Napi friss hírek

Is the room shrinks by everytime we apply one coat?

You're absolutely correct—each time we apply a coat of paint, the room's dimensions shrink slightly because the paint adds a thin layer to all interior surfaces. This means that both the surface area to be painted and the room's volume decrease with each coat. Let's refine the calculation by accounting for this effect.

### **Understanding the Shrinkage**

- **Thickness of one coat of paint (\( t \))**: 0.1 mm or 0.0001 meters.
- **Initial room dimensions**:
  - **Length (\( L_0 \))**: Approximately 4.472 meters (since \( \sqrt{20} \approx 4.472 \) for a square room).
  - **Width (\( W_0 \))**: Same as length.
  - **Height (\( H_0 \))**: 3 meters.

After each coat, the dimensions decrease:

- **Length after \( n \) coats (\( L_n \))**:
  \[
  L_n = L_0 - 2n t
  \]
- **Width after \( n \) coats (\( W_n \))**:
  \[
  W_n = W_0 - 2n t
  \]
- **Height after \( n \) coats (\( H_n \))**:
  \[
  H_n = H_0 - n t
  \]

### **Calculating the Volume of Paint Per Coat**

For each coat \( n \), the surface area to be painted is:

1. **Wall Areas**:
   - Two walls of area \( L_n \times H_n \).
   - Two walls of area \( W_n \times H_n \).
2. **Ceiling Area**:
   - Area \( L_n \times W_n \).

The total surface area \( A_n \) at the \( n \)-th coat is:

\[
A_n = 2(L_n H_n + W_n H_n) + L_n W_n
\]

The volume of paint needed for the \( n \)-th coat (\( V_n \)) is:

\[
V_n = A_n \times t
\]

### **Calculating the Cumulative Volume of Paint**

The cumulative volume of paint after \( N \) coats is:

\[
V_{\text{total}} = \sum_{n=1}^{N} V_n
\]

Our goal is to find the smallest integer \( N \) such that:

\[
V_{\text{total}} \geq V_{\text{room}}
\]

Where the initial room volume (\( V_{\text{room}} \)) is:

\[
V_{\text{room}} = L_0 \times W_0 \times H_0 = 4.472^2 \times 3 \approx 60 \, \text{m}^3
\]

### **Iterative Calculation**

Due to the complexity of the equations, we'll perform an iterative calculation using a spreadsheet or programming tool. Here's how:

1. **Initialize Variables**:
   - Set \( n = 0 \).
   - Set \( L_n = L_0 \), \( W_n = W_0 \), \( H_n = H_0 \).
   - Set \( V_{\text{cumulative}} = 0 \).

2. **Loop Until the Room is Filled**:
   - **Step 1**: Calculate \( A_n \) using the current dimensions.
   - **Step 2**: Calculate \( V_n = A_n \times t \).
   - **Step 3**: Update \( V_{\text{cumulative}} += V_n \).
   - **Step 4**: Reduce the room dimensions:
     - \( L_{n+1} = L_n - 2t \).
     - \( W_{n+1} = W_n - 2t \).
     - \( H_{n+1} = H_n - t \).
   - **Step 5**: Increment \( n \) by 1.
   - **Step 6**: Check if \( V_{\text{cumulative}} \geq V_{\text{room}} \). If so, stop the loop.

### **Sample Iteration**

To illustrate, let's perform the first few iterations manually:

#### **First Coat (\( n = 1 \))**:

- \( L_1 = L_0 - 2 \times 0.0001 = 4.4718 \, \text{m} \).
- \( W_1 = W_0 - 2 \times 0.0001 = 4.4718 \, \text{m} \).
- \( H_1 = H_0 - 0.0001 = 2.9999 \, \text{m} \).

Calculate \( A_1 \) and \( V_1 \):

- \( A_1 = 2(L_0 H_0 + W_0 H_0) + L_0 W_0 \approx 73.664 \, \text{m}^2 \) (same as before, since the dimensions haven't changed yet).
- \( V_1 = 73.664 \times 0.0001 = 0.0073664 \, \text{m}^3 \).

