# Project Log 12: Calculating more (incorrectly).

A project log for DIY Human-sized Mech

Mechs are not viable, nor cheap, so I will try to design and build one alone anyway.

FulanoDetail 12/07/2022 at 19:563 Comments
It is Wednesday my dudes² 16:53 07/12/2022

So, let's try again, and this time I wll make a list of what I want to calculate first, so I don't just give up on writting this damn thing:

1. Calculate the force generated by the High displacement artificial muscle.
2. Calculate the amount of fluid flow all these muscles will need.
3. Scratch the surface of power supply and liquid used, I feel like powering up 72 groups of muscles will require an absurd amount of liquid flow.

Edit: I think I didn't calculate anything correctly, so the values can be waaaay wrong or simply aren't meant to give the strength. I will only know if I actually make one of these personally.

So, I left the last Project Log giving up on calculating the High Displacement muscles, here is their article, and here is the image:

These need a smaller amount of pressure to work, producing 120 and 300 N at pressures of 35 and 105 kPa (5 and 15 PSI or 0,3 and 1 bar respectively).

So, I got the idea of using this online Circle Length Calculator to calculate the approximate length of the flat hose when deflated.

So, assuming the 2 inch description of the flat hose means the hose reaches 2 inches of diameter when round, the calculator says it has 15.9 cm of length.

Since the hose is folded on itself, I will round up its deflated length to ±7 cm. Like I said, I don't really need the hose anymore, as shown in the illustration of the article itself, I just need to make pouches.

• CR (%) = Δk/k0 x 100
"where k0 is the initial length of the actuator and Δk is the length it shortens during contraction."

Well, I can't tell how much it shortens because I don't have a prototype to test it out, so I will assume that there is 30% of contraction (the paper says it achieves 30-65%) and the full length is 30 cm and 20 cm of width (in the "Y" direction/axis).

But I will try to calculate the contraction rate of a single hose so I can have an idea of how many of these I will need on each muscle.

• CR% = 1.92/7 x 100
• Therefore: 27.4%

Oh, interesting, 27% for a single hose on the horizontal? I thought it would be 1% or less. Bruh

From the article:

"Outer Actuator Modeling
We first model the outer actuator, with the diagram and variables in Figure 6B. The inner actuator pushes on the middle portion of the outer actuator over a rectangular cross section, with a total outward force (F.out), where:

F.out = PhL

and P is the pressure inside the inner actuator, h is the width of the inner and outer actuators, and L is the height of the rectangle where the inner actuator connects to the outer actuator. If the actuator is made from a textile or other material that can bend, the simple actuator model in Figure 6B is not perfectly accurate at its ends. In this case, the region where the inner actuator connects to the outer actuator is not flat, but is a curved surface (Figures 6C,E,F), which results in an effective reduction in the inner actuator's height (L) and width (H) when the actuator is pressurized. Paulsen modeled the shape of a circular pouch (a Mylar balloon) when it is inflated, finding that the radius of an inflated pouch is about 0.7627 times the uninflated radius (Paulsen, 1994).

Let's calculate.

As said before, the pressure inside these are 15 PSI, I don't know if they are saying width in the direction it expands or the "depth" of the actuator, so I will go with the first: 2 inches/5,08cm and since this is a circle, the length will also be 2 inches.

• F.out = 15 x 5,08 x 5,08
• Therefore: 387.096

Is this newtons? They never say that, lol.

So this single hose actuator can lift 39,4kg/86,8lbs?!

Bruh, I feel like I'm mistaking something.

While our pouches are rectangular instead of circular, the center of each side matches the model closely: we performed measurements of sample pouches and found they match Paulsen's model within 1%. Therefore, the true outward force of the inner actuator is based on a reduced cross-sectional area, where the effective height (L) and width (h) of the inner actuator are then Lcorrection = 0.7627l, and hcorrection = 0.7627h. Thus, for inner actuators with flexible ends, the actual F.out highlighted in red (Figure 6F) is:

F.out = PhcorrectionLcorrection

Sooooo... I have to multiply or divide these decimal numbers by the numbers I have? hmm

I will try the following:

• F.out = 15 x (5,08 x 0.7627) x 1 x 0.7627)
• Therefore: 225.178113514

22kg/50,6lbs of force, still huge for a single cutaway of a hose, bruh.

The L is the height in wich the inner actuator connects to the outer layer, so maybe that is why the previous number was so high.

So, let's say it is a height of 6cm, so there is a space of 1cm above and below:

• F.out = 15 x (5.08 x 0.7627) x (6 x 0.7627)
• Therefore: 265.9

It got higher, lol

Well, what if I put it between parents instead of "x"?

