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### Topic: Impact measurement sensor (Read 671 times)previous topic - next topic

#15
##### Apr 20, 2019, 07:16 pm
What is the informed basis for your decision to use a 200 g acceleration sensor?
I haven't decided to use it. Hence the question. I'm trying to determine if it is the range sensor to use.

If you are not capable of making a reasonable estimate of the acceleration experienced by a soccer ball when it impacts a surface, how do you plan to interpret any measurements you might make?
Hence the questions. I'm trying to remember my physics classes from 22 years ago. That's not going so well. In the end the exact number doesn't matter. But for the design phase, I need to determine of the sensor is one of the correct range and sensitivity to make sure it will read something non-zero and the maximum value. It is not of much use if it only read 0 or max all the time.

#### jremington

#16
##### Apr 20, 2019, 07:31 pmLast Edit: Apr 20, 2019, 07:35 pm by jremington
If you are serious about this project, you will have to spend some time with the available literature. For example, this study
https://www.researchgate.net/publication/264275648_A_Study_of_Impact_Force_on_Modern_Soccer_Balls
shows that the acceleration of the ball, during a machine-driven kick, ranges from about 200 to 600 g.

The acceleration of a large surface impacted by the ball will be proportionally less by the ratio of the masses.

#17
##### Apr 20, 2019, 07:50 pmLast Edit: Apr 20, 2019, 08:26 pm by adwsystems
If you are serious about this project, you will have to spend some time with the available literature.
I always get a kick out of this assumption. But what needs to be evaluated is the impact of the backstop design on the readings or operation of the device), which is partially independent of the force from the soccer ball. It can be any force applied, but come back to how the design of the backstop affects the operation of the accelerometer.

It's all about the keywords. I just found something similar.

F=mA, F=0.5 * (200 * 9. / 4.448 = 220 pounds-force.

But the question remains, what is the mounting requirements for the sensor? How does the design of the backstop affect the readings (or what are is the best design characteristics for the backstop)? Should it be heavier, lighter, light and bottom weighted, or?

#### jremington

#18
##### Apr 20, 2019, 07:52 pm
Quote
I always get a kick out of this assumption.
Good to know, thanks!

#### polymorph

#19
##### Apr 20, 2019, 07:54 pm
Mount it on some kind of springs. If you hit it in the center, the deflection for a given soccerball velocity should have a high degree of repeatability.

The pressure in the soccer ball will have some effect. The mass of the plywood will affect it, the heavier the plywood, the smaller the acceleration.

The springs will of course affect the acceleration, too, if they are very stiff or are initially pulled tight.

This will be an elastic collision.
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8

#20
##### Apr 20, 2019, 08:14 pmLast Edit: Apr 20, 2019, 08:14 pm by adwsystems
The pressure in the soccer ball will have some effect. The mass of the plywood will affect it, the heavier the plywood, the smaller the acceleration.
This is what I was wondering. I can't figure out the exact physics, but mechanically it seems it would have an effect.

Mounting it from the center means making a stand and having to haul the standard around too. Also the stand would have be be large and heavy enough to withstand the impacts without falling over. But all soccer fields have goals with crossbars. ie., a good point from which to hang the target. Now I only have to lug the target around.

Would a 40g sensor be (physically) damaged if subjected to 200g acceleration?

#### MorganS

#21
##### Apr 20, 2019, 09:44 pm
Would a 40g sensor be (physically) damaged if subjected to 200g acceleration?
Maybe. Check the datasheet. Most of them will survive overloads but 5x overload is a lot.
"The problem is in the code you didn't post."

#### dougp

#22
##### Apr 20, 2019, 10:08 pm
What about letting the ball hit an air filled bladder driving a pneumatic pressure sensor?  The sensor then wouldn't be directly subjected to impact.  The compressibility of air would provide a bit of possibly undesired cushioning/integration.
Everything we call real is made of things that cannot be regarded as real.  If quantum mechanics hasn't profoundly shocked you, you haven't understood it yet. - Niels Bohr

No private consultations undertaken!

