# Project Log 11: How to calculate McKibben?

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.

It is Wednesday my dudes 08:25, 07/12/2022

Well, I'm having some difficulty finding a way of calculating the force of a McKibben artificial muscle.

I can't find a simplified model, nor a general guessing of what kind of force these muscles would be able to apply.
Like I did before in other logs "this video/image shows a muscle with X length Y diameter applying Z force, so I would need W amount of them to lift V kilograms".

But, since I'm not planning on using a conventional style of McKibben muscles, I'm in a little bit of predicament.

Taking into consideration that I'm planning on going on scrapyards and take as many materials as I possibly can, and that I'm also planning on using non-elastic flat hoses as the inner tubing, I don't think that the current McKibben style of muscle will be useful for me.

For example, there is a type of pneumatic muscle that has 65% of contraction that doesn't use inflatable materials, the mere change of shape from flat to round is enough to actuate with very low pressures, just like the flat hose I'm planning on using.

Here is the article, luckily, the article focuses on the modelling of the strength of such muscle. And as you can imagine, these can be made horizontally and vertically.
And they 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).

I was actually trying to avoid this type of muscle because they expand a lot, and since there will be 10 or even 50 of those in the same place, and maybe there won't be space left for all the muscles to contract.

Let's try the options I can find and see how well this car crash of a calculation this is.

Attempt 1:

I found this paper here that has the following formula:

I have no idea what the heck this means, but I will try to figure out by reading the article instead of jumping paragraph to paragraph like my ADHD ass normally does.

Maybe using the explanatory image the own article provides could be a good idea, right, myself?!

So... I will try to use "Letter/number+1" and "Letter/number+2" for decompressed and compressed. The problem is that I don't know what the measurements will be when compressed.

• P1 = 0.
• L1 = 30cm.
• D1 = 2 inches (5,08 cm).
• 01 (the weird zero) is about the braid angle, I don't know yet, I will just use the angle that normally is used on conventional braided sleeve, something between 18 and 11 degrees.
• These "Sin" and "Con" that you see can be added by the scientific calculator, dunno what they mean, but I was kinda puzzled when I first saw them on equations.

The compressed state I will assume it will be the following numbers:

• P2 = 100 PSI
• L2 = Since these normally compress to around 20%, it will be: 24cm
• D2= I also don't know, I will just throw 20%: 1,96 inches (4,97 cm)
• 02 = 45º angle?
• The triangle ∆L symbol (makes the letters bold on hackaday for some reason) is the difference between the initial length and the end length.

So, we can try and throw the following calculation on google scientific calculator:

F = (π × 4.97²2 × 100 ÷ 4 × sin(45)²2)(3(1 - 0.11 ÷ 24)²2 cos(45)²2 - 1)

(I'm using the pressurised numbers, since the result of the equation is the force applied, if the pressure is 0, the force will be 0)

So... the result is, accordingly to google:

-252.387549267

????????

Okay, let's try again, I added multiplication values where there's none, so I will put the number/value between (), so it is automatic multiplication, I think.

The result of:

((π(4.97))²(100 ÷ 4 (sin(45))²))(3(1 - 0.11 ÷ 24)² cos(45)²2 - 1)

Is

-4412.77987695

?!?!?!?!?!?!?!?!?!?!?!??!?!?!?!?!?!?!?!?!?!?!??!?!?!?!?!?!?!?!?!?!?!?

What if I took the "÷" out and added "/" instead, just like in the original equation?

(π × 4.97²(100/4 × sin(45)²))(3(1 - 0.11/24)² cos(45)²2 - 1)

It gives an error message, lol

If try "((π(4.97))²(100/4 (sin(45))²))(3(1 - 0.11/24)² cos(45)²2 - 1)" it gives the same -4412.

## Okay, this is hard.

I tried an online scientific calculators and the result was

19187.13008

And

6011.092667

For the first and second equations.

So I can assume that this is what? 19000 (or 6011) newtons of force? Lbs? Kg?

Because if it is Newtons, it will be 1956.5 (or 612) kilogram force and 43133 (or 1353,64)lbs respectively.

It is probably giving weird results because of minor mistakes I've made.

Now trying the High Displacement actuator I wrote at the beginning of this mess.

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

This is for contraction rate calculation.

Fully expanded is 2 inch, because it is what is said on the description of the flat hose. I don't know what is the diameter and length when flat...

Whatever, I feel like my brain has turned into a mush and I can't concentrate anymore, I will try to calculate it later.

I would need to have it in hands, but... I don't want to buy 50 meter of this stuff just to later let it accumulate dust because I changed my mind with a better idea.

## Discussions 