B The physics of flywheel launchers (like tennis ball shooters)

[Baluncore]

also, how does that work in a hooded shooter? there I highly doubt the de-compression will perfectly push the wheels.

The profile of the hood will determine the rate of compression and the rate of expansion, and they must balance.

After the wheel has pulled in the ball, the ball, as it exits, will push against the wheel and the hood. Since the gap widens towards the exit, that must return the compressive energy to the wheel.

The gas inside the ball will be heated by compression, it will then cool by expansion as it exits, before there has been time for significant heat loss. Perfection is impossible, there will be minor energy losses from the gas to the ball, and from the initial sliding friction to the ball.

 

[cardboard_box]

The gas inside the ball will be heated by compression, it will then cool by expansion as it exits, before there has been time for significant heat loss. Perfection is impossible, there will be minor energy losses from the gas to the ball, and from the initial sliding friction to the ball.

wait, I'm not sure I follow, what gas inside a ball? what if the ball is solid or has air or something else inside it?

 

[Baluncore]

wait, I'm not sure I follow, what gas inside a ball? what if the ball is solid or has air or something else inside it?

Air is a mixture of different gasses. All the balls I see in the launcher videos are typically air-filled. Gasses exhibit a thermal inefficiency when they are compressed and expanded, heated and cooled.

The ball may be made from a closed cell foam, in which case it will be as lossy as a gas filled ball. If the ball was filled with only one species of gas, it would probably be dry nitrogen, to avoid chemical reactions with the envelope during hot impacts.

If the ball is solid, maybe a polymer, then it will have elastic properties, and will probably bounce more efficiently than a gas filled ball.

A loaded, or a weighted ball, could be liquid filled, or a bean bag. I have trouble thinking of a sport that is played with a weighted ball, or why it would need a launcher. Weighted balls are more like weapons, likely to cause injuries.

 

[Lnewqban]

hi, I get that RPM is the cause of velocity and not torque, my problem is rather understanding why its so, when normal it is force that causes acceleration and therefore velocity. also, what exactly do you mean the torque is consumed in the impulse? would that not influence the momentum and therefore the velocity?

Yes, you are correct.
Certain amount of energy flows from the launcher to the projectile.
If the launcher is not fed with replenishing energy (like from an electric motor), and keeps giving impulse to several projectiles consecutively, it will simultaneously slow down more, and more and will deliver weaker and weaker torque, after each launch.

 

[cardboard_box]

A loaded, or a weighted ball, could be liquid filled, or a bean bag. I have trouble thinking of a sport that is played with a weighted ball, or why it would need a launcher. Weighted balls are more like weapons, likely to cause injuries.

it is used in tennis pitching machines to practice.

 

[cardboard_box]

Yes, you are correct.
Certain amount of energy flows from the launcher to the projectile.
If the launcher is not fed with replenishing energy (like from an electric motor), and keeps giving impulse to several projectiles consecutively, it will simultaneously slow down more, and more and will deliver weaker and weaker torque, after each launch.

but how can I measure this flow of energy/impulse without just looking at the finished results? and most importantly how exactly does it apply a torque while free spinning? objects with momentum must have the ability to exert a force by themselves to cause an impulse, so if we have 2 objects which we know the momentum of how do we find what force they will exert on one another during collision without knowing the end momentum of each object?
Also, wouldn't a faster spinning flywheel result in less time in the shooter and therefore lesser impulse? seems contradicting

 

[cardboard_box]

an updated question: does momentum exert a force? in the shooter the wheel has momentum but no active force on it, it interacts with the wheel via friction but friction is a "reactive" force, is friction effected by velocity or momentum the objects?
if we for example take 2 cubes, 1 is just standing in space and the other is moving towards it, when they impact there is an impulse and momentum change, so there must be a force. does it make sense the force is the reaction of friction and the atoms pushing back due to electric charge and resisting, and if so how can we measure the force an object will exert on another due to its momentum?

 

[Baluncore]

I do not think momentum will help you here. The ball is free until it is grabbed by the wheels and accelerated as a part of the wheels. It is then released at the final velocity of the wheels.

The acceleration impulse takes a finite time, while friction is slipping, until the ball is held against the wheels without slip. That may last a few milliseconds, it is not an instant. The friction and the acceleration happen, but the detail of the clutch is best ignored.

You know the mass of the ball, the RPM and the energy in the flywheels. Solve for the exit velocity of the ball, which will equal the final velocity of the wheels. That will be an energy balance equation. You can ignore the loss of energy to friction, unless you notice the ball gets hot when it is launched.

You can initially assume that two wheels, on the same shaft, form a cylindrical roller. Deeper analysis will show that the wheels do not contact the ball at its outer diameter, but closer to the ball's axis. That will not be important for two shafts with four identical wheels, two on each shaft. It will become important for a one-sided thrower, where one side of the ball is held against the hood.

