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Old 04-12-2013, 03:39 PM
grassmasterswilson grassmasterswilson is offline
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Prodiamine 65wg vs 4L

I've always used wg prodiamine but as the number of accounts has increases I've seen more clogging in my z spray.

So I'm thinking of switching over to 4L. I normally use 1.2 lbs per acre of wg with .8 lbs at the first app and .4 lbs at the second app.

I'm having trouble converting the units.... Can anyone tell me how many oz of 4L would equal .4 lbs per acre of 65wg?

For those who use 4L for all apps. Do you run the high end of the rate or can you get by with 30-35 oz or so.
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Old 04-12-2013, 06:18 PM
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turf hokie turf hokie is offline
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I switched from granular dimension to 4fl last year.

Using .5 oz per 1000. That is cool season grass so I don't know if that helps.

Spraying out of the z with no clogging or settling. I am using the grey tips without nozzle screens.

Didn't look into the conversion from the wdg. I never used it
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Old 04-12-2013, 06:49 PM
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WestGaPineStraw WestGaPineStraw is offline
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We use the wdg in our max at a rate of .017 lbs per 1000. We run at .5 gallons per 1000, red tips. Have not had a issue with clogging at all. We also pre mix in a skid.
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Old 04-12-2013, 11:27 PM
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RigglePLC RigglePLC is offline
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I think, 25.6 ounces per acre for the Prodiamine high rate and 12.8 for the low rate.
On a per thousand sqft basis about .6 and .3 ounces per thousand sqft.
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Old 04-13-2013, 05:55 AM
Skipster Skipster is offline
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I haven't seen anyone answer wilson's question yet, so I'll give it a shot.

This is an ai conversion exercise. When I talk about how much of a particular herbicide I use, I always refer to it in amount of ai per unit area (usually in lbs of ai per acre). Now, we need to read the product labels and do some math.

Prodiamine 65 WDG is comprised of 65% prodiamine and 35% other stuff. So, using 0.4 #/A of 65 WDG, we need to find how much ai we're using:

(.4# WDG/A) x (.65 #ai/1 # WDG) = 0.26# ai/A (or 0.26# prodiamine/A)

Now that we have something we can make an apples-to-apples comparison with, we need to do some math with the 4L product. The 4L product contains 4 lbs of prodiamine per gallon:

(.26# ai/A) x (1 gallon of 4L product/4# ai) x (128 fl oz of 4L/1 gallon 4L) = 8.32 fl oz of the 4L product/A.

So, to answer your question, .4# of 65 WDG/A = 8.32 fl oz of the 4L product/A.

You usually don't need to change your rate of ai when you switch between product formulations. If you were using 1.2# of the WDG/A (over the course of 2 apps) and having good results, you can get the same results from 24.96 fl oz of the 4L product/A (over the course of two apps).

Knowing how to do these ai conversion will seperate you from most other folks on this site and make you a more informed consumer of your herbicides. This type of knowledge seperates the men in this industry from the apes.
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Old 04-13-2013, 08:43 AM
grassmasterswilson grassmasterswilson is offline
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Skip - thanks. That was what I was looking for. I had been getting my calculations mixed up when figuring. I guess figureing lbs to fluid oz is the problem.

So my thoughts are I normally put down a slit app of 2/3 and 1/3 rates. I put down 1.2 lbs/ acre of 65 wg. The same control level equals 26 oz per acre of 4l.

So if be second round is at 1/3 rate I could just apply 7.8 oz per acre.

If dry oz and fluid oz are the same.... I put down a total of .78 lbs/acre/ai of wg at 1.2 lbs per acre..... So the 4l equivalent would be about 30.5 floz/acre of 4l to get the same ai.


I may just finish out the year with the wg and consider switching next year.


Quote:
Originally Posted by Skipster View Post
I haven't seen anyone answer wilson's question yet, so I'll give it a shot.