#### **Second Coat (\( n = 2 \))**:

- \( L_2 = L_1 - 2 \times 0.0001 = 4.4716 \, \text{m} \).
- \( W_2 = W_1 - 2 \times 0.0001 = 4.4716 \, \text{m} \).
- \( H_2 = H_1 - 0.0001 = 2.9998 \, \text{m} \).

Calculate \( A_2 \) and \( V_2 \):

- \( A_2 = 2(L_1 H_1 + W_1 H_1) + L_1 W_1 \).
- \( V_2 = A_2 \times 0.0001 \).

Continue this process until \( V_{\text{cumulative}} \geq 60 \, \text{m}^3 \).

### **Approximating the Total Number of Coats**

Given the minuscule change in dimensions per coat, the reduction in surface area and volume per coat is negligible in the early stages. However, as \( n \) becomes large, the cumulative effect becomes significant.

To approximate \( N \) without calculating each iteration:

1. **Assume Average Volume per Coat**:

   The average volume per coat decreases over time, but for an approximation, we can use the initial volume per coat \( V_1 \) and adjust for the average:

   \[
   V_{\text{average}} \approx \frac{V_1 + V_N}{2}
   \]

   Since \( V_N \) approaches zero as the room shrinks, \( V_{\text{average}} \approx \frac{V_1}{2} \).

2. **Estimate Total Coats Needed**:

   \[
   N \approx \frac{V_{\text{room}}}{V_{\text{average}}} = \frac{60}{0.0073664 / 2} \approx 16,290
   \]

This estimate suggests that accounting for the room's shrinkage approximately doubles the number of coats needed compared to our initial calculation.

### **Final Answer**

By considering the room's shrinkage with each coat, the total number of coats required to fill the room increases. The revised estimate is approximately **16,290 coats**.

### **Note on Practicality**

In practice, the difference between the initial estimate (~8,145 coats) and the revised estimate (~16,290 coats) illustrates the cumulative effect of the room shrinking. While the initial calculation provided a rough estimate, accounting for the room's shrinkage gives a more accurate (though still approximate) answer.

---

**Summary:**

- **Yes**, the room shrinks slightly with each coat of paint.
- **Adjusted Estimate**: Approximately **16,290 coats** are needed to completely fill the room when accounting for shrinkage.

---

**Disclaimer:** This is a theoretical exercise. In reality, painting a room this many times is impractical, and factors like paint drying, structural integrity, and material properties would significantly affect the outcome.

már benne van az optimizmussal töltött cselekvésre buzdító vágy. Te mész segíteni?

Ugyanitt keresnek arterben epitett luxusvillak homokzsakolasara embert.

És akkor a szagról még nem is beszéltünk. Az úgy beszívódik a falakba, h. azon a világ összes kifestése sem segít.

Ha nem nagy a szoba, ellenben nagy a költségvetés akkor esetleg a Wallkyd segíthet. Viszont ez nem diszperzites felületet ad.

Ez a számítás jónak tűnik.Érdekes belegondolni, hogy minden réteggel egyre csökken a fennmaradó felület (és a térfogat vagy légköbméter is). A vége felé már csak egy kis hasáb felületét kell kifestenünk, és annak kifestése után a maradék hasáb minden mérete két festékrétegnyivel kisebb lesz. Az eredetileg legrövidebb oldala a szobának fog előbb elfogyni, tehát utoljára két téglalapot festünk ki. De ha a legrövidebb oldal az elfogyás előtt csak egy festékrétegnyire zsugorodott - na akkor mi van? Akkor csak egy téglalapot festünk ki utoljára, és ezzel nullára csökkent a fennmaradó térfogat. Érdekes matematikai probléma, de mint láttuk a megoldása pofonegyszerű.

Tenyleg, ez egy ismert paradoxon (painter's paradox)

https://en.m.wikipedia.org/wiki/Gabriel's_horn