*1 pound of nitrogen per 1000 sq feet. *There are 43,560 square feet in a acre. *So 43,560 sq feet in an acre divide by 1000 square feet equals 43.56 × 1# of nitrogen. So we need 43.56 pounds of nitrogen per acre. * 6"×6" pot = 36 sq inches * 1 sq foot converted to sq inches is 12"x12" =144 sq inches. * So 36"÷144"= 0.25 sq feet * So 0.25 sq feet divided by 43560 sq feet in an acre = 0.0000057392 acres in that 6x6 pot. * We know we need 43.56 pounds of nitrogen in 1 acre. And our pot has 0.0000057392 acres in it. So 0.0000057392 acres × 43.56 pounds of nitrogen=0.0002499996 of nitrogen needed. * Alright back to the osmocote. It has 19% nitrogen in it which as a decimal is 0.19. * Alright from our math above we know our pot needs 0.0002499996# of nitrogen but our osmocote has 19% nitrogen in its container. So how much of the 19% osmocote do we need to dump out in the pot to give us the 0.0002499996# needed per pot. Just take 0.0002499996# ÷0 .19% osmocote = 0.0013157874 pounds of the osmocote per pot to give us the correct amount of fertilizer per pot.
What is the amout of fertilizer material for 7,000 using the diferent element sourse .21-0-0 ammonium sulfate,0-18- 0 (ordinary superphosphate) .0-0-60(mauriate of pktash orpotasium chloride
multiplied the recommended amount of the fertilizer to the amount of nitrogen that is in one pound of fertilizer. The N number of 20 in the NPK is a percentage. So, 20% of one pound of fertilizer is nitrogen. Hope that clears it up
I want to ask you about amount of fertilizer to put on plants. So Im working on this social project to teach local farmers about introducing a new type of fertilizer for them. I want to be able to teach them the basics of amount of N P K inside a fertile soil thats best for plants, specifically tomatoes, chillies, and other vegetables. For them to also cut their cost on fertilizers. I already know how to calculate the amount from your video but I havent been able to find the right amount of Nitrogen that is best for veggies. I read from you that is 1 lbs per 1000 sq feet for lawn. But I cant find the appropriate amount for veggies. Can you help? Thank you Regards Ren from Indonesia
It has been about 9 years so he has graduated by now. The instructor could have used the simplest method that was available but he chose the harder way. The target audience was backyard farmers or dummies like me who dont know a thing about farming and just getting a crash course in NPK. It took me a while to figure out what he wanted to teach and come up with a simplier method to teach the target audience. This is a more easier method ( this is not an easy topic): First, 1 lb of Nitrogen (N) = 1,000 s.f. of land. just switch it around and you get, 1,000 s.f. = 1 lb of N. Therefore 1 s.f. = 1/1000 = 0.001 lb of N. This applies if it is a 6 inch pot or a large 15,000 s.f. of land or even a 3 x 10 raised bed= 30 s.f. Second, calculate s.f of land , even if it is a 6 inch, 8 inch or 12 inch pot. When dealing with pots you deal in square inches. So you need to know how many square in = 1 s.f. Thus, 1 s.f. = 144 square inch. Switch it around you get 144 sq in = 1 s.f. Therefore 1 sq in =1/144= 0.007 s.f. So a 6 inch pot is 6 x 6 = 36 square inch. therefore 36 x .007 s.f. = 0.25 s.f. Now you know that a 6 inch pot = 0.25 s.f. Third, You then simply calculate the amount of N needed for a 6 inch pot. 1 s.f. of land = 0.001 lb of N. A 6 inch pot has only 0.25 s.f. of land (soil) Thus 0.001 x 0.25 s.f. = 0.00025 lb of N. Therefore a 6 inch pot only requires 0.00025 lb of Nitrogen. Fourth, you need to break down the 0.00025 lb of N into a smaller unit. You can use oz or grams. I prefer grams it is really easy to measure with grams. 1 lb = 453.592 grams. Thus a 6 inch pot requires 0.00025 lb of N x 453.592 grams = 0.1 gram of N. Lastly, now you are ready to factor in the NPK ratios. If you have a 100% 0% 0% fertlizer then all you need to do is to use 0.1 grams of it. He used 19. 6. 12. Therefore 1 / (0.19) = 5.26 lb of fertilizer to get 1 lb of N. Therefore since the 6 inch pot only requires 0.00025 lb of N. Then 0.00025 x 5.26 lb = 0.0013 lb of fertilizer. Convert into grams you get 0.0013 lb x 453.592 = 0.6 (0.59) grams of fertilizer.
