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PSA: Toss a rusty air compressor (infosec.pub)

submitted 1 week ago* (last edited 1 week ago) by Landless2029@lemmy.world to c/dull_mens_club@lemmy.world

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I borrowed some landscaping tools from my brother in law and he offered his old air compressor since he doesn't use it. Broken tube. Easy normal part to replace.

Got a air compressor accessories kit on the way home. Replaced the hose and tip. Plugged it in outside. Left it for 10 minutes...

Never turned off. Hmm. Might be one of those that have a release valve and not auto shutoff.

20 PSI? That's odd. Unplugged it. Hissing...
Air coming out the bottom. Maybe a puncture?

Found the spot. Got 80 grit sandpaper and 3300 PSI rated epoxy. Sanded and found rust.

Patched it and my spidey sense went off while looking at the epoxy...

Did a search:

A leaking air compressor tank—especially with a leak on the bottom where corrosion is likely—can be extremely dangerous, even if "repaired" with epoxy or other sealants. The primary risk is catastrophic rupture under pressure. If the integrity of the tank is compromised (for example, by internal rust or a patched-up hole), the tank can explode with explosive force, launching shrapnel and causing severe injury or death, as well as property damage

When tanks rupture, the velocity of air and shrapnel can be lethal. For example, a 60-gallon tank at 150 PSI can explode violently, creating 680 mph air blasts and extremely loud noise, both of which are highly hazardous for bystanders

Nope! Tossing it!
Dodged a nearly literal bullet there.

Bonus shots:

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[–] Ashelyn@lemmy.blahaj.zone 4 points 1 week ago (1 children)

Allow me to make a hackneyed code block diagram

 • <- rotational center
 |
 |
 +👈<<<  <- pounds (force applied at radius)
 ⇡
 feet (distance from center)

Because the final unit is a combined feet×lbs, it's subject to the commutative property of multiplication. For example, 89 foot-pounds of torque is equivalent to 89 pounds of force at 1 foot away from the rotational center, or 1 pound of force 89ft away from the rotational center.

I typically imagine it by putting a weight of x pounds at the end of a 1ft wrench held perfectly level, idk.

[–] ouRKaoS@lemmy.today 2 points 1 week ago

I understand how it works, but I don't get it like with other measurements.

Like how I can look at something and say "that's about 4 feet" or pick something up and say "that's about 2 pounds".

I can't twist the lid on a jar and say "damn! That took 10 foot pounds of torque!" I don't even know if that's a lot or normal; I can't estimate any kind of frame of reference.