 

[cardboard_box]

I do not think momentum will help you here. The ball is free until it is grabbed by the wheels and accelerated as a part of the wheels. It is then released at the final velocity of the wheels.

The acceleration impulse takes a finite time, while friction is slipping, until the ball is held against the wheels without slip. That may last a few milliseconds, it is not an instant. The friction and the acceleration happen, but the detail of the clutch is best ignored.

You know the mass of the ball, the RPM and the energy in the flywheels. Solve for the exit velocity of the ball, which will equal the final velocity of the wheels. That will be an energy balance equation. You can ignore the loss of energy to friction, unless you notice the ball gets hot when it is launched.

You can initially assume that two wheels, on the same shaft, form a cylindrical roller. Deeper analysis will show that the wheels do not contact the ball at its outer diameter, but closer to the ball's axis. That will not be important for two shafts with four identical wheels, two on each shaft. It will become important for a one-sided thrower, where one side of the ball is held against the hood.

but if the ball simply accelerates to the surface velocity of the wheel after a couple moments wouldn't it break the definition of energy as distance*force? because if the ball is in contact with the wheel for a distance of 5m or 5km it will still just be the surface velocity of the wheel, and then adding more wheels like a series would also not increase its velocity.

why would the ball start with kinetic friction that will then transform to static friction if it didn't start sliding on the wheel?

I am not sure what you mean in your final paragraph either.

 

[Baluncore]

but if the ball simply accelerates to the surface velocity of the wheel after a couple moments wouldn't it break the definition of energy as distance*force?

Once the ball is grabbed and reaches the speed of the wheels, the force becomes zero and there is no relative movement. Work is being done by the wheels on the ball, only while kinetic friction is operating.

A force without movement in the direction of the force, or a movement without a collinear force, does not constitute work being done. The pinching force, that holds the ball, is perpendicular to the movement, so does no work.

The acceleration and energy exchange is done, immediately the ball is locked to the wheels.

why would the ball start with kinetic friction that will then transform to static friction if it didn't start sliding on the wheel?

It will start sliding on the wheels, that kinetic friction will pull the ball further into the wheels, resulting in the ball being grabbed by the wheels.

I am not sure what you mean in your final paragraph either.

The final paragraph can be ignored for now. It is a subtlety that will mess up the numbers by throwing the ball faster than expected, from a one-sided thrower with a hood.

 

[cardboard_box]

Once the ball is grabbed and reaches the speed of the wheels, the force becomes zero and there is no relative movement. Work is being done by the wheels on the ball, only while kinetic friction is operating.

A force without movement in the direction of the force, or a movement without a collinear force, does not constitute work being done. The pinching force, that holds the ball, is perpendicular to the movement, so does no work.

The acceleration and energy exchange is done, immediately the ball is locked to the wheels.


It will start sliding on the wheels, that kinetic friction will pull the ball further into the wheels, resulting in the ball being grabbed by the wheels.


The final paragraph can be ignored for now. It is a subtlety that will mess up the numbers by throwing the ball faster than expected, from a one-sided thrower with a hood.

okay a couple of questions:
1. why does the ball start with kinetic and not static friction? that seems really dependent on the set up, since if it touches tangentially it should start rolling against the hood.

2. could you expand on relative velocity and why it makes the force turn to 0? I just am still not really sure what the source of the force is, it seems to be due to momentum since if for example 2 objects collide in a vacuum there is still a force exerted in the impulse, so does momentum exert a force during collision and how?

3. isn't a force without movement in the direction of the force paradoxical? since force causes acceleration there must be some sort of movement in the system.

4. if the ball is rolling, which in a lot of the mechanism I saw it those, then that should mean there is static and not kinetic friction.

 

[Baluncore]

1. why does the ball start with kinetic and not static friction?

Because there is a differential velocity throughout the period of acceleration. The relative movement of surfaces during acceleration must be kinetic, not static.

2. could you expand on relative velocity and why it makes the force turn to 0? I just am still not really sure what the source of the force is, it seems to be due to momentum since if for example 2 objects collide in a vacuum there is still a force exerted in the impulse, so does momentum exert a force during collision and how?

For this project, you must forget about the momentum of objects colliding in a vacuum. It is irrelevant here. Any initial ball momentum is lost because the ball is trapped for a finite time, trapped by the wheels spinning on a fixed axis. The ball is not free to preserve momentum. To preserve momentum, the ball and thrower would need to be floating free in space throughout their contact.

3. isn't a force without movement in the direction of the force paradoxical? since force causes acceleration there must be some sort of movement in the system.