Using your method, how do you calculate the amount of fertilizer you would need to add if you have 15 ppm N in your soil report & your plant requires 100 ppm of N??
*The simple math is:* -------------------------------- *N Application Rate =* 1 lb N / 1000 sf = *.001 lb/sf* *Pot Area =* 6*6/144 = *.25 sf* *N Required per pot =* .001 lb/sf * .25 sf = *.00025 lbs N per pot* *Fertizer Required per pot =* .00025 lbs N / .19 = *.0013 lbs fertilizer per pot* *Fertilizer Required per pot (in grams) =* .0013 lbs * 453.59237 = *.6 g fertilizer per pot*
@@zeffur7 I would thank you for taking the time out to actually read and criticize my post. I am a first time backyard farmer and like most people am easily lost in the math. Persons like me only want simple answers to our questions. We are not seeking to impress others, just to have a basic understanding to grow our veggies properly. The method I used is not my method, it is simple grade 6 math. I am only concerned with pots, barrels and less than 10,000 s.f. of land. As for your question I will be frank, I cant answer it. Using ppm (per million ..) is for scientist and others who do precise testing. The reality of the majority of folks out there is that they will shy away from these test and if they do get one all we would want to know is how much NPK to use in our soil.
@@geofferyromany4634 It's less difficult than you may think. To answer the question that I asked you, just take the amount needed by the plant minus the amount already in the soil (e.g. 100 ppm N - 15 ppm N)= *85 ppm N*
@@zeffur7 just looking at it hurts. My mind is blocking it. However, I do understand what you are teaching me and I would say thanks very much. Without this bit of info I would be lost. This was the final piece of the NPK puzzle. Thanks again.
Hi guys can someone help me pls. I hv 4-6-4 fert. Im using a 2 litre bottel to grow my plants based on this cal how much do i use or need? The math is very deep n its lbs...here v use grams n kgs.
Hi sir I have 100 trees I will fertilize with a liquid fertilizer 5 l per hectare (10,000 m²), what is the quantity of fertilizer and water to mix in the mother solution, I use the fertilization technique with ventury, the volume of solution tank is 300 liters. my greetings.
The fertilizer recommendation for corn is 90-60-60 kg per hectare. The available fertilizer materials are complete fertilizers (14-14-14) and urea (46-0-0). How many bags of fertilizer materials to be used to come up with the recommendation?
Thanks so much for the question! Our resident gardening expert Marianne would be delighted to assist. Please feel free to submit a question here: www.plantersplace.com/ask-the-gardening-expert.
Hi Francisco! Ryan Lee did weigh the Osmocote fertilizer, but it is not necessary for you to do. Osmocote products have application instructions that make it simple for applying the slow-release fertilizer such as one capful per 2 gallon pot or 4 sq. feet. For more information, visit www.plantersplace.com/osmocote-plant-food/
Wouldn't it be easier & more useful to your viewers if you left out the math shortcuts and provided them with a useful tool that they could use--You know.. like: 1. Define some plant nutrient requirement (e.g N=30 ppm or 60 lbs/acre). 2. Choose some typical soil report nutrient value (e.g. N=10 ppm or 20 lbs/acre). 