Give step-by-step solution with explanation and final answer:
Act like a helpful tutor and exlain me :
consider a = 2 , b = 6 , c = 1 , d = 4 , e = 8 , f = 1 , g = 5Consider your six-digit ID number

Question

Give step-by-step solution with explanation and final answer:
Act like a helpful tutor and exlain me :
consider a = 2 , b = 6 , c = 1 , d = 4 , e = 8 , f = 1 , g = 5Consider your six-digit ID number = a-b-c-d-e--g

Problem 1:

Afive-axis robot needs to perform a two-stage movement. In the first stage, the first joint
needs to go from initial angle of abe to a final angle of fg° in 5 seconds. In the second
stage, the first joint needs to go from fg° to acd? in 3 seconds.

The robot is at rest at the start of the whole process. The angular speed needs to be w/6
rad/s when passing fg. It will be fully stopped at acd. Find the corresponding third-order
polynomials for both stage 1 and stage 2.

VIPSolutions · Accepted Answer

Let's break the problem into clear steps and solve for the third-order polynomials (cubic polynomials) for each stage.

---

## **Step 1: Substitute Your ID Values**
Given:  
$ a = 2, \ b = 6, \ c = 1, \ d = 4, \ e = 8, \ f = 1, \ g = 5 $

Your six-digit ID = **2-6-1-4-8-1-5**  
But as per the problem, the format is $ a-b-c-d-e-f-g $.

---

## **Step 2: Find the Required Angles**

- **Stage 1:**  
  Initial angle: $ ab^\circ = 2 \times 6 = 12^\circ $  
  Final angle: $ fg^\circ = 1 \times 5 = 5^\circ $  
  Duration: $ t_1 = 5 $ s

- **Stage 2:**  
  Initial angle: $ fg^\circ = 5^\circ $  
  Final angle: $ acd^\circ = a \times c \times d = 2 \times 1 \times 4 = 8^\circ $  
  Duration: $ t_2 = 3 $ s

- **Speed at $ fg^\circ $:**  
  $ \omega = \frac{\pi}{6} $ rad/s at $ fg^\circ $

---

## **Step 3: Convert Degrees to Radians**

Because angular speed is given in **radians/sec**, convert all angles:
- $ \text{deg} \to \text{rad}: $ $ x^\circ = x \times \frac{\pi}{180} $ rad

| Angle         | Value (deg) | Value (rad)         |
|---------------|-------------|---------------------|
| $ ab^\circ $ | 12         | $ 12 \times \frac{\pi}{180} = \frac{\pi}{15} $ |
| $ fg^\circ $ | 5          | $ 5 \times \frac{\pi}{180} = \frac{\pi}{36} $  |
| $ acd^\circ $| 8          | $ 8 \times \frac{\pi}{180} = \frac{2\pi}{45} $ |

---

## **Step 4: Define the Cubic Polynomial**
The general cubic polynomial for joint angle:
$$
\theta(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3
$$
Boundary conditions:
- $ \theta(0), \ \dot{\theta}(0), \ \theta(T), \ \dot{\theta}(T) $

---

## **Step 5: Stage 1 Polynomial ($ t = 0 $ to $ t = 5 $ s)**

**Boundary Conditions:**
1. $ \theta_1(0) = ab^\circ = \frac{\pi}{15} $
2. $ \dot{\theta}_1(0) = 0 $    (starts at rest)
3. $ \theta_1(5) = fg^\circ = \frac{\pi}{36} $
4. $ \dot{\theta}_1(5) = \frac{\pi}{6} $ (required speed at $ fg^\circ $)

Let $ \theta_1(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 $.  
Take derivatives:
$$
\dot{\theta}_1(t) = a_1 + 2a_2 t + 3a_3 t^2
$$

**Plug in boundary conditions:**

1. $ t = 0: \ \theta_1(0) = a_0 = \frac{\pi}{15} $
2. $ t = 0: \ \dot{\theta}_1(0) = a_1 = 0 $
3. $ t = 5: \ \theta_1(5) = a_0 + a_1 \times 5 + a_2 \times 25 + a_3 \times 125 = \frac{\pi}{36} $
4. $ t = 5: \ \dot{\theta}_1(5) = a_1 + 2a_2 \times 5 + 3a_3 \times 25 = \frac{\pi}{6} $