#23
##### Apr 20, 2019, 10:37 pm
Maybe. Check the datasheet. Most of them will survive overloads but 5x overload is a lot.
LOL. Looked in the wrong spot. Abs Max= 5000g

#### Slumpert

#24
##### Apr 21, 2019, 06:01 amLast Edit: Apr 21, 2019, 06:02 am by Slumpert
Could you just use a radar speed gun (doable with Arduino) setup to measure ball speed?

I can see the MLB installing impact sensors in the bats to see how hard the ball was hit being a new stat for 2020.....

#25
##### Apr 21, 2019, 03:40 pm
Could you just use a radar speed gun (doable with Arduino) setup to measure ball speed?

#26
##### Apr 29, 2019, 12:36 am
Still working on the project. POC finished today and tried. Also found some more physics concepts.

For a free floating target (theoretical) of 1m2 weighing 4kg. If the accelerometer reads 1g (9.8m/s2), then the force is F=ma = 4 * 1* 9.8 = 39.2
Since the ball weighing 0.45kg is travelling at a constant speed when it hits F=mA does not apply, but instead use F=1/2mv2. making v = sqrt(2*F /m). = sqrt (2 *39.2 /0.45) = sqrt(174.2) = 13.2 m/s. [this comes straight from a physics example problem, just updated the numbers to my setup]

Problem is the target is not free floating. It is hanging from its top edge. I assume moment of inertia comes in to play here. But haven't found the nugget to explain how. I would expect the answer will affect line 1, to find that f=ma doesn't directly apply.

In testing. I hit the target and received a maximum reading of 568. The sensor zero reading is 400. The sensor is supposed to be idle at 1.65V (1.65/5*1024 = 338), but the reading of 400 shows an offset. 3.3V should be 675 counts. Not sure which way is correct (investigation still needed). The results of the two options are.

zero=400, maximum = 675, range =200g, 568 = 122g based on the math (568-400)/275 * 200
or
zero=400, maximum = 800, range =200g, 568 = 84g based on the math (568-400)/400 * 200

On the free floating target 122g = 146 m/s and 84g = 120m/s. Somehow I don't think I can kick that hard. Therefore I must conclude, I need information on the targets force equation. Anyone know the force equation for the hanging target described above?

#27
##### Apr 30, 2019, 12:26 amLast Edit: Apr 30, 2019, 12:35 am by adwsystems
18.9 m/s and 15.7 m/s if hit at the bottom edge with the sensor at the bottom edge. Dropping ratiometrically as the point of impact moves from the bottom edge to the top edge (distance from top edge to point of impact : distance from top edge to point of impact). If you assume that most shots are on center then most readings will be cut in half.

Assume hits will follow a similar pattern as rifle shots, and move away from the target in a sinusoidal or exponential drop off. then most reading will be most accurate by placing the sensor in the middle of the target.

#### MorganS

#28
##### Apr 30, 2019, 12:44 am
There are two concepts in the physics books: energy and momentum. You can use both, so long as you don't mix them.

The kinetic energy of a moving body is 1/2mv2.

The momentum of the same body is mv.

So when the ball hits the target it transfers energy and momentum. You could measure the energy by measuring the swing back on the target. It swings like a pendulum. The swing converts kinetic energy into potential energy because it goes back and up until all the kinetic energy is zero: meaning it stopped moving.
"The problem is in the code you didn't post."

#29
##### Apr 30, 2019, 12:50 am
There are two concepts in the physics books: energy and momentum. You can use both, so long as you don't mix them.

The kinetic energy of a moving body is 1/2mv2.

The momentum of the same body is mv.

So when the ball hits the target it transfers energy and momentum. You could measure the energy by measuring the swing back on the target. It swings like a pendulum. The swing converts kinetic energy into potential energy because it goes back and up until all the kinetic energy is zero: meaning it stopped moving.
That is a reasonable modification. A simple pot connected at the end of the pivot axis would note the angle of the swing. Since most pots turn 270 deg or +/-135 deg and the target will never swing more than 90 in either direction it would not break the potentiometer.

Still leaves me with needing the equation for the target to convert the angle into potential energy, already knowing the size and weight of the target.

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