Fish pond calculation force conversion

1. All of the above are not accurate enough. A good way to save money and improve quickly is to find someone to lead you up at level 1-10 (it's better to have one person in reality, or it will take time to find one). At the same time, you need to buy the "real vision scroll"
and then you can get double (actually you can get it when you start to learn from a teacher, but don't use it after you get it, First use the real vision)
in this way, if you continue to upgrade to level 20, you can go to the villa to buy two spirit runes and dig the strange corpse (you can double-click the bead to get experience) - new things

2. They asked about mercenaries. Are you blind? Level 10? It's a shame! It's not a pickaxe. Mercenary level 100

3. The above data are hard indicators, even if the difference, the winning rate will become 0%; On the other hand, if the defense and the king are higher, the three times of combat power sent by the boss is not enough, and the winning rate will become 0. However, everyone's attributes are not exactly the same, for example, 200 defense and 55 king can also pass. If you want to be accurate, you can calculate the boss's ability to reply to you twice by 60%, How many kings and defenses do you need to guarantee no more than 250% (the formula is the same as when calculating damage)

4. The upper limit of enhancement for mercenaries is level 30
the number of times of the upper limit of enhancement cannot be increased after full awakening. After full awakening, they can break through the four dimensions that can increase talent. You can see it in the upper left corner
if the enhancement reaches level 20, the treasure can be opened

I am a cosmic hero from [300 heroes and mining].

5. If the data is correct, the boss's four dimensions should be 108180, and the combat power is 1900000.
there are two kinds of skills. First, he will constantly release petrified gaze until he hits. The effect of this skill is to seal. The web version is to seal all the skills that can be sealed. However, the mobile game version is not clear at present. It may seal 15% or less of the skills that can not be confirmed
therefore, to be on the safe side, we use more than 20% skills for strategy
the second skill is 100% pure attack
OK, does it sound like a nightmare? Yes, it's a weakened version of the nightmare. He lacks the important characteristic of nightmare devil to return blood, so he can use fashion man, Gaia and other card groups, cooperate with Longma team, and use mining team to break it
however, the headache is the 180 king. In order to make the strategy stable (after all, I don't want to waste points or spend blood to revive), the key point is to improve the king. At the same time, it's better to attack 108 first. It's hard! On the contrary, there is no special requirement for combat power
PS: don't buy a dodge halo
the winning rate is given below, team: 6 Longma / qioba, 1 awakened Xun, 1 protagonist with no light strike, attack 108 first. Combat power and defense have little influence. Let's give 99999 combat power and 100 defense. The following are the winning rates given by different dodges and kings
it's better to have qioba. It doesn't matter if you don't have qioba. You may need to buy a halo to get the right winning rate. Wake up the smoke can not bring, 5% - 7% less winning. If the first attack is not enough, the winning rate will be 8% - 10%. Improving defense can improve some winning rate. After all, there is a seal skill in front of you. However, the priority of improvement is far less than that of dodge.

6.

In the feed self-sufficiency system, according to different farming systems, the planting area can be calculated according to the following methods

The results showed that: (1) non rotation planting was only one kind of feed in normal years, and could be fed by many kinds of herbivorous fish. For example, elephant grass, alfalfa, clover, green vegetables and so on, herbivorous fish can eat. This kind of situation should be calculated according to the following formula:

s = y · f / P · n (formula 1)

where:

S-the planting area (hectare) required for each hectare of fish pond

y -- the planned net yield of various herbivorous fish raised by the feed in the polyculture pond (kg / HA)

F -- the feed coefficient of the feed for herbivorous fish

P -- the average yield per hectare of the feed (kg / ha · stubble)

n -- the number of continuous cropping stubbles (n ≥ 1)

For example, in some areas of South China, one kind of cereal is planted all year round, and several crops are planted a year. Cereal can be eaten by all the fish, but the feed coefficients of different fish using the same feed are different. Then calculate the feed requirement according to the different feed coefficients of different fish, and then accumulate to get the molecular term of Formula 1

at the same time, plant several kinds of non rotation feed, or according to the different food habits of fish, use the above formula to calculate the planting area of each feed, and then add up; Or take one of them with the most dosage and the most reliable source; The molecular term of the above formula is calculated, that is, the feed requirement per unit area of fish pond:

m = y.f, and then the insufficient part is converted into another feed requirement according to the following formula:

