This is the magic of hydraulics!
While I can't be certain, let's assume a cylinder ID of d=100mm.
Based on the circle area being A=πr² (and r=½d), we get 2500π mm², or about 78.5398163397448309 cm².
The typical hydraulic operating pressure of a large excavator is 200-350 bar.
While technically being defined differently (specifically, one bar being 100,000 Pascal/100,000 N/m²), one bar can be approximated to 1 kg/cm². Now that's wrong on several levels because 10N / 9.80665m/s² ≠ 1kg, and also that's using kilograms as a unit of force, but this is napkin math.
200 kg/cm² × 78.5 cm² = 15,700 kg
350 kg/cm² × 78.5 cm² = 27,475 kg
So, depending on the cross-sectional area of the cylinder and the actual operating pressure, we can expect something between 15-30 tons of force between those jaws (but that may be an optimistic guess, because the cylinder is at an angle to the jaw, diminishing its effective force).
But the rocks in here also look like limestone, which is somewhat soft, and I'd expect it to be weaker under point loading like we see here than under bulk stress like as part of a wall.
So I decided to also go another route and just look up that model.
Going by image search, it looks like an Indeco IFP 28X (which you can also somewhat make out on the green sticker)
Their data sheet states a maximum operating pressure of 350 bar, and a maximum clamping force at the tip of 105 tons.
That's not a little error in guessing cylinder diameter or sth, that's an ORDER OF MAGNITUDE in difference.
@bisexual-engineer-enby any idea where my calculations went wrong so that I ended up with an order of magnitude less force than the manufacturer datasheet says?
Random thoughts i have
- The cylinder applies clamping pressure through a lever arm, so the force the rock experiences is not the same as what the cylinder applies
- Circle area scales by the square of the radius, so a small change in cylinder radius would lead to a comparatively big change in the force output
- Safety margin from the data sheet? You probably don’t want to be at the actual performance limits
So, the cylinder attaches about halfway between pivot and tip, so let's assume a 1:2 disadvantage. That puts needed cylinder force at around 200 tons.
Assuming 350 bar, that means the cylinder would need around 25cm ID, so 27-30cm OD. My eyecrometer isn't perfect but I don't think that's 30 cm diameter.
As for safety margin - that's the datasheet designed to sell that crusher, which means anything significantly less than that would be false advertising.
My guess would be that the cylinder sits at an angle in relation to the crusher jaw, but my gut feeling would be that that DECREASES force and not increase it.