When you rest a block on the table, gravity holds it on the table, but there is no movement, so no work is being done.

4. if the ball is rolling, which in a lot of the mechanism I saw it those, then that should mean there is static and not kinetic friction.

The ball may roll into the thrower, but then it is dynamically drawn in, accelerated, before being gripped by static friction. A one-sided thrower, accelerates the ball with kinetic friction, then employs static friction to roll the ball between the wheel and the hood. A ball rolls, with static friction, into a two-sided thrower, kinetic friction accelerates the ball, then static friction grips the ball between the wheels.

 

[cardboard_box]

Because there is a differential velocity throughout the period of acceleration. The relative movement of surfaces during acceleration must be kinetic, not static.


For this project, you must forget about the momentum of objects colliding in a vacuum. It is irrelevant here. Any initial ball momentum is lost because the ball is trapped for a finite time, trapped by the wheels spinning on a fixed axis. The ball is not free to preserve momentum. To preserve momentum, the ball and thrower would need to be floating free in space throughout their contact.


When you rest a block on the table, gravity holds it on the table, but there is no movement, so no work is being done.


The ball may roll into the thrower, but then it is dynamically drawn in, accelerated, before being gripped by static friction. A one-sided thrower, accelerates the ball with kinetic friction, then employs static friction to roll the ball between the wheel and the hood. A ball rolls, with static friction, into a two-sided thrower, kinetic friction accelerates the ball, then static friction grips the ball between the wheels.

I'm sorry but I don't really understand some of what you said, firstly assuming you mean that since the objects don't start with the same velocity but rather accelerate to it there must be some slippage to begin with, but I don't understand why it shouldn't be possible that it starts with static friction and roll against the hood of the flywheel, otherwise how exactly do you measure the speed it takes it to become static friction and be the same velocity of the wheel.

secondly, the ball can start with zero momentum on its own, but even if we fully ignore momentum and just use KE there still must be a force exerted by the wheel on the ball, what is the source of it?

thirdly, when I rest an object on the table it doesn't move because the table is also pushing back against the force of gravity and the object with the normal force, so there is a net force of zero.

and fourthly, I am not sure what you mean in the 3rd paragraph with switching between the 2 types of friction and what they do, isn't it impossible for an object to be experiencing both kinetic and static friction at the same time? and if they do switch that would mean the ball doesn't actually starts to roll until a certain point, which I'm not sure about.

 

[Baluncore]

why it shouldn't be possible that it starts with static friction and roll against the hood of the flywheel, otherwise how exactly do you measure the speed it takes it to become static friction and be the same velocity of the wheel.

You do not need to measure the speed, time, or force of acceleration. The ball will enter, accelerate, be held, and then be released. The numbers during that process are not important, it happens. The result will be the same, no matter which path is taken to the outcome.

secondly, the ball can start with zero momentum on its own, but even if we fully ignore momentum and just use KE there still must be a force exerted by the wheel on the ball, what is the source of it?

That force is the drag of kinetic friction, due to the contact differential velocity.

thirdly, when I rest an object on the table it doesn't move because the table is also pushing back against the force of gravity and the object with the normal force, so there is a net force of zero.

The two forces are not zero, they are equal and opposite.
Work is not being done, because there is no relative movement.

and fourthly, I am not sure what you mean in the 3rd paragraph with switching between the 2 types of friction and what they do, isn't it impossible for an object to be experiencing both kinetic and static friction at the same time?

Yes. That is why it must switch cleanly between the two types of friction.
The ball rolls into the thrower entry, (static friction).
The contact differential velocity pulls the ball in, (kinetic friction).
The ball is accelerated by the thrower (kinetic friction).
The ball reaches the speed of the thrower and is held (static friction).
The ball is released at the exit, no contact force, so no more friction.

 

[Lnewqban]

but how can I measure this flow of energy/impulse without just looking at the finished results?

You can’t.
Consider the shooter as a box, into which measurable energy flows (via electricity-motor) just to be transferred into the projectile (measurable kinetic energy).

and most importantly how exactly does it apply a torque while free spinning?

What is free spinning?

objects with momentum must have the ability to exert a force by themselves to cause an impulse, so if we have 2 objects which we know the momentum of how do we find what force they will exert on one another during collision without knowing the end momentum of each object?

You need to know the time duration of the collision.

Also, wouldn't a faster spinning flywheel result in less time in the shooter and therefore lesser impulse? seems contradicting

The final velocity does not depend on the rate at which the energy is transferred.
The final velocity of the ball will be higher for a faster spinning flywheel, which requires higher amount of energy that the flywheel must transfer (and loss) in a shorter period of time.

The flywheel will slowdown more drastically after each launch, which will require more time accumulating energy supplied by the motor between successive launches, or a more potent motor and higher consumption of electricity.