MB = ma · FB / FA (formula 2)

MB -- a certain feed quantity converted from the standard feed (kg)

MA -- the part to be converted in the standard feed (i.e. the insufficient part, kg)

FB -- feed coefficient of a certain feed

FA -- feed coefficient of standard feed

Both FB and FA must be the feed coefficients of the same fish. After each feed requirement is obtained, the area of each feed is obtained by dividing it by the denominator P · n of Formula 1, and then the total area is obtained. When the feed supply in the system (fish farm) is insufficient, formula 2 can also calculate the quantity of feed to be purchased

In a fish farm in southern China, it is planned to proce 10000 kg of swallowing fish, including 7500 kg of grass carp, 1500 kg of Megalobrama amblycephala and 1000 kg of common carp. Feed depends on planting elephant grass and two crops of corn. The annual average yield of elephant grass is 337500 kg / ha, and the feed coefficient of grass carp and Megalobrama amblycephala is 30; The yield of corn per hectare is 6000 kg, and the feed coefficient of carp is 4. What is the area of elephant grass and corn

Formula 1, weevil area (HA) = (7500kg + 1500kg) × thirty ÷ three hundred and thirty-seven thousand and five hundred ÷ L = 0.8 hA

formula 2, corn planting area (HA) = 1000 × four ÷ six thousand ÷ 2 = 0.33 ha (2) rotation planting

when taking more than two kinds of feed crop rotation in the whole year, we can use the following formula to calculate the feed demand in a rotation period:

MA = yfara (formula 3)

where:

ma-a rotation period

y -- the planned net yield of all kinds of swallowing fish raised in the feed (kg / HA)

The feed coefficient of the rotation period was

fa-a

RA is the proportion of fish consumption in the whole year (%). There is a lot of information on this. Or the proportion of accumulated temperature in the rotation period to the total accumulated temperature in the whole feeding period, but the accumulated temperature below 7 ℃ and above 35 ℃ should be removed for both. According to the author's research, before the end of may in the Yangtze River Basin, the net yield of herbivorous fish is about 40% of the annual net yield in the stage of feeding ryegrass, and 60% in the stage of feeding sudangrass

the demand of a certain feed in another rotation period is calculated as follows:

MB = (1-ra) or MB = yfbrb (formula 4)

MB, FB and Rb are the demand, feed coefficient and feed intake ratio of a certain feed in another rotation period

In the Yangtze River Basin, from October to the end of May every year, the area of ryegrass is about 0.3 ha (including the bank and its slope, the same below), and the area of Sudan grass or Pennisetum Hybrid is 0.5-0.6 HA from May to October. With proper water and fertilizer management, the above forage grass can be harvested for about 120 tons, and the net yield of herbivorous fish is more than 4500 kg Omnivorous fish about 1500 kg. In the first half of the year, the planting area should be appropriately expanded, and the surplus green fodder should be stored in cellar or fast drying for the second half of the year; In the second half of the year, if the forage with higher yield is selected, the area in the second half of the year can be reced, and the area in the two rotation periods can tend to be equal

7.

The number of livestock and poultry per unit area of fish pond can be calculated as follows:

n = (y1-ry2) × C ÷ M (equation 5)

where:

n -- number of livestock and poultry per unit area of fish pond (head / hectare)

Y1 - planned net yield of filter and omnivorous fish (kg / HA). According to the research of the author and some scholars, the net yield of filter feeding fish in high-yield fertilizer fish ponds is mostly 3000-6000 kg / ha. The omnivorous fish is about 1 / 4 of silver carp and bighead carp. Therefore, Y1 should be 3750-7500kg / ha

Y2 - planned gross yield of swallowing fish (kg / HA). Here is to consider swallowing fish feces fat and water, with filter, omnivorous fish, and the amount of feces of swallowing fish is related to their weight, so the gross yield is used here

R -- the ratio of the output of filter and omnivorous fish that can be brought out by the unit gross output of swallowing fish. This proportion is related to some ecological factors. According to the author's research results for many years, the proportion of grass carp that can be raised is about 21% - 34% depending on the manure and excreta of grass carp. The proportion of other swallowing fish was similar. It can be calculated by 30% in proction

