Monday, March 31, 2014

Proving Pascal’s Principle With Syringe Hydraulics

The Syringe Hydraulics Arm project on my website has been one of the most popular project articles. Lately I have been working on building a simulator to demonstrate Pascal’s Principle of fluids using syringes and plastic tubing. 

“Pascal's Principle states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container.”

What exactly does this mean in practice?  For the simulator I used a large syringe that has a piston cross section diameter of 34 mm and small syringe with cross section diameter of 13 mm.  Like other mechanical systems there is a mechanical advantage where distance moved and force tradeoff. When the smaller piston is pushed with a force, that force is distributed equally across the larger piston cross section causing a greater net force.   For the fluid to be spread across the larger cross section more fluid volume must be moved from the smaller cylinder.

For my first experiment I worked from the other direction and pushed the large cylinder a short distance of 8 mm which extended the small cylinder a much longer distance of around 45 mm until it could not move any farther. I calculated this also which was off somewhat from my observations which can happen when there are inaccuracies in the measured values.

Cylinder at Start Position

Large Cylinder Moved 8 mm Small Cylinder Moved 45 mm

For calculations we need the formula for the area of a circle :

Area of Circle = π x radius²
Large Piston   cross section area = 3.14 x (34/2)²  =   907 sq mm
Small Piston  cross section area = 3.14 x (13/2)² =  133 sq mm

Moving the large piston 8 mm will displace amount fluid = cross section x length of movement

Fluid Displaced = 907 sq mm x 8 mm = 7256 cu mm 
The movement of the small cylinder should be the fluid displaced / cross section area of small cylinder.
7256 / 133 = 54 mm   movement of small cylinder
Actual movement was recorded at 45 mm or 4.5 centimeters

I have not checked the amount of force generated but did check the amount required just to move the opposite cylinder.  Moving the small cylinder with the large cylinder took a large amount of force, 1250 grams or around 12 newtons. This is like pushing down on the short end of a lever.  Pushing the small cylinder took very little force.

Large Force Needed to Move Small Cylinder Pushing Large Cylinder

Small Amount of Force to Move Pushing Small Cylinder

Bill Kuhl

Tuesday, March 25, 2014

Progressing From Projects to Practical Application Garage Door Opener Repair

With all the projects on my website and the project building classes I have taught I sometimes wonder if the knowledge gained has directly benefitted anyone in their lives. The process of learning is always a good thing and the only feedback I have received from the students is that they enjoyed the experience. I feel that working with the various projects has improved my skills for home repair projects because the more items you struggle to put together and deal with any issues the better you will be in solving new challenges.

Gear That was in Bad Shape

This cold winter I was having problems with my automatic garage door opener not wanting to go down completely when the temperature was below about 15 degrees. Looking for the easy solution I found the proper spray to lubricate the hinges with, this might have helped but I was afraid there might be more to the problem. There was, a week ago the door went down about three feet and just stopped. The motor would run but nothing would move.

Pile of White Gear Dust Can be Seen

Searching for the symptoms on the Internet I soon found many people were having this problem because the plastic gears in the garage door opener were ground up. Opening the case of the garage door opened confirmed this was the problem. One good tutorial video for changing the gears was provided by Gate House Supplies so I ordered a part kit from them.   Video Link

New Gear in Place

The ideal way to change out the gears is to take the opener down and work on a workbench, I ignored the good advice and worked on it hanging from the garage ceiling in the cold. In the video the parts came off rather easily, I had a heck of a time getting a roll pin out and a bushing off a motor shaft. It also helps to have just the right tools as some of the screws and nuts are a little hard to reach. Another common problem is to not have the interrupter cup pushed in far enough and then the door only moves a small distance. Having read that on the Internet, I was able to fix that problem fairly easily.

Black Interrupter Cup Must be on Shaft Far 

Now I had the door going up and down but it would go right back up when it hit the bottom. After much adjusting with no success I noticed a contact in the limiting device had fallen out.  With the contact back in place the garage door was working wonderful, now I just need to put the case back together.

Some people might read this and think what an awful frustrating experience, but I had so much fun with the challenge and feel such a sense of accomplishment. It also gave me an appreciation for the experienced technicians that could no doubt do the task in a fraction of the time.

Bill Kuhl

Thursday, March 20, 2014

Torque and Lever Arm Calculations for a Mousetrap Car

Torque has been defined either as a twisting force or the tendency to rotate around an axis.  A common example of torque is tightening a bolt with a wrench.  To know how much torque is being applied to a bolt a mechanics will often use a special type of wrench known as a “torque wrench” so a specified amount of torque can be applied to a bolt.

The formula for torque is very simple if the force is applied perpendicular to the lever: Torque = radius x force.  Normally units of torque are foot pounds or newton meters.  This equation gives the torque applied to the pivot point, this is the mechanical advantage concept.

I have been thinking more about how math could be used to predict an outcome. To start with a mousetrap car if you could calculate what force is available at the end of the lever arm based on the torque at the axis it would give you some idea on how much force is available to propel the mousetrap car.  It is also interesting to see how the force decreases as the spring unwinds. 

For a mousetrap car that will go a long distance the lever arm needs to be longer to pull more line that is wrapped around the driving axle.  From these calculations it can be easily seen that the amount of force available rapidly diminishes as the length of the lever arm increases. 

To me it is also interesting if you can measure calculated outcomes and think about the reasons for inaccuracies.  For this experiment inaccuracy was related to the spring scale I was using and how I was using it.  Doc Fizzix sells a torsion wheel to measure the torque of the mousetrap spring.  Link to Torsion Wheel product.

For my experiment I measured the force in grams at 4 centimeters from the axis at 25, 90, and 180 degrees. I then measured at 28 centimeters from the axis and then calculated what the force should be based on the measurements at 4 centimeters.