3. If 2 is less than 1, then subtract 2 from 1 and get the amt of nutrient that must be added to the soil (e.g 20 ppm or 40 lbs/acre). If 2 is greater than 1, then explain that no N needs to be added to the soil for optimal plant growth. *Then show the math for:* *1. Farmers* (who plant acres) so that they have a tool that will assist them before they buy nutrients (e.g. Lbs of Fertilizer Reqd = lbs of nutrient reqd/acre * acres / % of nutruent per lb of fertilizer (e.g.: Lbs Fert Reqd = 40 lbs N/acre * 40 acres / .46 = *~3,478 lbs. of fert* reqd (e.g.: 40 * 40 * / .46 = *~3,478 lbs. of fert* reqd *Note:* ".46" (above) is urea nitrogen fertilizer (46-0-0). You could use any type of organic or commercial fertilizer that you wish (e.g.: human urine (1 fluid oz over 1 square ft adds ~13.2 ppm N to the soil), ammonium nitrate fertilzer (34-0-0), etc. *2. Gardeners* (who plant by the pot/row/patch) so that they have a tool that will assist them before they buy nutrients (e.g. Lbs of Fertilizer Reqd = lbs of nutrient reqd/acre * (sf * 1 acre/43560 sf) / % of nutrient per lb of fertilizer (e.g.: *For a 6" x 6" pot:* Lbs Fert Reqd = 40 lbs N/acre * ((6"/12" * 6"/12") * 1 acre/43560 sf) / .46 = *~0.0005 lbs of fert* = *~0.23g of fert* reqd (i.e. 0.0005 lbs * 453.59237 g/lb). (e.g.: 40 * (.5*.5/43560) / .46 = *~0.0005 lbs of fert* = *~0.23g of fert* reqd (i.e. 0.0005 lbs * 453.59237 g/lb). *For a 4' x 25' row:* Lbs Fert Reqd = 40 lbs N/acre * (4' * 25' * 1 acre/43560 sf / .46 = *~0.2 lbs fert* reqd = *~91g of fert* (i.e. 0.2 lbs * 453.59237 g/lb). (e.g.: 40 * (4*25/43560) / .46 = *~0.2 lbs of fert* reqd = *~91g of fert* reqd (i.e. 0.2 lbs * 453.59237 g/lb). *For a 10' x 10' area:* Lbs Fert Reqd = 40 lbs N/acre * (10' * 10' * 1 acre/43560 sf) / .46 = *~.2 lbs of fert* reqd = *~91g of fert* (i.e. 0.2 lbs * 453.59237 g/lb). (e.g.: 40 * (10*10/43560) / .46 = *~0.2 lbs of fert* reqd = *~91g of fert* reqd (i.e. 0.2 lbs * 453.59237 g/lb). *Note:* ".46" (above) is urea nitrogen fertilizer (46-0-0). You could use any type of organic or commercial fertilizer that you wish (e.g.: human urine (1 fluid oz over 1 square ft adds ~13.2 ppm N to the soil), ammonium nitrate fertilizer (34-0-0), etc. Then show visually how little fertilizer each one of those quantities actually represents, so that people can see just how little fertilizer they actually have to add to their growing area..
+hulmil 0.32 grams is a fraction of a gram. ~1/3 of a gram (which is a very small amount of fertilizer). To determine exactly how much that is, weigh out 1 gram of your fertilizer. Then divide that 1 gram pile into 3 equal parts. 1 of those parts is the amount that you would spread evenly across your 6" x 6" soil area. *Important NOTE:* Remember, the example calculation in my original comment is for *urea nitrogen fertilizer (46-0-0)*. If you plan to use a different fertilizer, you would change the .46 in the example formula to whatever your fertilizer bag indicates (for example if your fertilizer is 20-0-0, you would use .20 in the example formula rather than .46). Also note that volume amount (e.g. teaspoons) will vary with the density of the fertilizer. Different fertilizer formulations have different densities. You will need to weigh out a specific volume & weigh it to determine the density of your fertilizer.
+zeffur7 Thanks for your answer.I don't have a soil report, I just want to mix up a batch of some balanced 12-12-12 fertilizer I have on hand for my tomatoes. Your method shows I need .06 gram of 12-12-12 fertilizer for each 6 inch pot (.5 *12 *454)/43560. Now if want to dissolve the fertilizer in water to make up a gallon of liquid fertilizer, how much do I add to a gallon? Do I calculate how much water I would pour on the plant? For example: If I would normally dissolve .06 gram in 1/c cup of water, and a gallon is 16 cups, then I need (16 * .06)/.5 or 19 g per gallon. (Assuming it will dissolve!) Does this make sense? Luckily I have a scale that measures in grams. Now a teaspoon of salt is 5.69 grams, so if these are comparable, 19 grams of fertilizer is roughly 3 teaspoons or one tablespoon. When I measure out the 19 grams, I will see if it is roughly a tablespoon. Hope this helps hulmil.