C -- the fertilizer conversion coefficient of a certain livestock and poultry manure for fish farming is also related to some ecological factors, which can be referred to the previous paper

m -- manure yield of a livestock or poultry per year or cycle (kg / head). Livestock urine can be calculated as 1 / 5-1 / 7 of its quantity (1 / 5 for high nitrogen and phosphorus content, and 1 / 7 for low nitrogen and phosphorus content)

if more than two kinds of manure are used, one kind of manure can be selected as the standard, and the ratio of fertilizer conversion coefficient of two kinds of manure can be used for mutual conversion. For example, if 1000 kg chicken manure is deficient, the conversion coefficients of chicken manure and pig manure are 4 and 10 respectively. The amount of pig manure to be replaced is 1000 × ten ÷ 4 = 2500 kg. For example, if the proction cycle of livestock and poultry is less than one year (meat livestock and meat poultry), the number of livestock and poultry heads should be calculated, and then they should be distributed according to the fertilizer requirement in different seasons and raised at an appropriate time. If the bracket of equation 5 is zero, theoretically speaking, it means that the planned gross yield of the swallowing fish can bring the planned net yield of the filtering and omnivorous fish, and the ratio of the two yields is appropriate without fertilization. In practice, it depends on the amount of feeding fish and whether the amount of defecation is enough to fertilize water. Generally, fish ponds with a water depth of 1 m to 2.5 m need 900-1500 kg / ha for swallowing fish. In addition, it depends on the bottom material of the fish pond and the amount of mud in the pond. If the bracket of equation 5 is negative, it means that the net yield of the fish that can be fed by swallowing fish is greater than the planned yield. When the loading capacity of fish pond is allowed, the planned yield of filter and omnivorous fish can be increased without fertilization. Of course, in practice, we should also consider factors such as the amount of swallowing fish and fish pond conditions

In example 3, a fish farm in the Yangtze River Basin has 10 hectares of fish ponds, with a net yield of 6000 kg per hectare, of which 2250 kg are herbivorous fish (450 kg are released), and the rest are filter food fish. The ratio of forage fish to omnivorous fish was 30 kg, and each ck proced 50 kg feces per year. The fertilizer conversion coefficient of ck feces was 10. Find the number of cks to be raised

number of cks needed:

solve the problem according to formula 5:

net output of filter and omnivorous fish (Y1) = 6000-2250 = 3750 kg / ha

yield of eating fish hair (Y2) = 2250 + 450 = 2700 kg / ha

then the number of cks needed in the farm = (3750 kg / ha - 30% × 2700 kg / HA) × ten ÷ 50 kg / year of ck × 10 ha = 5880 cks

the results of comprehensive experiment and investigation showed that the net yield per hectare in the Yangtze River Basin was 6000 kg (60% - 65% of which were filter food fish, and the rest were grass carp and black carp). Besides grass carp and black carp feeding, 1350-1500 layers, 900-1125 Kangbeier cks, 750 breeding geese, or about 60 meat pigs were provided per hectare, Or 6 alt cows weighing 450-500 kg, which can meet the fertilizer demand of fish pond. If only filter and omnivorous fish are raised and the net yield per hectare is 3750 kg (among which omnivorous fish account for 1 / 4-1 / 3), the number of livestock and poultry per hectare in fish pond should be increased by about 20% (because there is no feeding fish)

8. Mu is the unit of area, and cubic meter is the unit of volume (volume). There is no conversion between them.

if you want to calculate the volume of a fish pond, you must know how many meters the depth of the fish pond is,

1 mu ≈ 667 square meters. Multiply 667 by the depth of the fish pond to calculate the volume of the fish pond,

the weight of 1 cubic meter of water is equal to 1 ton or 1000 kg,

the weight of water can be calculated according to the volume of fish pond.

9. Don't be so careful. The big fish tank is two spoons, and the small one is half a spoonful. If it's tropical fish, it's another matter

10. 60 cm = 0.6 m
3 × one point three × 0.6 = 2.34
2.34 cubic meters of water, that is 2.34 tons Because the specific gravity of water at normal temperature is 1)