
# shelter: jacket: design



## AI convo


Objective: To build a smart winter jacket with a lithium battery and 320 cm of heating wire.

> Imagine an adult person who is wearing a normal jacket in a -10-celsius  winter. To make them feel warm, how many watts of energy should we add, so to speak, inside the jacket? This is the heat that is generated by this device.

<details>
<summary>watts</summary>

To determine the required power for a smart jacket to maintain body temperature in a -10°C environment, here's a practical breakdown: 1. **Body Heat Loss**: An adult at rest loses about **100–150 watts** of heat in extremely cold conditions (-10°C). This is the baseline power needed to maintain body temperature. 2. **Jacket Efficiency**: A smart jacket with heating elements (like wires) is not 100% efficient. Assume **30–50% efficiency** for the system (due to losses in the heating elements, insulation, and materials). This means only a fraction of the input power becomes usable heat. 3. **Required Input Power**: To offset the 100–150 watts of heat loss, the jacket must input **more than that** to account for inefficiency. - At **30% efficiency**: 

$$ \text{Input Power} = \frac{100\ \text{W}}{0.3} \approx 333\ \text{W} $$

 - At **50% efficiency**: 

$$ \text{Input Power} = \frac{100\ \text{W}}{0.5} = 200\ \text{W} $$

 4. **Practical Consideration**: You'll need a lithium battery capable of supplying **200–300 watts of continuous power**. However, real-world variables (e.g., jacket insulation, ambient wind, user activity) might increase this requirement. **Answer**: The jacket needs **approximately 200–350 watts of input power**, depending on efficiency and environmental factors.


</details>

- an adult at rest in extremely cold conditions (-10°C) loses about 100–150 watts of heat.
- 30 – 50% efficiency -> 200-350 watts

> If we use a 3 x 26650, 5000 mAh 5C, 3.7 V/4.2V li-ion battery, connected in series, what is the nominal current that is drawn from the battery?

<details>
<summary>current</summary>

To determine the **nominal current** drawn from the battery, we need to clarify what you mean by "nominal current." This could refer to: 1. **Maximum sustainable current** the battery can supply (based on its **C-rate**). 2. **Actual current** drawn by the jacket, which depends on its power requirements and voltage. ### 1. **Maximum Sustainable Current** The battery pack is 3 cells in **series**, each with a **5C rating**. - **5C** means the battery can charge or discharge at **5 times its capacity**. - Each cell has a capacity of **5000 mAh (5 Ah)**. - At **5C**, each cell can supply **5 × 5 Ah = 25 A**. - Since the cells are in series, the **total current capacity** of the pack remains **25 A** (the same as one cell, as series connection does not add current capacity). **Result**: - **Maximum sustainable current** from the pack: **25 A**. ### 2. **Actual Current Drawn by the Jacket** If the jacket requires **200 W** of power (as calculated earlier), and the battery pack supplies **3.7 V** (nominal voltage), the current drawn is: 

$$ I = \frac{P}{V} = \frac{200\ \text{W}}{3.7\ \text{V}} \approx 54\ \text{A} $$

 **Result**: - **Actual current drawn** (if the jacket requires 200 W): **~54 A**. ### Key Notes: - The **5C rating** defines the **maximum current the battery can supply** (25 A in this case). - The **actual current** depends on the jacket's power draw (e.g., 200 W) and the voltage (3.7 V). - If the jacket draws **54 A**, it would exceed the battery's **25 A maximum sustainable current**, risking overheating or damage. **Final Answer**: - **Maximum sustainable current**: **25 A** (based on 5C rating). - **Actual current drawn**: **~54 A** (if the jacket requires 200 W). Let me know if you'd like to explore safety margins or battery lifespan impacts.


</details>

- 5C x 5000 mAh -> each cell can supply 25 A
- 200 W == 18 A

> For how long can we keep this device on?

<details>
<summary>operation-time</summary>

To calculate **runtime** (how long the jacket can stay on), we need to consider: - **Battery energy storage** (in watt-hours, Wh). - **Power consumption** of the jacket (in watts, W). ### Step-by-Step Calculation: 1. **Battery Capacity**: - 3 cells in series: **3 × 5 Ah = 15 Ah** total capacity. - Nominal voltage: **3.7 V × 3 = 11.1 V**. - **Energy storage**: 

$$ \text{Energy} = \text{Capacity} × \text{Voltage} = 15\ \text{Ah} × 11.1\ \text{V} = 166.5\ \text{Wh} $$

 2. **Jacket Power Draw**: - Assumed earlier: **200 W** (to offset heat loss and inefficiency). 3. **Runtime**: 

$$ \text{Runtime} = \frac{\text{Energy}}{\text{Power}} = \frac{166.5\ \text{Wh}}{200\ \text{W}} \approx 0.83\ \text{hours} $$