• F.out = 15((5.08(0.7627))(6(0.7627)))
• Therefore: 265.9

It is the same result, lol

Let's get this bread, I will try to calculate a single muscle with multiple hoses one in front of other and above of other, so it has more area to work with.

However, the article doesn't that that into consideration (I think), so I will maybe need to multiply the result of a single horizontal line by the number of muscle "cells" on each layer.

Another matter is that the illustration is just a illustration, I will try to fit everything at 30 cm as close as possible.

I will try to get the width of each deflated flat hose to be 2cm of width and 7cm of length, inflated it will be 5,08 x number of hoses in line = 20.32cm.
I will assume a height of 6cm, so 1cm above the "bone" and 1cm below the "shoulder bone".

• F.out = PhL
• 15 x (20.32 x 0.7627) x (6 x 0.7627)
• Therefore: 1063.83360715
• If there is 5 of them: 5319.16803575
• Newton to Kg = 542.4
• Newton to Lb = 1195.7

So I would need 2 of those in order to lift 1 ton, which means I would need 10 of those in an arm to apply a force of 5 ton.

But wait, this doesn't take into consideration the "depth" of the pouches, nor that you're not limited by 15 PSI.

If I multiply the final result by the length on something around 2 or 20cm, I can achieve 1 ton easily. Not to mention that if you multiply by, let's say, 35 or 100 PSI, you will get:

• 35 x (20.32 x 0.7627) x (6 x 0.7627) = 2482.27841669 = 253kg = 558lbs
• 100 x (20.32 x 0.7627) x (6 x 0.7627) = 7092.22404768 = 723kg = 1594lbs

Almost at 1 ton, I think it can do it.

## How much hydraulic oil this would need?

Well... Sigh... I will try to figure out how much this would need.

I will try to find a calculator for that.

And guess what? There is one, lol.

I will insert the diameter to be 5.08cm, the length to be 20cm and the density to be 0.9 g/ml accordingly to google.

The result is 0.405 liters and 0.36kg (0,79lb) of mass.

So, since I need to feed 20 of those for each ±30cm muscle, so I would need to feed 8.1 liters of oil in a single muscle, which would weight 7.02 kg (15,4 lbs) when fully inflated.

And since I will have 72 groups of thos with at least 5 of them in each, I would have to flow and weight:

• (8.1 x 5) x 72 = 2916 liters of volume.
• (7.02 x 5) x 72 = 2527.2 kg (5571 lbs) of mass.
• This doesn't take pressure into consideration.

I'm completly fine.

Well, I can make other numbers into consideration.

For example, I could simply reduce the number of muscles per group.

like a single one for each 72 muscles.

• 8.1 x 72 = 583.2 liters
• 7.02 x 72 = 505.44 kg (1113,33 lbs)

Concerning, but not that extreme as the previous one.

Recalculating. Again.

Ok, I believe I will need to recalculate again. Because the paper says a lot about width, length and height, which got me confused. So I will take the following image into consideration:

• F.out = PhL (of single cell, then it will multiply it by 5 cells and then more 5 cells, so a 5x5 grid)
• 15 x (20 x 0.7627) x (6 x 0.7627) = 1047 Newton = 106.7kg = 235.3lbs
• 100 x (20 x 0.7627) x (6 x 0.7627) = 6980 Newton = 711 kg = 1569 lbs
• Since there is 5 of them in line horizontally at 15 PSI = 5235 Newton = 533kg = 1176lbs
• At 100 PSI = 34900 Newton = 3558 kg = 7845 lbs
• And since there is 5 lines of them one above the other (15 PSI) = 26175 Newton = 2669kg = 5884lbs.
• At 100 PSI = 174500 Newton = 17794kg = 39229lbs.
• Now calculating the volume: ((0.405 x 5) x 5) x 72 = 729 liters
• Mass: ((0.36 x 5) x 5) x 72 = 648kg = 1428,6 lbs
• Okay, this is way better.

Also, I just noticed that if it was bars on in Mpa instead of PSI, it would be a completly different result, I wonder if this mess of equations I've made are even remotely correct...

Edit²:
So, I just noticed that normally they refer to pressure as "Kpa", so the value of "100 PSI" that I've been inputing is probably 100 Kpa, which is just 14,5 PSI, so If I was to actually apply 100 PSI, I would be reaching values of 689,4 Kpa. Which would reach 1 203 095 Newtons = 122 681 kg = 270 466 lbs.

Of course, assuming nothing gets rips apart by the sheer forces involved.

Edit³:

I forgot to calculate the volume and mass of the muscles if these were conventional McKibben muscles.

One tube with 2 inches (5,08cm) diameter and 30 cm length would have 0.608 liters of hydraulic oil and 0.55 kg of mass in a single McKibben hydraulic muscle.

Since there is ten of those for each one of 72 groups of muscles, we would have 437.76 liters of oil and 396 kg of mass.

