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Extremely simple question but I can't find the answer anywhere. What is the average maximum weight of a whole potato plant during it's growth cycle? If you know the average weight percentage of the all the tubers compared to the whole plant that would be great too :)

Also, if anyone knows a site which states the mass distribution of different plants that would be very helpful too :)

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It is a difficult question. Plants size and weight vary so match, and it depends on cultivation methods. You may want to use less potatoes (as "seed") and more place per potato, or much more potatoes, and so smaller plants. The yield is not so different. For this reason, you may find more often the yield per square meter (or square foot, or also larger surfaces).

And this is what you should look: the yield (so the potatoes). The rest of the plant weight nothing compared to the potatoes. Note: tomatoes plants seems to weight much more (per surface unit), and pumpkin plants (without the fruits) are heavy: a lot of water. And these are smaller than soil weight (e.g. to calculate static forces on "vertical gardening").

I never seen sites with such numbers. You may find "dry mass", which is often more useful, e.g. to calculate fertilizers. This is often split by fruits and rest (ev. further split in green and non-green parts).

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    Not to be disrespectful, but... Why answer if you don't even have an answer? :S I've already looked around and found dry mass and such, but I need the typical/ average weight of the whole plant. Feb 10 '20 at 16:22
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    Because there is no answer, and I explain few thing why the question cannot be really answered. Plants are not like animals which have a typical size and weight. (for me it is the same as asking "what is it the weight of an office building?"). Maybe if you explain why do you need the weight of the whole plant, we could find a good way to answer. Note: we are "gardening" and not "biology", so expect practical answers. Feb 10 '20 at 16:46
  • I'm going to Mars and I need to know the weight of my food production plant! ;D Also, obviously there's an average weight of potato plants, it's not like asking how long a string is. Feb 10 '20 at 17:37
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    As I wrote, no, there is no average weight. It depends on how you plant them (distances between plants). Remember: we cultivate everything in a very different situation than wild plants. If you are going to Mars, you would take seeds (if weight is so important) or potatoes to split and plant. [and the water is calculated in total spacecraft, because it will be recycled, so an other unexpected place where dry weight is useful]. And the weight depends on age. It begin with one shot and (maybe just) a 1/4 of a small semi-dried potato. But you can approximate as the weigh of potatoes, in all ages. Feb 11 '20 at 8:08
  • I realized when reading the other answer I had not specified exactly what type of weight I was looking for, so I updated the question a bit. The water estimation is actually one of the main reasons I am interested in the weight of the plant itself, since you have to bring the water to make sure the plant can grow, so the dry mass is only useful if you know how much water you need for the "wet mass". Also, this is concerning food production on the ship in transit to Mars. Feb 11 '20 at 22:33
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Let's break this puzzle down into pieces. We need some additional assumptions, such as the age of the plant - a fully mature plant consists of only the tubers since the haulm and root will have dried up to nothing since their job is done. A very young plant will weigh about half a potato which will be the chitted piece with sprouting eyes. So probably what you are looking for is a plant at full green maturity where the tubers are almost finished sizing up and the green part has not started to die down - a maxo plant. I just coined that word to describe the biggest size of the package.

So this maxo plant consists of about 7-10 tubers of various sizes, the green top and the root; as a rule of thumb you can say that the root volume is about the same as the green top otherwise the green top could not be supported in terms of water and nutrient. So if we can find the weight of the top then we also have approx. the weight of the root.

Potato haulms consist of thick and watery stems, one from each eye that decides to expand. A reasonable number of expanded eyes might be 10-12, and these will branch a little as they extend, but not much, certainly not as much as a tomato given its head. So imagine that potato stem sitting in front of you and assess the weight. Most of it is water - scrunch it down into a cube and how many cc do you have, let's say 5x5x5 = 125 cc.

So in kg.:

  • 10 tubers (medium Russet) ....... 1.0
  • Green top x 12 @ 125cc water .... 1.5
  • Root same as top ................ 1.5

Total 4 kilograms.

Edit: to describe the weight over time as the plant grows we need a function which evaluates to half a potato at t=0, 4 kg at t=maxo, and 1 kg at t=maturity. A sine wave won't serve because it grows too fast at the beginning. A sigmoid/exponential would be better for the first part to maxo, but then it needs damping down as the top dries out and the water is recaptured and stored for re-use with the following crop.

There are some articles on line discussing modelling of plant growth, see for example "Dynamical Models of Plant Growth" by Bessonov and Volpert which will give you keys into other work.

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  • The numbers seem a bit off here, especially the ratio of crop weight to root weight. FWIW you would expect the crop weight to be about 25 times the weight of the planted tuber (or more accurate, part of the tuber) - and a single potato cut up and planted has produced over 50kg of crop. There is no way the "roots plot tops" would have weighed another 150kg. Of course if you are growing a potato indoors in a plant pot to get a few "fresh new potatoes" for Christmas dinner, you won't get anywhere near that productivity.
    – alephzero
    Feb 10 '20 at 22:43
  • Thank you, this was the kind of answer I was looking for :) I especially liked the "maxo plant weight" term, since this is actually what I am interested in I'll update my question to accurately reflect that. Feb 11 '20 at 22:27

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