 - **0.83 hours ≈ 50 minutes**. ### Key Notes: - This is a **theoretical maximum**. Real-world runtime is shorter due to: - **Battery inefficiency** (e.g., 90% efficiency, reducing stored energy by ~10%). - **Voltage drop** during discharge (cells may not maintain 3.7 V fully). - **Jacket power draw** (actual draw could be higher due to temperature fluctuations or system overhead). ### Adjusted Estimate: - With **90% efficiency** and a **10% voltage drop**, runtime drops to **~40 minutes**. - If the jacket draws **250 W** (e.g., due to higher ambient temperatures or system inefficiency), runtime drops to **~30 minutes**. Would you like to refine this further with specific assumptions (e.g., battery efficiency, jacket power draw, or voltage)?


</details>

- at 200 W: ~ 1 hour

> What is the range of resistance per meter for heating wires?

<details>
<summary>resistance</summary>

The resistance per meter of heating wire depends on the **material**, **diameter**, and **alloy composition**. Here's a breakdown of typical ranges: ### 1. **Resistivity (ρ)** Resistivity (Ω·m) determines the material's inherent resistance to current. For heating wires: - **Nichrome (common alloy)**: ~1.0 × 10⁻⁶ Ω·m (1.0 × 10⁻⁶ ohm-meters). - **Kanthal (another common alloy)**: ~1.5 × 10⁻⁶ Ω·m. - **Carbon fiber (used in some applications)**: ~1.0 × 10⁻⁵ Ω·m (higher than Nichrome). ### 2. **Resistance per Meter (R/m)** Resistance per meter is calculated using the formula: 

$$ R = \rho \times \frac{L}{A} $$

 Where: - $ \rho $ = resistivity (Ω·m), - $ L $ = length (m), - $ A $ = cross-sectional area (m²). #### Example Calculation: For a Nichrome wire with a **diameter of 0.5 mm (0.0005 m)**: - Radius $ r = 0.25 \, \text{mm} = 0.00025 \, \text{m} $, - Cross-sectional area $ A = \pi r^2 \approx 1.96 \times 10^{-7} \, \text{m²} $. 

$$ R = 1.0 \times 10^{-6} \, \Omega \cdot m \times \frac{1 \, \text{m}}{1.96 \times 10^{-7} \, \text{m²}} \approx 5.1 \, \Omega/m $$

 ### 3. **Typical Ranges** - ** Nichrome**: ~ **5–10 Ω/m** (for common diameters like 0.5 mm). - ** Kanthal**: Slightly higher (e.g., ~6–12 Ω/m). - **Thinner wires** (smaller diameter): Higher resistance per meter. - **Thicker wires**: Lower resistance per meter. ### 4. **Practical Considerations** - Heating wires are often manufactured with **specific resistance values** tailored for their intended use (e.g., 10 Ω/m for a jacket). - Always consult manufacturer specs for precise values, as alloys can vary (e.g., Nichrome-80 vs. Nichrome-90). If you need help calculating the total resistance for a specific wire length or power requirements, let me know!


</details>

common alloys:
- Nichrome: ~1.0 × 10⁻⁶ Ω·m
	- 5–10 Ω/m, for diameter=0.5 mm
- Kanthal: ~1.5 × 10⁻⁶ Ω·m.
	- ~6–12 Ω/m, for diameter=0.5 mm
- Carbon fiber: ~1.0 × 10⁻⁵ Ω·m

## summary

- n parallel heating wires, totaling $L$ meters -> each is $L/n$ long and with a resistance of $R$ = $L\rho/n$ (Ω).
- total power: P (w)

$$P=n\frac{v^2}{\frac{L\rho}{n}}=n^2\frac{v^2}{L\rho}$$

or,

$$\rho=\frac{n^2v^2}{LP}$$


at $P=100$ W, $L=3.2 m$, $v=3*3.7=11.1 V$, $n=4$ -> $\rho=6$ Ω/m ⭐️

## build

```bash
@sbc build_README \
	ai,root=shelter/jacket/design

@upload - jacket-design-v3
@upload public,zip jacket-design-v3
```

[jacket-design-v3.tar.gz](https://kamangir-public.s3.ir-thr-at1.arvanstorage.ir/jacket-design-v3.tar.gz)


---
👾 Xerxes-1 | bluer_agent-5.463.1 | bluer_ai-12.633.1-main | bluer_objects-6.582.1 | bluer_options-5.345.1 | blueness-3.122.1 | torch-2.2.1 | numpy-1.26.4 | Python 3.12.9 | Darwin 23.6.0 | dev | kamangir | Sion | access: cloud, web, pypi