+JennyJardin re: "Your method shows I need .06 gram of 12-12-12 fertilizer for each 6 inch pot (.5 *12 *454)/43560." No, you did your math incorrectly. The proper calculation according to the example that I provided would be as follows *for the fertilizer that you plan to use*: lbs fert reqd = 40 lbs fert/acre * ((6"/12" * 6"/12") * 1 acre/43560 sf) / .12 = *~0.002 lbs of fert* = *~0.87g of fert* reqd e.g.: 40 * (.5*.5/43560) / .12 = *~0.002 lbs of fert* = *~0.87g of fert* reqd (i.e. 0.002 lbs * 453.59237 g/lb). *Important Information to Remember:* --------------------------------------------------------------- 1. Adding 40 lbs of N/acre will cause the soil to increase by 20 ppm of N. 2. A tomato plant requires 60 ppm of N to produce well (as well as other nutrients too!--but, where I live I never have to add P or K to my soil as it is always above the proper levels). 3. The soil N + your fertilizer N should = 60 ppm of N for the whole growing period (usually less than 4 months). 4. It's best to ensure that your growing medium has 40 ppm of N when the seed or seedling is put into the growing medium (or soil) & then add the remaining 20 ppm of N after the plant blooms. 5. The calculation that I showed above will only provide 20 ppm of N to your soil (it was just an example problem that wasn't setup for tomato plants). If your soil has 0 ppm of N in it, then your tomato plants will have insufficient N to grow properly. You can take the computed fertilizer weight & divide it by 2. In your case that would be ~.87g/2 = .43g of fertilizer = 10 ppm. So, your tomatoes require 60 ppm & let's assume your soil has 10 ppm of N already in it, you would then calculate 60ppm - 10ppm = 50 ppm of N will need to be added to your growing medium. Since you know .43g of fert=10 ppm of N, you would calculate 50ppm / 10ppm = 5. Therefore: 5 * .43g fert = ~2.8g of fert = 50 ppm of N that you would add to one 6" square pot during the growing season (usually 2/3rds of that amount would be added when the seed or seedling is planted & the other 1/3rd of that amount would be added when the plant blooms (no addition N will need to be added for the rest of the growing period). Note: If you intend to grow your tomato plants in pots during the whole growing season (like I do), I would suggest that you use at least a 3 gallon pot for each plant. *Making a Solution of Liquid Fertilizer:* .43g of 12%/wt N fert will provide a 6" square pot with 10ppm of N. You can mix that amount (or more to get a higher N ppm concentration) with any amount of water that the fertilizer will dissolve in completely. Apply that whole solution amount to one pot. *To make a batch of liquid fertilizer for multiple plants do this:* 1. Plants = 16 (you choose this number) 2. N ppm to add to each plant = 33.33 (this is 2/3 of 50 ppm from above) 3. *Fertilizer amt* = 33.33ppm N/10ppm = 3.33 * .43 g Fert * 16 plants *= ~23g of fert*. 4. *Water amt* = 8 fl oz for each plant * 16 plants = 128 fl oz *= 1 gallon* 5. Dissolve the fertilizer completely in the water & then water each plant with 8 fl oz of the liquid fertilizer solution. Happy Gardening!
6 inch x 6 inch = 36 sq inches. Since 12" x 12" = 144 square inches, we convert 36 sq in to sq ft as follows 36 sq in * (1 sf / 144 sq in) = .25 sq ft (which is what he properly calcuated in the video.
You never EXPLAINED where 0.2 came from... I ASSUME you mean it's 20% in decimal form. Please be more thorough and explain things. You also can't up with 50 something on your next calculation. Again, where did you get 50 from?
By the time I get the formula right, I will be a mathematician not a gardener.. cheers.
*1 pound of nitrogen per 1000 sq feet.
*There are 43,560 square feet in a acre.
*So 43,560 sq feet in an acre divide by 1000 square feet equals 43.56 × 1# of nitrogen. So we need 43.56 pounds of nitrogen per acre.
* 6"×6" pot = 36 sq inches
* 1 sq foot converted to sq inches is 12"x12" =144 sq inches.
* So 36"÷144"= 0.25 sq feet
* So 0.25 sq feet divided by 43560 sq feet in an acre = 0.0000057392 acres in that 6x6 pot.
* We know we need 43.56 pounds of nitrogen in 1 acre. And our pot has 0.0000057392 acres in it. So 0.0000057392 acres × 43.56 pounds of nitrogen=0.0002499996 of nitrogen needed.
* Alright back to the osmocote. It has 19% nitrogen in it which as a decimal is 0.19.
* Alright from our math above we know our pot needs 0.0002499996# of nitrogen but our osmocote has 19% nitrogen in its container. So how much of the 19% osmocote do we need to dump out in the pot to give us the 0.0002499996# needed per pot. Just take 0.0002499996# ÷0 .19% osmocote = 0.0013157874 pounds of the osmocote per pot to give us the correct amount of fertilizer per pot.
