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bwhite82
May 10th, 2010, 02:38 AM
This should be simple, but it is frustrating me. First a little background info, I am studying for a Private Applicator License to be able to use restricted-use pesticides on ornamental turf grass. The exam is heavy on pesticide labeling. Here is my dilemma and the math issue:

I am working with two 1-gallon containers of a certain pesticide, both contain the SAME active ingredients, but one merely has higher percentages of active ingredients than the other (one is marketed toward joe homeowner, the other to a lawn care company). The actual labels are below for your reference:

Trimec Classic (http://www.pbigordon.com/pdfs/TrimecClassic-SL.pdf)

Trimec "Lawn Weed Killer" (http://www.pbigordon.com/pdfs/TrimecLawnWeedKiller-SL.pdf)

Trimec Classic contains 1.98 lbs (21.54%) of 2-4-d per gallon
Trimec "Lawn Weed Killer" contains .54 lbs (6.30%) of 2-4-d per gallon

Trimec Classic says to mix 1.2 to 1.5 fl oz of product into .5 to 6 gal of water (depending upon spray volume) per 1000 sq ft of lawn

Trimec "Lawn Weed Killer" says to mix 2 fl into 2 gallons of water per 500 sq ft of lawn

My question is, will both have the same "weed killing power"? ie do they both contain the same amount of active ingredient (2-4-d) after mixed into their respective 2-gallon spraying containers?

Pulling my hair out here! Should be simple, I mean I'm not dealing with calculus here, just basic mathematics.

NovaAesa
May 10th, 2010, 02:51 AM
You obviously want to find out the amount of 2-4-d per square foot of lawn. So amount_per_sq_ft = concentration(%) * amount_of_product_going_on_the_total_of_the_lawn / size_of_lawn.

With that formula you should be able to work it out.

bwhite82
May 10th, 2010, 03:20 AM
You obviously want to find out the amount of 2-4-d per square foot of lawn. So amount_per_sq_ft = concentration(%) * amount_of_product_going_on_the_total_of_the_lawn / size_of_lawn.

With that formula you should be able to work it out.

Thanks! I'll keep plugging away. Should have paid more attention in math class! :/

Ebere
May 10th, 2010, 03:41 AM
This should be simple, but it is frustrating me. First a little background info, I am studying for a Private Applicator License to be able to use restricted-use pesticides on ornamental turf grass. The exam is heavy on pesticide labeling. Here is my dilemma and the math issue:

I am working with two 1-gallon containers of a certain pesticide, both contain the SAME active ingredients, but one merely has higher percentages of active ingredients than the other (one is marketed toward joe homeowner, the other to a lawn care company). The actual labels are below for your reference:

Trimec Classic (http://www.pbigordon.com/pdfs/TrimecClassic-SL.pdf)

Trimec "Lawn Weed Killer" (http://www.pbigordon.com/pdfs/TrimecLawnWeedKiller-SL.pdf)

Trimec Classic contains 1.98 lbs (21.54%) of 2-4-d per gallon
Trimec "Lawn Weed Killer" contains .54 lbs (6.30%) of 2-4-d per gallon

Trimec Classic says to mix 1.2 to 1.5 fl oz of product into .5 to 6 gal of water (depending upon spray volume) per 1000 sq ft of lawn

Trimec "Lawn Weed Killer" says to mix 2 fl into 2 gallons of water per 500 sq ft of lawn

My question is, will both have the same "weed killing power"? ie do they both contain the same amount of active ingredient (2-4-d) after mixed into their respective 2-gallon spraying containers?

Pulling my hair out here! Should be simple, I mean I'm not dealing with calculus here, just basic mathematics.

First, let's assume that you are going to mix the "Classic" into 2 gallons of water.

And the "Lawn Weed KIller" will be mixed into 4 gallons of water.

It takes four gallons of the latter to equal the coverage of the former.

So, you are mixing 1.5 ounces of the classic into the amount of water needed to cover 1000 sq ft.

And 4 ounces of the homeowner stuff into the amount of water needed, to cover 1000 sq ft.

A gallon is 128 ounces.

If the percentage is 21.54% per gallon, for the classic, that would be 0.16828125% per ounce. Or 0.252421875% of the "2-4-D" to cover the 1000 sq ft.

If the percentage is 6.30% per gallon for the homeowner stuff, that would be 0.04921875% per ounce. Or 0.196875% of the "2-4-D" to cover the 1000 sq ft.

Short answer... No. They do not have the same amount of 2-4-D.

bwhite82
May 10th, 2010, 03:50 AM
First, let's assume that you are going to mix the "Classic" into 2 gallons of water.

And the "Lawn Weed KIller" will be mixed into 4 gallons of water.

It takes four gallons of the latter to equal the coverage of the former.

So, you are mixing 1.5 ounces of the classic into the amount of water needed to cover 1000 sq ft.

And 4 ounces of the homeowner stuff into the amount of water needed, to cover 1000 sq ft.

A gallon is 128 ounces.

If the percentage is 21.54% per gallon, for the classic, that would be 0.16828125% per ounce. Or 0.252421875% of the "2-4-D" to cover the 1000 sq ft.

If the percentage is 6.30% per gallon for the homeowner stuff, that would be 0.04921875% per ounce. Or 0.196875% of the "2-4-D" to cover the 1000 sq ft.

Short answer... No. They do not have the same amount of 2-4-D.

Thank you thank you thank you. You deduced something in 2 min (probably) that I still haven't gotten in an hour! Hate math. Anyways, my suspicions are correct then: not only is the big box store crap diluted, it is also weaker after mixed! One would be led to believe that because you have to mix more per gallon of water, that it would have the same potency as the other. Tsk.

Ebere
May 10th, 2010, 04:45 AM
Thank you thank you thank you. You deduced something in 2 min (probably) that I still haven't gotten in an hour! Hate math. Anyways, my suspicions are correct then: not only is the big box store crap diluted, it is also weaker after mixed! One would be led to believe that because you have to mix more per gallon of water, that it would have the same potency as the other. Tsk.

Believe it or not, I am math-phobic.