#### ceaman

##### Senior Member

- Location
- Central Indiana

- Thread starter ceaman
- Start date

- Location
- Central Indiana

- Location
- SE CT

Dino

- Location
- Central Indiana

- Thread Starter
- #3

Is dino the only one who has any info on this..... i am looking for some sugestions to use as starting points for temperature/ rates. i understand there is a learning curve but why cant anyone give sugestions about how it works for them to help someone learn the curve faster? I know there are some taboo subjects on here that no one will talk about but I have to say I was hoping this wasnt one of them.

ill take info on any deicer... not just Iceban.

ill take info on any deicer... not just Iceban.

- Location
- NJ

http://www.plowsite.com/showthread.php?s=&threadid=4111

......which is 17 threads down from the one you just started, will give you some answers.

Gotta love that search feature........

~Chuck

- Location
- NJ

30°---------------------46 lbs. of ice

25°---------------------14 1/2 lbs. of ice

20°-----------------------8 1/2 lbs. of ice

15°-----------------------6 1/2 lbs. of ice

10°-----------------------5 lbs. of ice

5°------------------------4 lbs. of ice

0°------------------------3 1/2 lbs. of ice

-6°-----------------------3 lbs. of ice

- Location
- Grand Rapids, MI

Chuck, how are we supposed to measure how much the ice weighs? LOL

- Location
- Central Indiana

- Thread Starter
- #7

I wondered too but didnt ask..... figured i would get the.

"You ask a dumb question.... You get a dumb answer." Answer!

"You ask a dumb question.... You get a dumb answer." Answer!

- Location
- Somerset, NJ

1,000 sq. ft. of pavement with 1\8" ice = approximately X pounds,

1,000 sq. ft. of pavement with 1\4" ice = approximately X pounds

Etc.

that would make the other chart's info useful to us?

- Location
- NJ

Oh man, you want ME to do the math, LOL.

~Chuck

~Chuck

- Location
- Somerset, NJ

I can do the math, I just don't know what formula to do. I think your busting my chops, but I may have written the question unclearly.

1,000 sq. ft. of 1\8" ice = approximately X pounds of ICE

Now once I figure that out I can say for example, if I have 1\8" of ice in my 10,000 sq ft lot and the temperature is 25 degrees, if x=116 pounds then I know I'll need about 80 pounds of salt to melt it. Now how do we figure out X to make that Salt Institute chart useful? Or am I missing something here? Otherwise, instead of making that chart, they could just say "as the temperature drops, it takes a whole lot more salt to melt the same amount of ice" and be done with it LOL. Please help me figure out how that chart actually helps me, the poor old plow dude with no snow to plow.

Disclaimer: I'm dumber than Mark & ceaman?

- Location
- Central Indiana

- Thread Starter
- #11

I dicided that it never going to snow again anyway so i quit worrying about it...... HAHAHA

- Location
- Somerset, NJ

Yea, but it might snow next year, so I'll keep trying to learn about this stuff.

- Location
- Western CT

Next year is only a little over a week away. I am confident it will snow next year too.

- Location
- Southwestern Pa.

Well, I'm not mentally prepared to do much math right now either, but I found a website that will do conversions of lots of different stuff. The one for the depth of water (in a square foot) to pounds of water gave this result:I can do the math, I just don't know what formula to do. I think your busting my chops, but I may have written the question unclearly.

.02 feet of water calculates to (rounded off) 1.25 lbs.

.02 would be about 1/4 inch--figure there are 96 1/8s of an inch in one foot (12 x 8=96), so .01 (1/100th) is tolerably close to 1/8. (Right?)

Therefore, 1/4" of ice on 1000 sq. ft. would weigh roughlyt 1250 lbs. So, 10,000 sq. ft. would be 12,500 lbs. Of course, the density of water is a little different than ice, but this ain't rocket science, at least not the way I'm capable of doing it....