What is the amout of fertilizer material for 7,000 using the diferent element sourse .21-0-0 ammonium sulfate,0-18- 0 (ordinary superphosphate) .0-0-60(mauriate of pktash orpotasium chloride
I'm confused. How did you get 5lb fertilizer per 1000 square ft?
multiplied the recommended amount of the fertilizer to the amount of nitrogen that is in one pound of fertilizer. The N number of 20 in the NPK is a percentage. So, 20% of one pound of fertilizer is nitrogen. Hope that clears it up
Great Presentation
Very helpful and nice
So could you save money buying 20-30-20 vs 1-2-1? Kinda like buying in bulk and just cutting the dose in 1/4th, for example. Thanks in advance.
Im wondering if you could estimate the nitrogen & phosphate rate in one litre of urine?
I want to ask you about amount of fertilizer to put on plants.
So Im working on this social project to teach local farmers about introducing a new type of fertilizer for them. I want to be able to teach them the basics of amount of N P K inside a fertile soil thats best for plants, specifically tomatoes, chillies, and other vegetables. For them to also cut their cost on fertilizers. I already know how to calculate the amount from your video but I havent been able to find the right amount of Nitrogen that is best for veggies. I read from you that is 1 lbs per 1000 sq feet for lawn. But I cant find the appropriate amount for veggies. Can you help?
Thank you
Regards
Ren from Indonesia
what is the 50lbN / AC , which is at 4:34 on this video, from where we got 50 ?
How do I prepare a Hydrophonic Solution using this method of Solid Fertilizers?
It has been about 9 years so he has graduated by now. The instructor could have used the simplest method that was available but he chose the harder way. The target audience was backyard farmers or dummies like me who dont know a thing about farming and just getting a crash course in NPK. It took me a while to figure out what he wanted to teach and come up with a simplier method to teach the target audience.
This is a more easier method ( this is not an easy topic):
First, 1 lb of Nitrogen (N) = 1,000 s.f. of land. just switch it around and you get, 1,000 s.f. = 1 lb of N. Therefore 1 s.f. = 1/1000 = 0.001 lb of N. This applies if it is a 6 inch pot or a large 15,000 s.f. of land or even a 3 x 10 raised bed= 30 s.f.
Second, calculate s.f of land , even if it is a 6 inch, 8 inch or 12 inch pot. When dealing with pots you deal in square inches. So you need to know how many square in = 1 s.f. Thus, 1 s.f. = 144 square inch. Switch it around you get 144 sq in = 1 s.f. Therefore 1 sq in =1/144= 0.007 s.f. So a 6 inch pot is 6 x 6 = 36 square inch. therefore 36 x .007 s.f. = 0.25 s.f.
Now you know that a 6 inch pot = 0.25 s.f.
Third, You then simply calculate the amount of N needed for a 6 inch pot.
1 s.f. of land = 0.001 lb of N. A 6 inch pot has only 0.25 s.f. of land (soil) Thus 0.001 x 0.25 s.f. = 0.00025 lb of N.
Therefore a 6 inch pot only requires 0.00025 lb of Nitrogen.
Fourth, you need to break down the 0.00025 lb of N into a smaller unit. You can use oz or grams. I prefer grams it is really easy to measure with grams. 1 lb = 453.592 grams. Thus a 6 inch pot requires 0.00025 lb of N x 453.592 grams = 0.1 gram of N.
Lastly, now you are ready to factor in the NPK ratios. If you have a 100% 0% 0% fertlizer then all you need to do is to use 0.1 grams of it. He used 19. 6. 12. Therefore 1 / (0.19) = 5.26 lb of fertilizer to get 1 lb of N. Therefore since the 6 inch pot only requires 0.00025 lb of N. Then 0.00025 x 5.26 lb = 0.0013 lb of fertilizer.
Convert into grams you get 0.0013 lb x 453.592 = 0.6 (0.59) grams of fertilizer.
Using your method, how do you calculate the amount of fertilizer you would need to add if you have 15 ppm N in your soil report & your plant requires 100 ppm of N??
*The simple math is:*
--------------------------------
*N Application Rate =* 1 lb N / 1000 sf = *.001 lb/sf*
*Pot Area =* 6*6/144 = *.25 sf*
*N Required per pot =* .001 lb/sf * .25 sf = *.00025 lbs N per pot*
*Fertizer Required per pot =* .00025 lbs N / .19 = *.0013 lbs fertilizer per pot*
*Fertilizer Required per pot (in grams) =* .0013 lbs * 453.59237 = *.6 g fertilizer per pot*
@@zeffur7 I would thank you for taking the time out to actually read and criticize my post. I am a first time backyard farmer and like most people am easily lost in the math. Persons like me only want simple answers to our questions. We are not seeking to impress others, just to have a basic understanding to grow our veggies properly. The method I used is not my method, it is simple grade 6 math. I am only concerned with pots, barrels and less than 10,000 s.f. of land. As for your question I will be frank, I cant answer it. Using ppm (per million ..) is for scientist and others who do precise testing. The reality of the majority of folks out there is that they will shy away from these test and if they do get one all we would want to know is how much NPK to use in our soil.
