, -1]

# Plot the shapes and points
plt.figure(figsize=(10, 6))

# Half-ellipses
theta = np.linspace(-np.pi/2, np.pi/2, 100)
x1 = 3 * np.cos(theta)
y1 = 1 * np.sin(theta)
x2 = 1.5 * np.cos(theta)
y2 = 1 * np.sin(theta)
cx1, cy1 = 0, 0
cx2, cy2 = 2, 0
plt.plot(cx1 - x1, cy1 + y1, label='Half-Ellipse 1 (a=-3, b=1)', color='blue')
plt.plot(cx2 + x2, cy2 + y2, label='Half-Ellipse 2 (a=1.5, b=1)', color='green')

# Rectangle
plt.plot(rect_x, rect_y, label='Rectangle', color='red')

# Centroid
plt.scatter(bar_x, bar_y, color='purple')
plt.text(bar_x, bar_y, 'Centroid', fontsize=12, ha='right', color='purple')

# Points A and B
plt.scatter(point_A[0], point_A[1], color='orange')
plt.text(point_A[0], point_A[1], 'Point A', fontsize=12, ha='right', color='orange')
plt.scatter(point_B[0], point_B[1], color='orange')
plt.text(point_B[0], point_B[1], 'Point B', fontsize=12, ha='right', color='orange')

# Formatting
plt.xlim(-4, 4)
plt.ylim(-1.5, 2)
plt.axhline(0, color='black', linewidth=0.5)
plt.axvline(0, color='black', linewidth=0.5)
plt.grid(color='gray', linestyle='--', linewidth=0.5)
plt.xlabel('X-axis')
plt.ylabel('Y-axis')
plt.title('Graph of Two Half-Ellipses with a Rectangle, Centroid, and Cut Points')
plt.legend()
plt.show()

This Python code generates the visual representation of the two half-ellipses, the rectangle, the calculated centroid, and the cut points. It ensures the correct understanding of the composite shape and its centroid.

By following these steps and using both analytical and graphical methods, we can accurately determine the centroid of a composite shape consisting of multiple geometric figures. This approach is essential in various engineering and design applications where precision is crucial.

Long story short, Point A (rear hole) can be found (−0.6194, 0) and Point B at (1.3806,
0). Now just to cut those it should be working! (Well, apparently when rotating these 2D shapes around the Y-axis to form 3D solids, the centroid positions might change due to the added dimension. So we can be off a little, but shouldn’t be too far and we can still compensate with ballast.)