This is an ai conversion exercise. When I talk about how much of a particular herbicide I use, I always refer to it in amount of ai per unit area (usually in lbs of ai per acre). Now, we need to read the product labels and do some math.

Prodiamine 65 WDG is comprised of 65% prodiamine and 35% other stuff. So, using 0.4 #/A of 65 WDG, we need to find how much ai we're using:

(.4# WDG/A) x (.65 #ai/1 # WDG) = 0.26# ai/A (or 0.26# prodiamine/A)

Now that we have something we can make an apples-to-apples comparison with, we need to do some math with the 4L product. The 4L product contains 4 lbs of prodiamine per gallon:

(.26# ai/A) x (1 gallon of 4L product/4# ai) x (128 fl oz of 4L/1 gallon 4L) = 8.32 fl oz of the 4L product/A.

So, to answer your question, .4# of 65 WDG/A = 8.32 fl oz of the 4L product/A.

You usually don't need to change your rate of ai when you switch between product formulations. If you were using 1.2# of the WDG/A (over the course of 2 apps) and having good results, you can get the same results from 24.96 fl oz of the 4L product/A (over the course of two apps).

Knowing how to do these ai conversion will seperate you from most other folks on this site and make you a more informed consumer of your herbicides. This type of knowledge seperates the men in this industry from the apes.
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Last edited by grassmasterswilson; 04-13-2013 at 08:53 AM.
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Old 04-13-2013, 10:24 AM
Skipster Skipster is offline
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Quote:
Originally Posted by grassmasterswilson View Post
Skip - thanks. That was what I was looking for. I had been getting my calculations mixed up when figuring. I guess figureing lbs to fluid oz is the problem.



The liquid s dry is very simple if we remember what the formulations and ai loads mean. The 65 WDG is a dry product consisting of 65% prodiamine. So, 1# of product had 0.65# of prodiamine.

The liquid, however, contains 4# of prodiamine per gallon.

This is why I always refer to how much ai I'm using. That way, when comparing products, I can be sure I'm getting the same impact.



Quote:
Originally Posted by grassmasterswilson View Post
So my thoughts are I normally put down a slit app of 2/3 and 1/3 rates. I put down 1.2 lbs/ acre of 65 wg. The same control level equals 26 oz per acre of 4l.

Maybe our math isn't lining up right:

(1.2# WDG/A) * (0.65# ai/ 1# WDG) = 0.78# ai/A
(0.78# ai/A) * (1 gal 4L / 4# ai) * (128 oz 4L / gallon 4L) = 24.96 fl oz of 4L

So, 1.2 #/A of 665WDG = 24.96 fl oz of 4L

26 fl oz of 4L (your figure from above) = 1.25# WDG/A

It may not make a huge difference for you, but I want to be sure we get the numbers right. I know someone on this board will put us both through the gauntlet even for a small rounding error.

Quote:
Originally Posted by grassmasterswilson View Post
So if be second round is at 1/3 rate I could just apply 7.8 oz per acre.

Back to our previous example, if you want ot match what you put down before, you would need 8.32 fl oz of the 4L/A.

Quote:
Originally Posted by grassmasterswilson View Post
If dry oz and fluid oz are the same.... I put down a total of .78 lbs/acre/ai of wg at 1.2 lbs per acre..... So the 4l equivalent would be about 30.5 floz/acre of 4l to get the same ai.

Most of this math is presented above. Others here will rake you over the coals for stating that dry and fl oz are the same. Remember, they're not the same and they measure different things.

1 oz of the WDG = 0.04# ai
1 fl oz of 4L = 0.03125# ai
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Old 04-14-2013, 11:59 AM
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humble1 humble1 is offline
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just curious do you think you can get enough coverage at 1/3 gal per K or do you feel min of .5 gal per K on the prodiamine?
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Old 04-14-2013, 04:39 PM
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i did great with 1/3 gal nozzles last yr. i had great control with split applications
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