@@geofferyromany4634 It's less difficult than you may think. To answer the question that I asked you, just take the amount needed by the plant minus the amount already in the soil (e.g. 100 ppm N - 15 ppm N)= *85 ppm N*
@@zeffur7 just looking at it hurts. My mind is blocking it. However, I do understand what you are teaching me and I would say thanks very much. Without this bit of info I would be lost. This was the final piece of the NPK puzzle.
Thanks again.
very helpful thanks
Very interesting.
Hi guys can someone help me pls. I hv 4-6-4 fert. Im using a 2 litre bottel to grow my plants based on this cal how much do i use or need? The math is very deep n its lbs...here v use grams n kgs.
Can you give some good questions about this topic,,,please
Hi sir
I have 100 trees I will fertilize with a liquid fertilizer 5 l per hectare (10,000 m²), what is the quantity of fertilizer and water to mix in the mother solution, I use the fertilization technique with ventury, the volume of solution tank is 300 liters. my greetings.
How do you measure the fertilizer? What weighing scale did you use please?
they normally come in a 50 pound bag
The fertilizer recommendation for corn is 90-60-60 kg per hectare. The available fertilizer materials are complete
fertilizers (14-14-14) and urea (46-0-0). How many bags of fertilizer materials to be used to come up with the
recommendation?
If i dissolve urea (46-0-0) in a gallon of water, how much do i need to fertilize my 75 potted(8x8x16) plants. Pls help
1.21 gigawatts
@@Dr-TH-cam And a flux capacitor.. lol
Where did you "just fine up with 50lb of Nitrogen per acre? I needed to know in order to understand. Send like you just pulled it out of thin air...
Wow this is great. But the calculation is kinda difficult for me sir
I am a layman. Is it good enough to measure the fertilizer using the EC tester?
Thanks so much for the question! Our resident gardening expert Marianne would be delighted to assist. Please feel free to submit a question here: www.plantersplace.com/ask-the-gardening-expert.
How to compute for fert application rate to a 30kilogram of soil?
The first step is to convert 30kg to lbs = 30 kg * ~2.2046 lb/kg = ~66.14 lbs (note: ~ means approximately)
Hi! iu grad 1996. Now horticulturalist!
Hii am also horticulturist india
But fertlizer calculation is different ...i dont know how to calculate ferilizer in europe countries
5 pounds of nitrogen per 1000 sq feet in what time period
Thanks for the info ;)
doing containers use grams instead of lbs cause most scales weigh by pounds don't weigh up that small amounts.
Should have gone to Purdue 😂 Boiler Up!
5 Square yards, how many square inch?!
Let me just Measure 0.0013 LB with my eye
Yea Luigi? You do that and it'll burn ya.
no sound
Do you literally weigh the fert?
Hi Francisco! Ryan Lee did weigh the Osmocote fertilizer, but it is not necessary for you to do. Osmocote products have application instructions that make it simple for applying the slow-release fertilizer such as one capful per 2 gallon pot or 4 sq. feet.
For more information, visit www.plantersplace.com/osmocote-plant-food/
You didn’t even cover phosphorus or potassium
Wouldn't it be easier & more useful to your viewers if you left out the math shortcuts and provided them with a useful tool that they could use--You know.. like:
1. Define some plant nutrient requirement (e.g N=30 ppm or 60 lbs/acre).
2. Choose some typical soil report nutrient value (e.g. N=10 ppm or 20 lbs/acre).
3. If 2 is less than 1, then subtract 2 from 1 and get the amt of nutrient that must be added to the soil (e.g 20 ppm or 40 lbs/acre). If 2 is greater than 1, then explain that no N needs to be added to the soil for optimal plant growth.
*Then show the math for:*
*1. Farmers* (who plant acres) so that they have a tool that will assist them before they buy nutrients (e.g. Lbs of Fertilizer Reqd = lbs of nutrient reqd/acre * acres / % of nutruent per lb of fertilizer
(e.g.: Lbs Fert Reqd = 40 lbs N/acre * 40 acres / .46 = *~3,478 lbs. of fert* reqd
(e.g.: 40 * 40 * / .46 = *~3,478 lbs. of fert* reqd
*Note:* ".46" (above) is urea nitrogen fertilizer (46-0-0). You could use any type of organic or commercial fertilizer that you wish (e.g.: human urine (1 fluid oz over 1 square ft adds ~13.2 ppm N to the soil), ammonium nitrate fertilzer (34-0-0), etc.