Well... This seems way better than the high displacement artificial muscle, and the best part is that I don't need to actively and completely pump the oil out of the muscles like the High Displacement one.

But just now I remembered that I would need to use 5 tons for each ton I need to lift at the tip of the limps.
If it was 50 muscles for each one of the 72 groups, it would be 2188 liters and 1980 kg of mass.

Well... sh*t.

## How to calculate the amount of flow?

Seriously, I don't know.

I will try by taking these 729 liters for a third of a second, the time/speed I want these things to contract. So 2187 liters per second.

So, since the flow rate is calculated as liter per minute, I would need 131 220 liters per minute.

Obviously, not all muscles would be contracted at same time, but if they were, this would be the maximum value they would need.

So, accordingly to this flow rate calculator, Inside a 25mm diameter hose, with maximum length of 4 meters, I would need 5-6 bars/87 PSI/0,6 Mpa of pressure in order to achieve 131 220 ltrs/min.

Which is not that much actually (I hope), not to mention that I hope to have a fluid pump for every limb (torso+arm1+arm2+leg1+leg2), so I would need 5 to 6 pumps with 12v that can achieve 60 PSI.

But oil pump extractors with 12 volt can achieve 0.8 Mpa way above what each limp will need. I found a random one that uses 45 watts, so 5 or even 6 of these would suck 225-270 watts (maybe per hour).

I could easily power that with 4 or 5 small battery packs used for electric car toys for each pump  would be able to supply them for a few hours, accordingly to this google result, batteries normally lasts to 50-100 hours. lol

Again, some things may not work properly in real life, so...

This doesn't take the progressively magnetic active valves energy consumption into consideration (the ones I talked about on Project Log 4.

Edit⁴:

I forgot to take into consideration that hydraulic oil has a different viscosity than water, so the oil pumps would need to work in a different manner.

Accordingly to this flow rate calculator that I think it is more precise, if I started with a flow of 0,8 Mpa (produced by the oil pump I talked about), with a pressure of 0,1Mpa at the end of a 4 meter long 25mm diameter wide tube, I would reach 1438,1 l/min.

Which is 91 times less of what I need.

So 72 Mpa/10 442 PSI at the start.

I would need a really heavy duty hydraulic pump to fill a hydraulic accumulator with 700 liters to actuate every muscle once.

... Or not, I could let the muscles partially inflated and use the pressurized flow to draw more fluid to the hoses.

But I don't know how to calculate all of that, so the High Displacement muscle is back to the mind archive.

If it was the McKibben muscles, thus needing to fill up 1313 l/second, this setup would only provide 13 liters per second.

HOWEVER, this would mean that the muscles are completely empty (which isn't the case) and would need to be filled instantaneously, but they wouldn't.
In fact, I would only need them to be filled around 3 times its initial volume to fully actuate them.

So, let's see.

There are 5 limbs, I just need to actuate 3 of them, but let's actuate just one:

437 liters in total divided by 5 = 87.4 liters per limb (more or less).

So in order to provide them with 3 times its initial volume, I would need 262.2 liters/second...

In order to achieve such volumes, I would need a 4mpa oil pump (that uses 220V) and a 2 meter hose, for each limp. And each would consume 1300 W of energy per hour, so 6500 watts per hour in total.

# At least hydraulically, I will still be researching for the electromagnetic counter part like I talked about in Project Log 1.

That's it, bois, more wasting time! :D

## Discussions

dekutree64 wrote 12/08/2022 at 15:21 point

*No low pressure hydraulic mechs :)

5MPa as in that video I posted before means 1/7 the fluid mass and flow rate compared to operating at 0.7MPa. Still not great though, and the kevlar braided sleeve is indeed expensive.

It looks like excavators operate around 5000PSI/35MPa, so about 1/50 the fluid mass and flow rate. Much more reasonable.

Are you sure? yes | no

FulanoDetail wrote 12/08/2022 at 15:31 point

Oh, I see. But still, could you help me out with these calculations? I can't make heads of the amount of fluids and pressures I need to operate that.
It is really confusing to me...

Are you sure? yes | no

dekutree64 wrote 12/09/2022 at 03:07 point

I'm not so good at calculations either, but I think you're doing ok. Moving a heavy and powerful mech takes a lot of power, and low pressure would need a lot of fluid volume to do it. Definitely better to use air instead of liquid in that case. That also has the advantage that you can use compressed air as your energy source instead of needing a pump and something to power it. Though it is bulky and still won't have very good run time.

Probably the biggest advantage of hydraulic is that it can run on a combustion engine. Typical lawn mower engines are 10hp/7.5kW, and can run for a long time on a few kg of fuel. I hate working with combustion engines, but if you like them then it's a good option.

Are you sure? yes | no