FulanoDetailWednesday, 16:41, 01/03/2023
Edit⁴:
Ok, this is starting to get annoying.
I actually asked around, because I don't trust ChatGPT that much, and yet again, the ChatGPT was incorrect.
A couple of engineers actually gave a look and showed me an equation, and basically, I would need 30,000 watts of power (around 40hp) to make a piston pump to work at 3000 rpm and output 1179 Liters per minute.
So, If I calculated it again correctly, I would only produce 117 liters per minute at 3000 watts of power if I tried to "tune it down".
Which kinda puts into perspective the real power necessary for making the goddamn overpriced hydraulic pumps to work at full fluid flow and 4-8 bars of pressure.
Sorry again guys, but it seems I hit another brick wall.
So...
I know that on Project Log 30 I said "I was going to take a while for a new Project Log because I was sourcing materials and components to be bought" and all that stuff...
But I started to make so many post-editions on Project Log 29 that I decided to actually make an update.
But after this Project Log it is for real, ok? I will take a while to post new stuff.
So... I was trying to calculate something here and I don't know if it could allow for the 5 ton McKIbben muscles, but it seems promising.
Basically, I was trying to calculate a piston pump again, just like in Project Log 28, simply because I felt like I didn't do a good enough job.
So, I was looking at real combustion piston engines and how many times those pistons move up and down per revolution.
In this website is said:
Internal combustion engines (with rare exceptions) operate on either a 2-stroke or 4-stroke cycle.
In both types, each full rotation of the crankshaft produces two strokes of the piston, one moving inwards towards the head of the cylinder, followed by a second which withdraws the piston back to its initial position.
So, with that said, a piston pump would have a pulling fluid action and a pushing fluid action per revolution.
So, basically only in one stroke the pump is doing useful work.
If you somehow isolated the down part of the piston, you could make twice the work.
Like in this picture.
With that in mind, we could assume that every revolution gives the amount of fluid in the "working area" of the piston.
And since actual combustion engine pistons normally have 50mm stroke lengths and diameters of aroun 90-100mm, I will keep that way, just like I did on Project Log 29 (funny how I was correct with that even though I was working blindly).
Anyway, a 100mm (10cm) diameter piston with 50mm (5cm) stroke gives around 0.393 liters of working area per revolution, more or less double of that if you make the down action to push fluid too.
So, assuming we are at 3000 revolutions per minute, this means that a single stroke hydraulic pump pushes around 1179 liters per minute.
Meaning that in order to achieve 400 liters per minute like to fill up all those McKibben muscles on Project Log 29, I would only need a single piston rotating at 900-1000 rpm.
Ok, this seems interesting.
This means that in order to achieve 34k or 17k liters per minute, I would need... 28 pistons at 3000 RPM.
If I were to make double stroke hydraulic piston pump, I would only need more or less half of it, around 14-18 pistons.
Now I only need to find out how much torque I would need for this thing to work.
Well, good news guys, I think the ChatGPT finally gave an useful question regarding mathematics/physics.
Me:
What should be the torque necessary for a piston hydraulic pump with 100mm diameter and 50mm stroke?
So, I couldn't find anything that answered me that, so I hope I can make enough information available here.
The density and viscosity of the oil can be generic hydraulic oil o.9~0.8 g/ml.
The viscosity is the same as conventional hydraulic mineral oil CST 46.
The oil will be pumped at maximum 1 meter of distance vertically.
The pressure is between 4-8 bars.
I intend on using pistons and cylinders from broken cars (or out of PVC), that uses metal, with a cylinder diameter of 100mm and a stroke of 50mm.
The connecting rod distance is kinda hard to find, but let's say it is 150mm.
The rotation speed will be 3000 rpm.
The diameter of the input and output holes/hoses is 12mm, but if this represents too much resistance, I can change it.
I read that a piston engine goes up and down every 1 full revolution, and since the working area of this piston pump would fit 0.393 liters of volume, just by this information alone, it would output 1179 liters per minute at 3000 rpm.
But, as you can guess, I don't know if this is completely correct or if the torque necessary to make this rotate this fast would even be enough.
So, what should be the torque necessary for this pump to achieve 3000 rpm with 4~8 bar of pressure?
ChatGPT:
Calculating the required torque for a hydraulic pump involves several factors, including the pump's displacement, pressure, and rotational speed. Based on the information provided, we can estimate the torque required for your piston hydraulic pump as follows:
Calculate the pump's displacement:
The working volume of the piston pump can be calculated as:
V = A * S
where A is the area of the piston (pi/4 * D^2), and S is the stroke length (50mm). Substituting the given values:
V = pi/4 * (100mm)^2 * 50mm = 0.393L
Calculate the flow rate:
The flow rate of the pump can be estimated as:
Q = V * n
where n is the rotational speed (3000 rpm / 60 sec/min = 50 revolutions per second). Substituting the values:
Q = 0.393L * 50 rev/sec = 19.65 L/sec
Calculate the pressure drop:
The pressure drop across the pump depends on the pump's efficiency and the resistance in the system. Let's assume an efficiency of 90% and a pressure drop of 4 bar.
Calculate the required torque:
The torque required to drive the pump can be estimated as:
T = (P * Q) / (eta * n)
where P is the pressure drop (4 bar), eta is the efficiency (0.9), and n is the rotational speed (50 rev/sec). Substituting the values:
T = (4 bar * 19.65 L/sec) / (0.9 * 50 rev/sec) = 1.74 Nm
Therefore, the estimated torque required to drive your piston hydraulic pump at 3000 rpm with a pressure drop of 4 bar and an efficiency of 90% is approximately 1.74 Nm. Please note that this is only an estimate, and actual torque requirements may vary depending on several other factors.