*2. Gardeners* (who plant by the pot/row/patch) so that they have a tool that will assist them before they buy nutrients
(e.g. Lbs of Fertilizer Reqd = lbs of nutrient reqd/acre * (sf * 1 acre/43560 sf) / % of nutrient per lb of fertilizer (e.g.:
*For a 6" x 6" pot:* Lbs Fert Reqd = 40 lbs N/acre * ((6"/12" * 6"/12") * 1 acre/43560 sf) / .46 = *~0.0005 lbs of fert* = *~0.23g of fert* reqd (i.e. 0.0005 lbs * 453.59237 g/lb).
(e.g.: 40 * (.5*.5/43560) / .46 = *~0.0005 lbs of fert* = *~0.23g of fert* reqd (i.e. 0.0005 lbs * 453.59237 g/lb).
*For a 4' x 25' row:* Lbs Fert Reqd = 40 lbs N/acre * (4' * 25' * 1 acre/43560 sf / .46 = *~0.2 lbs fert* reqd = *~91g of fert* (i.e. 0.2 lbs * 453.59237 g/lb).
(e.g.: 40 * (4*25/43560) / .46 = *~0.2 lbs of fert* reqd = *~91g of fert* reqd (i.e. 0.2 lbs * 453.59237 g/lb).
*For a 10' x 10' area:* Lbs Fert Reqd = 40 lbs N/acre * (10' * 10' * 1 acre/43560 sf) / .46 = *~.2 lbs of fert* reqd = *~91g of fert* (i.e. 0.2 lbs * 453.59237 g/lb).
(e.g.: 40 * (10*10/43560) / .46 = *~0.2 lbs of fert* reqd = *~91g of fert* reqd (i.e. 0.2 lbs * 453.59237 g/lb).
*Note:* ".46" (above) is urea nitrogen fertilizer (46-0-0). You could use any type of organic or commercial fertilizer that you wish (e.g.: human urine (1 fluid oz over 1 square ft adds ~13.2 ppm N to the soil), ammonium nitrate fertilizer (34-0-0), etc.
Then show visually how little fertilizer each one of those quantities actually represents, so that people can see just how little fertilizer they actually have to add to their growing area..
true!
its hard to comprehend especially if your a beginner.
+hulmil
0.32 grams is a fraction of a gram. ~1/3 of a gram (which is a very small amount of fertilizer).
To determine exactly how much that is, weigh out 1 gram of your fertilizer.
Then divide that 1 gram pile into 3 equal parts. 1 of those parts is the amount that you would spread evenly across your 6" x 6" soil area.
*Important NOTE:*
Remember, the example calculation in my original comment is for *urea nitrogen fertilizer (46-0-0)*. If you plan to use a different fertilizer, you would change the .46 in the example formula to whatever your fertilizer bag indicates (for example if your fertilizer is 20-0-0, you would use .20 in the example formula rather than .46).
Also note that volume amount (e.g. teaspoons) will vary with the density of the fertilizer. Different fertilizer formulations have different densities.
You will need to weigh out a specific volume & weigh it to determine the density of your fertilizer.
+zeffur7 Thanks for your answer.I don't have a soil report, I just want to mix up a batch of some balanced 12-12-12 fertilizer I have on hand for my tomatoes. Your method shows I need .06 gram of 12-12-12 fertilizer for each 6 inch pot (.5 *12 *454)/43560. Now if want to dissolve the fertilizer in water to make up a gallon of liquid fertilizer, how much do I add to a gallon? Do I calculate how much water I would pour on the plant? For example: If I would normally dissolve .06 gram in 1/c cup of water, and a gallon is 16 cups, then I need (16 * .06)/.5 or 19 g per gallon. (Assuming it will dissolve!) Does this make sense? Luckily I have a scale that measures in grams. Now a teaspoon of salt is 5.69 grams, so if these are comparable, 19 grams of fertilizer is roughly 3 teaspoons or one tablespoon. When I measure out the 19 grams, I will see if it is roughly a tablespoon. Hope this helps hulmil.
