GROUP ASSIGNMENT REPORT Group OneChen Yongtao Kuang Yun Zhu ZhenxuanWang Gaonan Cheng Yuchen

2010.6.6

AGENDA1. Abstract2. Structure of the Catapult3. The Fish-bone Diagram4. Design of the Experiment5. Analysis of the Experimental Data6. Optimization7. Test of the Results and Conclusion8. Weakness and Possible Improvement9. References

ABSTRACT

•A study on the factors that influence the ejection range of the catapult

Topic: •Elucidate the influence of the factors and try

to find the best combination of the parameters which make the ejection range closest to 2.7 m

Goal:

•We found 3 combinations of the parameters and chose the best combination with std.=6.4

Experimental Result:

STRUCTURE OF THE CATAPULT

A 是橡皮筋支点的位置，有 4 个 levelB 是挡板的位置，有 6 个 levelC 是投石杆发射的位置，是个连续变量

The fish-bone diagram

DESIGN OF THE EXPERIMENT (1)• Detail of variables:

控制变量（单位）control variable (units)

正常水平与可操作范围normal level & range

测量精度与方法meas.

precision & setting error

How known?

推荐试验设置proposed settings, based on predicted

effects

对响应变量的预期效果

predicted effects (for

various responses) 橡皮筋支点

水 平 4 个位置

标号记录 1 号位置，此时弹力最大 支点越高，飞得越远

挡杆的位置

依 次 6 个位置

标号记录 投石杆与挡杆水平位置越接近越好，但投石杆不得超过挡杆

挡杆位置越靠前，球飞得越远

投 石 杆 发 射 的位置 ( 角度 )

120—180° 刻度记录精度： 0.1°

投石杆与挡杆水平位置越接近越好，但投石杆不得超过挡杆

投石杆发射位置越接近水平位置，发射距离越远

DESIGN OF THE EXPERIMENT (2)• Detail of fixed variables:

 固定变量（单位） factor (units)

期望试验设置值及允许变动范围desired

experimental level & allowable range

测量精度与方法

measurement precision How known?

变量控制方法how to

control (in experiment)

预期效果anticipated effects

水 平 放 置角 度（）

水平 仪 （ 手机自带）

使用胶带固定 实验 过 程 中车身一直保持水平

弹球 / 车身抛射角（俯视）

直尺圆规 用力方向控制 弹球 飞 出 方向平行于车身的朝向

固 定 张 力橡 皮筋（ N/cm ）

使用一根 无 不 要 将橡 皮 筋弄裂或弄断

一直 使 用 一根橡皮筋

弹球质量（ g ） 弹球质量差别小于 平 均 质 量 的5%

电子秤 剔除质量过重或者过轻的弹球

弹球质量基本均等

DESIGN OF THE EXPERIMENT (3)• Parameters.

干扰参数（单位）nuisance

factor (units)

测量精度及方法measurementprecisionHow known?

试验应对策略strategy (e.g.,

randomization,

blocking, etc.)

预期效果anticipated effects

试验地风速（ m/s ） 风速计 / 人体感觉 尽量在室内、空气流速慢的地段进行试验

避免风速对弹球的飞行距离造成影响

测距卷尺位移（ cm ） 另一个卷尺测量 固定卷尺起点于地面

避免试验过程中卷尺的位移

DESIGN OF THE EXPERIMENT (4)• Take two levels of A, B and C, we denote:• A: The hole at the highest level is -1, and the lowest be +1.

• B: The second hole is -1, and the fifth one is +1.• C: Make ， and as +1.

• Then we conduct a full factorial design.• There are 8 combinations of parameters, each of which we do 5 replications, and totally we get 40 groups of data.

ANALYSIS OF THE EXPERIMENTAL DATA (1)

• Here are the data:

Our response values are Mean and std. So totally we have 8 data.

ANALYSIS OF THE EXPERIMENTAL DATA (2)

• Main effects chart and interaction chart of Mean distance:

1-1

400

300

200

1-1

1-1

400

300

200

A

平均值

B

C

Mean 主效应图数据平均值

1- 1 1- 1

500

300

100500

300

100

A

B

C

-11

A

-11

B

Mean 交互作用图数据平均值

ANALYSIS OF THE EXPERIMENTAL DATA (3)

• Main effects chart and interaction chart of std.:

1-1

12. 5

10. 0

7. 5

5. 01-1

1-1

12. 5

10. 0

7. 5

5. 0

A

平均值

B

C

Standard devi ati on 主效应图数据平均值

1- 1 1- 1

15

10

5

15

10

5

A

B

C

-11

A

-11

B

Standard devi ati on 交互作用图数据平均值

ANALYSIS OF THE EXPERIMENTAL DATA （ 2 ）• For mean distance:

