93EXCivic
93EXCivic MegaDork
10/23/12 8:46 a.m.

I need a 4 pound cylindrical weight with a 1" diameter hemispherical striking surface. I have no idea where to even begin looking for this....

DrBoost
DrBoost PowerDork
10/23/12 8:58 a.m.

Uh, can you tell us what it's for? That might help me picture what you want/need. 4 pounds huh? How much to the weights that are used in older double-hung windows to balance the lower sash?

I think they are around 4 pounds.

93EXCivic
93EXCivic MegaDork
10/23/12 9:01 a.m.

It is for impact testing. I am just using a description from MSHA.

GameboyRMH
GameboyRMH PowerDork
10/23/12 9:02 a.m.

Sex shop?

stumpmj
stumpmj Dork
10/23/12 10:32 a.m.

Hit up a machine shop. They can turn the end of a 1" steel bar to form the hemisphere you need. Just make it the right length to hit 4 lbs.

mthomson22
mthomson22 Reader
10/23/12 10:44 a.m.

Pi x .5 squared x Length (x.283) = 4lbs

.283 is the weight multiplier for mild steel

You need 17.997 inches of 1 inch round mild steel.

If my math is correct

Conquest351
Conquest351 SuperDork
10/23/12 10:47 a.m.

This forum rocks!!

DrBoost
DrBoost PowerDork
10/23/12 10:49 a.m.
mthomson22 wrote: Pi x .5 squared x Length (x.283) = 4lbs .283 is the weight multiplier for mild steel You need 17.997 inches of 1 inch round mild steel. If my math is correct

There are many childish things I could say about this post. But for once, I'll be grown up.

yamaha
yamaha Dork
10/23/12 11:13 a.m.

In reply to DrBoost:

I've hit one of those window counterweights with a mower before......steel beat cast iron.

93EXCivic
93EXCivic MegaDork
10/23/12 11:55 a.m.
mthomson22 wrote: Pi x .5 squared x Length (x.283) = 4lbs .283 is the weight multiplier for mild steel You need 17.997 inches of 1 inch round mild steel. If my math is correct

I am thinking that is what I am going to end up doing.

pilotbraden
pilotbraden Dork
10/23/12 12:03 p.m.
DrBoost wrote: Uh, can you tell us what it's for? That might help me picture what you want/need. 4 pounds huh? How much to the weights that are used in older double-hung windows to balance the lower sash? I think they are around 4 pounds.

Off topic, those are some of the finest anchors for small boats in rocky streams.

DrBoost
DrBoost PowerDork
10/23/12 12:05 p.m.
pilotbraden wrote:
DrBoost wrote: Uh, can you tell us what it's for? That might help me picture what you want/need. 4 pounds huh? How much to the weights that are used in older double-hung windows to balance the lower sash? I think they are around 4 pounds.
Off topic, those are some of the finest anchors for small boats in rocky streams.

Or nunchucks for Chuck Norris.

JohnInKansas
JohnInKansas HalfDork
10/23/12 12:18 p.m.
93EXCivic wrote:
mthomson22 wrote: Pi x .5 squared x Length (x.283) = 4lbs .283 is the weight multiplier for mild steel You need 17.997 inches of 1 inch round mild steel. If my math is correct
I am thinking that is what I am going to end up doing.

Keep in mind that if you go with a hemispherical end, the length for 4 pounds will be slightly longer.

Weight = [(2/3)Pi x r^3+Pi x r^2 x L] x rho

where r = .5", rho = .283 lb/in^3, Weight = 4lb.

Comes out to 17.663" cylinder length, plus half an inch for the radius of the hemispherical end, or 18.163" overall length.

I suppose you want some kind of attachment point at the opposite end though, right? That'll jack up the calculation again.

ultraclyde
ultraclyde Dork
10/23/12 2:33 p.m.

if you drill a simple through-hole neat the top for attachment, you should be able to calculate the weight loss of the cylindrical void, then add that back to the total target weight to get your new target length.

Just curious, but most dictated methods have suppliers for the recommended equipment. Did this one not? Or was the price a lubeless shag?

93EXCivic
93EXCivic MegaDork
10/23/12 2:38 p.m.

In reply to ultraclyde:

To be honest, it is only a semi-scientific test to shut some salesmen up.

93gsxturbo
93gsxturbo Dork
10/23/12 6:28 p.m.

I would consider making it slightly underweight, then filling it with some lead from melted tire weights, then drilling that lead to get it dead nuts on to 4 lbs, depending on how scienteriffic you want to be.

Would suck to have the exact part you want in steel and it weighs 3.98 lbs instead of 4.00 lbs. If its overweight you can always file a little bit off.

Appleseed
Appleseed PowerDork
10/23/12 7:15 p.m.

Wouldn't it be easier to make it slightly oversize and file down the mild steel itself?

Woody
Woody MegaDork
10/24/12 10:53 a.m.
93EXCivic wrote: In reply to ultraclyde: To be honest, it is only a semi-scientific test to shut some salesmen up.

In that case, just find something that's close to four pounds and smack him in the head with it.

93EXCivic
93EXCivic MegaDork
10/24/12 10:59 a.m.
Woody wrote:
93EXCivic wrote: In reply to ultraclyde: To be honest, it is only a semi-scientific test to shut some salesmen up.
In that case, just find something that's close to four pounds and smack him in the head with it.

That sounds like an excellent plan to me.

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