+JennyJardin
re: "Your method shows I need .06 gram of 12-12-12 fertilizer for each 6 inch pot (.5 *12 *454)/43560."
No, you did your math incorrectly. The proper calculation according to the example that I provided would be as follows *for the fertilizer that you plan to use*:
lbs fert reqd = 40 lbs fert/acre * ((6"/12" * 6"/12") * 1 acre/43560 sf) / .12 = *~0.002 lbs of fert* = *~0.87g of fert* reqd
e.g.: 40 * (.5*.5/43560) / .12 = *~0.002 lbs of fert* = *~0.87g of fert* reqd (i.e. 0.002 lbs * 453.59237 g/lb).
*Important Information to Remember:*
---------------------------------------------------------------
1. Adding 40 lbs of N/acre will cause the soil to increase by 20 ppm of N.
2. A tomato plant requires 60 ppm of N to produce well (as well as other nutrients too!--but, where I live I never have to add P or K to my soil as it is always above the proper levels).
3. The soil N + your fertilizer N should = 60 ppm of N for the whole growing period (usually less than 4 months).
4. It's best to ensure that your growing medium has 40 ppm of N when the seed or seedling is put into the growing medium (or soil) & then add the remaining 20 ppm of N after the plant blooms.
5. The calculation that I showed above will only provide 20 ppm of N to your soil (it was just an example problem that wasn't setup for tomato plants). If your soil has 0 ppm of N in it, then your tomato plants will have insufficient N to grow properly. You can take the computed fertilizer weight & divide it by 2. In your case that would be ~.87g/2 = .43g of fertilizer = 10 ppm. So, your tomatoes require 60 ppm & let's assume your soil has 10 ppm of N already in it, you would then calculate 60ppm - 10ppm = 50 ppm of N will need to be added to your growing medium. Since you know .43g of fert=10 ppm of N, you would calculate 50ppm / 10ppm = 5. Therefore: 5 * .43g fert = ~2.8g of fert = 50 ppm of N that you would add to one 6" square pot during the growing season (usually 2/3rds of that amount would be added when the seed or seedling is planted & the other 1/3rd of that amount would be added when the plant blooms (no addition N will need to be added for the rest of the growing period).
Note: If you intend to grow your tomato plants in pots during the whole growing season (like I do), I would suggest that you use at least a 3 gallon pot for each plant.
*Making a Solution of Liquid Fertilizer:*
.43g of 12%/wt N fert will provide a 6" square pot with 10ppm of N.
You can mix that amount (or more to get a higher N ppm concentration) with any amount of water that the fertilizer will dissolve in completely. Apply that whole solution amount to one pot.
*To make a batch of liquid fertilizer for multiple plants do this:*
1. Plants = 16 (you choose this number)
2. N ppm to add to each plant = 33.33 (this is 2/3 of 50 ppm from above)
3. *Fertilizer amt* = 33.33ppm N/10ppm = 3.33 * .43 g Fert * 16 plants *= ~23g of fert*.
4. *Water amt* = 8 fl oz for each plant * 16 plants = 128 fl oz *= 1 gallon*
5. Dissolve the fertilizer completely in the water & then water each plant with 8 fl oz of the liquid fertilizer solution.
Happy Gardening!
+zeffur7 Thanks. I am printing this out to keep handy.
This dude is making it WAAAY more confusing than it needs to be.
Hey buddy could u elaborate some more on that statement and maybe shoot me some links to a easier explanations
You're absolutely incorrect. He's making easy! You need to understand the calculation and this is very simple.
I don't agree. I like seeing the math involved. He made it very clear to me. And I suck at math.
Good presentation. But basic math is required at least a knowledge of ratio and proportion,
6inch x 6inch = 36 inch , not 6 square inch ...
6 inch x 6 inch = 36 sq inches. Since 12" x 12" = 144 square inches, we convert 36 sq in to sq ft as follows 36 sq in * (1 sf / 144 sq in) = .25 sq ft (which is what he properly calcuated in the video.
You never EXPLAINED where 0.2 came from... I ASSUME you mean it's 20% in decimal form. Please be more thorough and explain things. You also can't up with 50 something on your next calculation.
Again, where did you get 50 from?
C'mon man, jeez, I'm reaching for a oxygen mask!
video is cool,but u need to solve the math and explain in a simplest way.
Indian englis is really bad😡😡😡
well this was a waste
Stupid explaination, u cant calculate for P & K same way like N
Yes, actually, you can calulate all 3 nutrients using the same method.
Sorry but not much help to me.
You need to study more... review before you post..