We can obtain the following model ：10. 07. 55. 02. 50. 0- 2. 5- 5. 0

99

9590

80706050403020

105

1

标准化效应

百分比

A AB BC C

因子 名称

不显著显著

效应类型

C

A

标准化效应的正态图 Mean Al pha = . 05)（响应为 ，

ANALYSIS OF THE EXPERIMENTAL DATA （ 3 ）• For standard deviation:

43210- 1- 2- 3

99

9590

80706050403020

105

1

标准化效应

百分比

A AB BC C

因子 名称

不显著显著

效应类型

C

标准化效应的正态图 Standard devi at i on Al pha = . 05)（响应为 ，

The model of std.:

OPTIMIZATION(1)• As our goal is set the range closest to 2.7m and get the

smallest std., we define our strategy as “setting one parameter to make the std. as small as possible, then adjust the other two parameters to get mean of 2.7m”.

• Solution 1: • According to the main effect

chart, we should set C to be -1, that is x3=150°.

1-1

12. 5

10. 0

7. 5

5. 01-1

1-1

12. 5

10. 0

7. 5

5. 0

A

平均值

B

C

Standard devi at i on 主效应图数据平均值

OPTIMIZATION(2)

A

B

1. 00. 50. 0- 0. 5- 1. 0

1. 0

0. 5

0. 0

- 0. 5

- 1. 0

C - 1保持值

> – – – – – < 100

100 120120 140140 160160 180180 200

200

Mean

Mean B, A 与 的等值线图

We can see that the range never reaches 2.7m when C take the smallest value, so we should adjust A and B and try to find a combination.

Solution 1 is A=1,B=5,C=158°.

Set mean with 270, then

To make x3 small, we can set x1=-1 and x2=1.

OPTIMIZATION(3)• Solution 2:• We set A with the largest value, then follow the previous

steps.

• Solution 2 is A=4,B=3,C=176.4°.B

C

1. 00000. 3333- 0. 3333- 1. 0000

1. 0

0. 5

0. 0

- 0. 5

- 1. 0

A 1保持值

> – – – – – < 2

2 44 66 88 10

10 1212

devi at i onStandard

Standard devi ati on C, B 与 的等值线图

OPTIMIZATION(4)• Solution 3:We set B with the smallest value, then follow the previous

steps:

Solution 3 is A=3,B=2,C=171°.

A

C

1. 00000. 3333- 0. 3333- 1. 0000

1. 0

0. 5

0. 0

- 0. 5

- 1. 0

B - 1保持值

> – – – < 4

4 66 88 10

10

devi at i onStandard

Standard devi ati on C, A 与 的等值线图

TEST OF THE RESULTS AND CONCLUSION(1)

• Test of solution 1• Using the model, we can obtain:• Mean distance=288.6cm, std.=12.74cm• Then we got 12 distances under this experimental condition, and here

is the histogram of these data:

280270260250

4

3

2

1

0

Resul t1

频率

均值 266. 2标准差 9. 600N 12

Resul t1 直方图（包含正态曲线）

TEST OF THE RESULTS AND CONCLUSION(2)

• Test of solution 2• Using the model, we can obtain:• Mean distance=270.9cm, std.=7.98cm• Then we got 10 distances under this experimental condition, and here

is the histogram of these data:

280275270265260255

3. 0

2. 5

2. 0

1. 5

1. 0

0. 5

0. 0

resul t2

频率

均值 266. 2标准差 6. 529N 10

resul t2 直方图（包含正态曲线）

TEST OF THE RESULTS AND CONCLUSION(3)

• Test of solution 3• Using the model, we can obtain:• Mean distance=275.7cm, std.=7.29cm• Then we got 10 distances under this experimental condition, and here

is the histogram of these data:

280275270265260255250245

3. 0

2. 5

2. 0

1. 5

1. 0

0. 5

0. 0

resul t3

频率

均值 262. 4标准差 7. 382N 10

resul t3 直方图（包含正态曲线）

TEST OF THE RESULTS AND CONCLUSION(4)

• Conclusion• After compared the performance of the above three solutions, we

can see the best solution is solution 2——A=4, B=3, C=176.4°

WEAKNESS AND POSSIBLE IMPROVEMENTS

• We didn’t consider the second-order relationship in the model.

• In our analysis, we treat the mean and std. of a five-value group as the response variables, so we only have 8 response values, and we have no replications. As a result, the model we obtained has relatively large error.

References• Design and Analysis of Experiments(Sixth Edition),

Douglas C.Montgomery, 2005

THANK YOU!

Q&A