Using $ a_0, a_1 $:
- $ a_0 = \frac{\pi}{15} $
- $ a_1 = 0 $

Now, plug into 3 and 4:
$$
\frac{\pi}{15} + 25a_2 + 125a_3 = \frac{\pi}{36}
$$
$$
0 + 10a_2 + 75a_3 = \frac{\pi}{6}
$$

Solve these two equations:

**Equation 1:**  
$ 25a_2 + 125a_3 = \frac{\pi}{36} - \frac{\pi}{15} = \frac{5\pi - 12\pi}{180} = -\frac{7\pi}{180} $

So,
$$
25a_2 + 125a_3 = -\frac{7\pi}{180}
$$

**Equation 2:**  
$ 10a_2 + 75a_3 = \frac{\pi}{6} $

---

**Solve the system:**

Multiply Equation 2 by 2.5 to match coefficients for $ a_2 $:

- $ 2.5 \times 10a_2 = 25a_2 $
- $ 2.5 \times 75a_3 = 187.5a_3 $
- $ 2.5 \times \frac{\pi}{6} = \frac{2.5\pi}{6} = \frac{5\pi}{12} $

So:
$$
25a_2 + 187.5a_3 = \frac{5\pi}{12}
$$
Subtract Equation 1:
$$
(25a_2 + 187.5a_3) - (25a_2 + 125a_3) = \frac{5\pi}{12} + \frac{7\pi}{180}
$$
$$
(187.5 - 125)a_3 = \frac{5\pi}{12} + \frac{7\pi}{180}
$$
$$
62.5a_3 = \frac{5\pi}{12} + \frac{7\pi}{180}
$$

Find common denominator (180):

- $ \frac{5\pi}{12} = \frac{75\pi}{180} $
- $ \frac{7\pi}{180} $

So,
$$
62.5a_3 = \frac{75\pi + 7\pi}{180} = \frac{82\pi}{180}
$$
$$
a_3 = \frac{82\pi}{180 \times 62.5} = \frac{82\pi}{11250}
$$

Now, substitute $ a_3 $ into Equation 2:
$$
10a_2 + 75 \times \frac{82\pi}{11250} = \frac{\pi}{6}
$$
$$
10a_2 + \frac{6150\pi}{11250} = \frac{\pi}{6}
$$
$$
10a_2 = \frac{\pi}{6} - \frac{6150\pi}{11250}
$$

Convert $ \frac{\pi}{6} $ to denominator 11250:
- $ \frac{\pi}{6} = \frac{1875\pi}{11250} $

So,
$$
10a_2 = \frac{1875\pi - 6150\pi}{11250} = \frac{-4275\pi}{11250}
$$
$$
a_2 = \frac{-4275\pi}{112500} = -\frac{57\pi}{1500}
$$

**Stage 1 Polynomial:**
$$
\boxed{
\theta_1(t) = \frac{\pi}{15} - \frac{57\pi}{1500} t^2 + \frac{82\pi}{11250} t^3
}
$$
Where $ t $ in seconds, valid for $ t \in [0,5] $.

---

## **Step 6: Stage 2 Polynomial ($ t = 0 $ to $ t = 3 $ s)**

**Boundary conditions:**
1. $ \theta_2(0) = fg^\circ = \frac{\pi}{36} $
2. $ \dot{\theta}_2(0) = \frac{\pi}{6} $ (matching speed from stage 1's end)
3. $ \theta_2(3) = acd^\circ = \frac{2\pi}{45} $
4. $ \dot{\theta}_2(3) = 0 $ (stopped at final position)

Let $ \theta_2(t) = b_0 + b_1 t + b_2 t^2 + b_3 t^3 $, $ t $ from 0 to 3.

- $ \dot{\theta}_2(t) = b_1 + 2b_2 t + 3b_3 t^2 $

Plug in conditions:
1. $ b_0 = \frac{\pi}{36} $
2. $ b_1 = \frac{\pi}{6} $
3. $ b_0 + b_1 \times 3 + b_2 \times 9 + b_3 \times 27 = \frac{2\pi}{45} $
4. $ b_1 + 2b_2 \times 3 + 3b_3 \times 9 = 0 $

So,
- $ b_0 = \frac{\pi}{36} $
- $ b_1 = \frac{\pi}{6} $

**From condition 3:**
$$
\frac{\pi}{36} + \frac{\pi}{6} \times 3 + 9b_2 + 27b_3 = \frac{2\pi}{45}
$$
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