What Is The Most Effective Helicopter Lab Hypothesis
Helicopter Lab
November, 2012
Kevin, Lucy, Charlotte
Objective or Question
What is the most effective helicopter design for landing the slowest?
Hypothesis
The most effective helicopter design that allows itself land the slowest would be a cylindrical cone with many small, rectangular wings attached to its base.
Equipment & Materials
Materials: Equipment:
Index cards Scissors
Scotch tape Timer
Paperclips Measuring tape Meter stick
Procedure:
1). Fold an index card into a cone shape.
2). Take another index card and cut out rectangle shaped wings
3). Tape them onto the base of the cone
4). Measure the height from the balcony to the floor
5). Drop the helicopter from the balcony and time it.
Data …show more content…
& Observations
# Wings:
#Wings Time (sec) Average time (sec) Height of drop (m)
7 2.46, 2.69, 3.11 2.75 5.36 m
6 3.21, 2.76, 2.26 2.74
5 3.17, 3.08, 2,70 2.90
Mass (with seven wings):
Mass (paperclips) Time (sec) Average Time (sec) Height of drop (m)
0 2.46, 2.69, 3.11 2.75 5.36 m
1 2.40, 3.27, 3.23 2.90
2 2.89, 2.44, 2.06 2.46
Calculations
#Wings: Mass:
7 wings: (2.46+2.69+3.11)/(3) = 2.75 0 paperclips: (2.46+2.69+3.11)/(3) = 2.75
6 wings: (3.21+2.76+2.26)/(3) = 2.74 1 paperclip: (2.40+3.27+3.23)/(3) = 2.90
5 wings: (3.17+3.08+2.70)/(3) = 2.90 2 paperclips: (2.89+2.44+2.06)/(3) = 2.46
Discussions & …show more content…
Conclusions
Couple days ago, we started a lab on making a paper helicopter that will land the slowest.
We had couple different designs. We first had the original paper helicopter design. We cut a line through 2/3 of the index card and fold them. Then we had a different design which is more effective: a cylindrical cone that has rectangles of wings taped to its base. We predicted that more wings the cylindrical cone has would cost it faster time to land. First, we tested the drop with the helicopter that has seven wings. It ended up with an average of 2.75 seconds. We were confused, caused that didn’t follow the rule of “the greater the mass, less the acceleration.” Then, we took away one wing and tested then it ended up with an average of 2.74 seconds. In the end, we took away one more wing and tested. It ended up with an average of 2.90 seconds. Then we tested it out by adding paperclips onto a seven-winged helicopter. Without any paperclips, the result is 2.75 seconds. Then we added a paperclip, the result was 2.90 seconds.In the end, we added two paperclips to the helicopter which was 2.46 seconds. By then we realized because of our helicopter’s unique design, the changes of small masses didn’t affect as much it should have, however the way you drop helicopter could really affect its flight. Therefore following the rule of F(force)=Ma(mass times acceleration), the more mass you added onto the helicopter, the longer time it will take. We didn’t do a good job on dropping it and our
helicopter had a unique design, so the datas weren’t perfect.
November, 2012
Kevin, Lucy, Charlotte
Objective or Question
What is the most effective helicopter design for landing the slowest?
Hypothesis
The most effective helicopter design that allows itself land the slowest would be a cylindrical cone with many small, rectangular wings attached to its base.
Equipment & Materials
Materials: Equipment:
Index cards Scissors
Scotch tape Timer
Paperclips Measuring tape Meter stick
Procedure:
1). Fold an index card into a cone shape.
2). Take another index card and cut out rectangle shaped wings
3). Tape them onto the base of the cone
4). Measure the height from the balcony to the floor
5). Drop the helicopter from the balcony and time it.
Data …show more content…
& Observations
# Wings:
#Wings Time (sec) Average time (sec) Height of drop (m)
7 2.46, 2.69, 3.11 2.75 5.36 m
6 3.21, 2.76, 2.26 2.74
5 3.17, 3.08, 2,70 2.90
Mass (with seven wings):
Mass (paperclips) Time (sec) Average Time (sec) Height of drop (m)
0 2.46, 2.69, 3.11 2.75 5.36 m
1 2.40, 3.27, 3.23 2.90
2 2.89, 2.44, 2.06 2.46
Calculations
#Wings: Mass:
7 wings: (2.46+2.69+3.11)/(3) = 2.75 0 paperclips: (2.46+2.69+3.11)/(3) = 2.75
6 wings: (3.21+2.76+2.26)/(3) = 2.74 1 paperclip: (2.40+3.27+3.23)/(3) = 2.90
5 wings: (3.17+3.08+2.70)/(3) = 2.90 2 paperclips: (2.89+2.44+2.06)/(3) = 2.46
Discussions & …show more content…
Conclusions
Couple days ago, we started a lab on making a paper helicopter that will land the slowest.
We had couple different designs. We first had the original paper helicopter design. We cut a line through 2/3 of the index card and fold them. Then we had a different design which is more effective: a cylindrical cone that has rectangles of wings taped to its base. We predicted that more wings the cylindrical cone has would cost it faster time to land. First, we tested the drop with the helicopter that has seven wings. It ended up with an average of 2.75 seconds. We were confused, caused that didn’t follow the rule of “the greater the mass, less the acceleration.” Then, we took away one wing and tested then it ended up with an average of 2.74 seconds. In the end, we took away one more wing and tested. It ended up with an average of 2.90 seconds. Then we tested it out by adding paperclips onto a seven-winged helicopter. Without any paperclips, the result is 2.75 seconds. Then we added a paperclip, the result was 2.90 seconds.In the end, we added two paperclips to the helicopter which was 2.46 seconds. By then we realized because of our helicopter’s unique design, the changes of small masses didn’t affect as much it should have, however the way you drop helicopter could really affect its flight. Therefore following the rule of F(force)=Ma(mass times acceleration), the more mass you added onto the helicopter, the longer time it will take. We didn’t do a good job on dropping it and our
helicopter had a unique design, so the datas weren’t perfect.