This time I actually made the equations on a calculator to check if these were correct, because there was a time that ChatGPT couldn't even make 2x2 equations for some reason.
Kinda messed up how a chat AI could code a calculator, but couldn't calculate. õ-o
Anyway, using the horsepower equation to find out how much horsepower I would need:
By the way, the torque is in lb/foot, not kgfcm or newton meter.
So, 3000x1,28335814/5252 = 0.733068244478294
Let's just say it is 0.7-0.8 horsepower.
This mean that in order to power that crap of 5 ton hydraulic muscles using 28 pistons at 3000 RPM, I would need...
20.5 horsepower.
I mean... Not bad, not good.
But much more feaseable than 400hp, no?
Edit¹:
I just found out an hydraulic pump that can output f*cking 580 liters per minute (and costs 3000 reais [577,38 dollars]), and a lot of the designs I thought are suddenly viable now.
Anyway, I think the mech is possible now.
If I used aramid braided sleeves I could use 3-4 MPa (30-40 bars) of pressure, so every muscle would apply 1 ton of force (these can reach even 1.6 ton).
So, in total, I would need 230 bundle of muscles (since the legs need to be 3 times stronger than the arms), and actual total of 690 muscles (nice).
Edit²:
I forgor how I got to the 230 bundles in total, so I guess from now on I need to write things down.
Basically, 6 muscles on the thigh to move around (or just 3), +2 for the thighs go up, +2 for the thighs to move down, +2 for the knees going up, +2 for the knees going down, +6 for the feet, so 20 in each leg, total 40 bundles.
I decided to add 2 bundle of muslces on the movements of the legs for the sake of redundancy, so it would be 32 in total.
So, since the legs have the triple of muscles, and each bundle of muscle is 3 muscles, one would need 9 McKIbben muscles per bundle of muscle, so 360 muscles in total.
So, 1425.6 liters per minute if actuated in 1/3 of a second, 712.8 LPM if only half is fully actuated.
Now, the top of the body, +6 for the torso, +6 for a neck, +6 for the shoulder, +1 for moving the arms up, +1 to move the arms down, +1 for the biceps, +1 for the triceps, +6 for the wrist, +3 per finger.
31 muscles for each arm alone, +6 (the neck doesn't need to be that strong, so I won't count it), 68 bundle of muscles in total.
So, since each muscle bundle has 3 McKibben muscles, then it is 204 muscles in total, 807.84 LPM if actuated in 1/3 of a second, 400LPM if only half is supplied.
So, 2233.44 LPM in total at 8 bars (because it will lose pressure), so 0.8 hp per pump, so 4 pumps in total, 3.2hp.
If it was half, 1112.8 LPM, 2 pumps at 0.8 bars, so 1.6hp of power.
Of course, *at least* these values, in reality, it would change. A lot.
If it was at 40 bars, one would need 20hp or 10hp in the case for the second.
Not bad. I suppose.
But goddang, why this LPM keeps changing so much? Why I'm so bad at math?
Edit³:
I also tried to look up on piston hydraulic cylinders, basically, I reduced the number to just 40 in total, and that is relevant because hydraulic cylinders can achieve the two types of action (push and pull), unlike the McKibben muscles, that can only do so much.
The hydraulic pistons can also act as a skeleton and dumper, unlike the muscles, that would need
I would need around 14 horsepower for all the muscles, and a single 580 LPM hydraulic pump.
Besides, the cylinders have like, 3cm of inner diameter and 20cm of stroke, carbon fiber tubes are sold for a few bucks, specially on these dimensions.
Also, since these are hydraulic pistons, I can diminish the fluid flow requirements even more doing two things:
- Since the top needs to be 3 times weaker than the bottom, this means that I can put a limiter on the top part of the exoskeleton to 50bars of pressure, so the pump would only work at full throttle ont he bottom half.
- One interesting thing about hydraulic/pneumatic cylinders is: if you pressurize them and then connect both ends (in a double action cylinder), they will automatically move to the middle, because the parts of the piston have different areas of contact.
So if I play my cards right, I can cut the fluid flow even more.
And since every muscle needs around 0.022 liters to actuate, and actuate at a third of a second, I would need 2732.4 liters per minute (this is also valid to the 10mm nylon braided sleeve).
Of course, like I said: not all muscles would be contracted, at least half wouldn't be.
So, around 1488.96 liters per minute.
And thus, I would need 2.5 (reallistically, 3, since I can't cut one in half, lol) of these hydraulic pumps at 3hp to output 30 bars, and thus, in total, I would need 9 hp to run it.
In the case of the nylon and smaller (and weaker) nylon braided sleeves, I would need 0.4-0.8 hp for the 4-8 bars of pressure. Meaning I would need 1.2-2.4 horsepower.
Well, Way better than 1000-400 horsepower of energy required.
The most expensive part is the pump, so I guess I would just buy everything in aramid (or basalt).
I would need 207 meters in total, and thus, 4505.355 reais (867,10 dollars) with 217 reais (40 dollars) per 10 meter.
The pumps on the other hand, would cost 9000 reais (1732,13 dollars).
So, just this two will cost 13500 reais (2598,20 dollars).
... But I don't think I will take this path to be honest.
I feel like the McKibben muscle has too many limitations for a 2,55 meter Mech that can lift 1 ton.
Meaning:
The way I was planning on doing the McKibben muscles may not be as useful in the end, simply becase the steel cables would offer too much resistance to the inflation action.
And on top of that, the way I was planning on making the braided sleeve would let square-ish voids the size of a thumb, which could let to the rapture of the inner bladder. :/
However, I think that other designs may be more feaseable now, like using 50mm braided sleeves and so on.
Maybe the pleated artificial msucles?
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