Above are the comparisons of the measurements and the calculations which appear in red.

Measured Forces on Lever Arm at 25 Degrees

Measured Forces on Lever Arm at 90 Degrees

Measured Forces on Lever Arm at 180 Degrees

Bill Kuhl

New Mousetrap Car Article on Website

For an explanation of many aspects of simple aerodynamics with math problems check out my article Basic Aerodynamics With a Lesson

Monday, March 17, 2014

Judging Sail Car Event at MN Renewable Energy Challenge

For the second time I had the opportunity to volunteer as a judge for the Minnesota Renewable Energy Challenge at the School of Environmental Studies in Apple Valley Minnesota. This event is collaboration between High Tech Kids and the Kid Wind Project.

 On the previous tournament I was one of the judges for documentation, this year I was a judge for the Sail Car event.  There were two teams running at the same time but it was not a race, Asia Ward was the judge for the A team and I for the B.

Materials to Build Sail Cars

This event which might appear very simple but there is so much opportunity for creative thinking, working as a team, and applying principles of physics. With a time limit, it is important to think about how to divide the tasks during the time limit. If too much time is spent in discussion of a design idea and construction, there might be little time left for testing and tweaking the car to get the longest possible distance.

Hitting the Wall

The rules for the event went something like this: car chassis made from coroplast that has two axles made from bamboo skewers with four wheels made from wood balls are available in sufficient number that team members can make their own cars or more than one car can be attached together. There was no limit on how many runs of the cars, only the overall time limit 20 minutes. The furthest distance obtained by placing cars in front of two fans was the score for the team along with some judged construction points.

Asia at left was the Other Judge

Fans Provided the Wind

Construction was in building of the sail structure from bamboo skewers and construction paper. Many ideas were tried and it became fairly obvious what was not working but less obvious what the optimum design might be.  The car had to go straight without turning into a wall or tipping over which was not easy for some designs.  Early distances were in the 6 meter range but by afternoon an amazing distance of almost 9 meters was made by one happy young man.

Side by Side Design

Long and Narrow Design

What was rather amazing was that a car might stop but if you let it set for several seconds it might start moving again. This really added to the drama of the event.  I had measured the wind speed before the event started at a couple of distances and had noticed it was varying. If anyone has theories on this, please comment below.

Bill Kuhl

Order Sailcar Kits from Kidwind
Model Wind Turbine Project Article on My Website

Friday, March 14, 2014

Math for Calculating Area of Vertical Fin

To keep an airplane flying straight normally a vertical fin is needed, the position is usually above the horizontal stabilizer. For a sport rubber model 10% of the wing area is good proportion to try. What would be the area for the vertical fin using the 36 square inches of wing area?    [3.6 sq. in.]  After computing the square inches of the vertical fin, what function would you use on this area to get equal length sides?   [square root]   

Side View of Rubber Powered Model Airplane

You decide that for the shape of the vertical fin part of the area to the front will be a right triangle. You decide the rear part of the fin will now be a square with sides 1.5” long.  How much area will this be? [ 2.25 sq. in].  What do you think the next step will be?   Yes, [ 3.6” – 2.25” = 1.35 sq. in.] Now we need a right triangle of 1.35 sq. in. with the opposite side 1.5” long to match up with a side of the square with sides that are 1.5" long. How long will the adjacent side of this triangle be?

I started looking for a formula to compute the adjacent side given the area and the opposite side but did not find one. My reasoning was to create a rectangle by drawing in another triangle and doubling the area. This would make it easy to compute the adjacent side. 1.35 sq. in. x 2 = 2.7 sq. in for rectangle area.  Divide this by 1.5” which equals  1.8“ for the bottom side (adjacent side).    

Entire Vertical Fin

To check my method I divided the area of rectangle 1.5” x 1.8” /2 = 1.35 square inches. I could now put the triangle and square together to form a vertical fin with 3.6 square inches of area.  

If anyone has another solution to find the missing dimension of the triangle comment below.

Bill Kuhl

For an explanation of many aspects of simple aerodynamics with math problems check out my article Basic Aerodynamics With a Lesson


Thursday, March 6, 2014

Climb of an Airplane Maybe Not Intuitive

During the process of working on an article for the new website on aerodynamics I ran across a couple of simple formulas regarding the climb of an airplane.

Lift = Weight x Cosine Angle of Climb
Thrust = Drag + (Weight x Sine of Angle of Climb)

I proceeded to test the lift formula with a hypothetical model airplane weighing 10 grams for climb angles of 30, 45, and 75 degrees. The results were rather shocking to me as I thought of the four forces of flight and how lift and weight are exactly equal in level flight. It would seem that lift would keep increasing more than weight as the airplane climbs. The cosine of the angle decreases as the angle increases however.  Here are my sample results:

30 degree climb      Lift = 10 grams x .866  = 8.66 grams     Cosine 30 = .866
45 degree climb      Lift = 10 grams x .707 = 7.07 grams    Cosine 45 = .707
75 degree climb      Lift = 10 grams x .259 = 2.59 grams     Cosine 75 = .259 

Sine of an angle increases with greater angles which means the thrust will be increasing as the airplane climbs steeper. That makes sense to me now and then when you think of the situation where the airplane is hovering the airplane will be completely supported by the thrust and there will be no lift from the wing. It also means that the thrust is pulling some of the weight upward, think of vectors.

To validate this I read further articles on the Internet and posed the question on a model airplane listserve.  It is true.

Embrace the Challenge of Learning 

Bill Kuhl

For an explanation of many aspects of simple aerodynamics with math problems check out my article Basic Aerodynamics With a Lesson