GROUP ASSIGNMENT
REPORT
Group One
Chen Yongtao
Wang Gaonan
2010.6.6
Kuang Yun Zhu Zhenxuan
Cheng Yuchen
AGENDA
1. Abstract
2. Structure of the Catapult
3. The Fish-bone Diagram
4. Design of the Experiment
5. Analysis of the Experimental Data
6. Optimization
7. Test of the Results and Conclusion
8. Weakness and Possible Improvement
9. References
ABSTRACT
Topic:
Goal:
Experimental
Result:
• A study on the factors that influence the
ejection range of the catapult
• 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
• We found 3 combinations of the
parameters and chose the best
combination with std.=6.4
STRUCTURE OF THE CATAPULT
A是橡皮筋支点的位置，有4个level
B是挡板的位置，有6个level
C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
橡皮筋支点
水平4个位
置
挡杆的位置
依次6个位
置
投 石 杆 发 射 的 120—
180°
位置(角度)
测量精度
与方法
meas.
precision
& setting
error
How known?
标号记录
标号记录
刻度记录
精度：0.1°
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)
1号位置，此时弹力最大
支点越高，飞得
越远
投石杆与挡杆水平位置越 挡 杆 位 置 越 靠
接近越好，但投石杆不得 前，球飞得越远
超过挡杆
投石杆与挡杆水平位置越 投石杆发射位置
接近越好，但投石杆不得 越接近水平位置，
超过挡杆
发射距离越远
DESIGN OF THE EXPERIMENT (2)
• Detail of fixed variables:
期望试验设置值
固定变量（单位） 及允许变动范围
factor
desired
(units)
experimental
level &
allowable
range
水平放置角度
（°）
±2°
弹球/车身抛射
角（俯视）°
±2°
测量精度与方
法
measurement
precision
How known?
变量控制方法
how to
control (in
experiment)
水平仪（手机
自带）
使用胶带固定
直尺圆规
固 定 张 力 橡 皮 使用一根
无
筋（N/cm）
弹球质量（g）
弹球质量差别小 电子秤
于平均质量的5%
用力方向控制
预期效果
anticipated
effects
实验过程中车
身一直保持水平
弹球飞出方向
平行于车身的朝
向
不要将橡皮筋 一直使用一根
弄裂或弄断
橡皮筋
剔除质量过重或 弹球质量基本均
者过轻的弹球
等
DESIGN OF THE EXPERIMENT (3)
• Parameters.
干扰参数（单位）
nuisance
factor (units)
试验地风速（m/s）
测距卷尺位移（cm）
测量精度及方法
measurement
precision
How known?
风速计/人体感觉
另一个卷尺测量
试验应对策略
strategy (e.g.,
randomization,
blocking, etc.)
预期效果
anticipated effects
尽 量 在 室 内 、空 避免风速对弹球的飞
气流速慢的地段进 行距离造成影响
行试验
固 定 卷 尺 起 点于 避免试验过程中卷尺
地面
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 150° as − 1，and 180° as +1.
• Then we conduct a 23 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:
Mean 主效应图
Mean 交互作用图
数据平均值
数据平均值
A
B
-1
1
-1
1
400
500
300
300
A
平均值
200
100
500
-1
1
C
-1
1
300
B
400
100
300
200
C
-1
1
A
-1
1
B
-1
1
ANALYSIS OF THE EXPERIMENTAL
DATA (3)
• Main effects chart and interaction chart of std.:
Standard deviation 主效应图
Standard deviation 交互作用图
数据平均值
数据平均值
A
-1
B
1
-1
1
12.5
15
10.0
10
A
7.5
平均值
A
-1
1
5
15
5.0
-1
1
C
12.5
-1
1
10
B
5
10.0
7.5
C
5.0
-1
1
B
-1
1
ANALYSIS OF THE EXPERIMENTAL
DATA （2）
• For mean distance:
标准化效应的正态图
（响应为 Mean，Alpha =
.05)
99
效应类型
不显著
显著
95
90
C
80
百分比
70
60
50
40
30
20
A
10
5
1
-5.0
-2.5
0.0
2.5
标准化效应
5.0
We can obtain the following model：
7.5
10.0
因子
A
B
C
名称
A
B
C
ANALYSIS OF THE EXPERIMENTAL
DATA （3）
• For standard deviation:
标准化效应的正态图
（响应为 Standard deviation，Alpha =
.05)
99
效应类型
不显著
显著
95
90
因子
A
B
C
C
80
百分比
70
60
50
40
30
20
10
5
1
-3
-2
The model of std.:
-1
0
1
标准化效应
2
3
4
名称
A
B
C
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°.
Standard deviation 主效应图
数据平均值
A
B
12.5
10.0
平均值
7.5
5.0
-1
1
C
12.5
10.0
7.5
5.0
-1
1
-1
1
OPTIMIZATION(2)
Mean 与 B, A 的等值线图
1.0
100
120
140
160
180
B
0.5
0.0
保持值
C -1
-0.5
-1.0
-1.0
-0.5
0.0
A
0.5
1.0
Mean
<
–
–
–
–
–
>
100
120
140
160
180
200
200
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.
Set mean with 270, then
To make x3 small, we can set x1=-1
and x2=1.
Solution 1 is A=1,B=5,C=158°.
OPTIMIZATION(3)
• Solution 2:
• We set A with the largest value, then follow the previous
steps.
Standard deviation 与 C, B 的等值线图
1.0
Standard
deviation
<
2
2 –
4
4 –
6
6 –
8
8 – 10
10 – 12
>
12
C
0.5
0.0
保持值
A 1
-0.5
-1.0
-1.0000
-0.3333
0.3333
B
• Solution 2 is A=4,B=3,C=176.4°.
1.0000
OPTIMIZATION(4)
• Solution 3:
We set B with the smallest value, then follow the previous
steps:
Standard deviation 与 C, A 的等值线图
1.0
Standard
deviation
<
4
4 –
6
6 –
8
8 – 10
>
10
C
0.5
保持值
B -1
0.0
-0.5
-1.0
-1.0000
-0.3333
0.3333
A
Solution 3 is A=3,B=2,C=171°.
1.0000
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:
R e s u l t 1 直方图（包含正态曲线）
均值
266.2
标准差 9.600
N
12
4
频率
3
2
1
0
250
260
270
Result1
280
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:
result2 直方图（包含正态曲线）
均值
266.2
标准差 6.529
N
10
3.0
2.5
频率
2.0
1.5
1.0
0.5
0.0
255
260
265
270
result2
275
280
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:
result3 直方图（包含正态曲线）
均值
262.4
标准差 7.382
N
10
3.0
2.5
频率
2.0
1.5
1.0
0.5
0.0
245
250
255
260
265
result3
270
275
280
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