It seems contradictory only because we don’t see the flows of energy.

 

[cardboard_box]

You can’t.
Consider the shooter as a box, into which measurable energy flows (via electricity-motor) just to be transferred into the projectile (measurable kinetic energy).

What is free spinning?

You need to know the time duration of the collision.

The final velocity does not depend on the rate at which the energy is transferred.
The final velocity of the ball will be higher for a faster spinning flywheel, which requires higher amount of energy that the flywheel must transfer (and loss) in a shorter period of time.

The flywheel will slowdown more drastically after each launch, which will require more time accumulating energy supplied by the motor between successive launches, or a more potent motor and higher consumption of electricity.

It seems contradictory only because we don’t see the flows of energy.

but if I know everything that is inside the box shouldn't I be able to determine the end results and how each component effected it? for example if we just have 2 blocks collide in a vacuum and we know their mass their velocity their size ect ect. there should be one universal result which will happen again and again no matter how many times I repeat the exact same interaction.

free spinning means that the motor isn't actively applying a torque to it, so imagine I spin up the wheel with a motor then just pop the motor out of existence, the wheel will keep rolling and will still be able to exert a force on the projectile.

the question is WHY it would be faster for a higher RPM flywheel, the wheel having more energy to transfer to the projectile doesn't necessarily mean it WILL transfer more energy. and even if we use either momentum conservation or energy conservation we could simply increase the mass of the wheel to increase instead of velocity.

why exactly will it decrease more after every shot if the projectiles are all the same?

and regarding not being able to see the energy transfer that isn't the problem, the problem is there any way to calculate how much energy/momentum will be transfered and how every variable in the system effects it?

 

[russ_watters]

and regarding not being able to see the energy transfer that isn't the problem, the problem is there any way to calculate how much energy/momentum will be transfered and how every variable in the system effects it?

If you have the dimensions and weights of the ball and flywheels, the rotation rate, and the location where the ball is captured on the wheels, you can calculate the final speed with a simple energy balance and get pretty close without knowing the fine details of the interaction. As @Baluncore said in post #38.

You can always go deeper for any problem, but adding in complexities like the loss due to frictional heating and compression of the ball(how about air resistance? The bearings of the flywheel shaft? Etc.) will make the problem VERY complicated without changing the answer much.

 

[A.T.]

but if I know everything that is inside the box shouldn't I be able to determine the end results and how each component effected it

Yes, with a complex numerical simulation, as I already told you in post #26.

 

[Lnewqban]

but if I know everything that is inside the box shouldn't I be able to determine the end results and how each component effected it? for example if we just have 2 blocks collide in a vacuum and we know their mass their velocity their size ect ect. there should be one universal result which will happen again and again no matter how many times I repeat the exact same interaction.

Yes, you will be able.

free spinning means that the motor isn't actively applying a torque to it, so imagine I spin up the wheel with a motor then just pop the motor out of existence, the wheel will keep rolling and will still be able to exert a force on the projectile.

Yes, that is the principle of a flywheel.
Your flywheel is solidly connected to a motor that can't be just pop out of existence.
The flywheel will rotate a little slower after shooting the ball, it will try to slowdown the motor, which will react applying a torque to speed up the flywheel until reaching the motor nominal rpm's.

the question is WHY it would be faster for a higher RPM flywheel, the wheel having more energy to transfer to the projectile doesn't necessarily mean it WILL transfer more energy. and even if we use either momentum conservation or energy conservation we could simply increase the mass of the wheel to increase instead of velocity.

The case is similar to a car trying to reach a high speed in first gear: the wheels receive lots of torque, but are unable to spin faster.
If we increase the mass of the wheel instead of velocity, our machine will be able to shoot heavier balls at the same velocity.

why exactly will it decrease more after every shot if the projectiles are all the same?

If no connected to a source of energy (the electrical motor in this case), the flywheel will have less momentum after giving some of it away to each launched ball.
As the mass of the flywheel can't be reduced, a reduction in rotational velocity will be the result of a free spinning flywheel launching several balls successively.

and regarding not being able to see the energy transfer that isn't the problem, the problem is there any way to calculate how much energy/momentum will be transferred and how every variable in the system effects it?

Yes.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#c2

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#am

 

[cardboard_box]

Yes, with a complex numerical simulation, as I already told you in post #26.

yes sorry, let me be more precise. how exactly do I go about building such a function? specifically how does the velocity of the wheel change the velocity of the ball?

even if I do find the coefficient of restitution of the projectile I still don't fully understand how I pick the wheel variables based on it (since I want to treat it as an engineering problem where I am given the projectile and the velocity&rate at which I want to fire). I don't really want to just go over every possible wheel and make a spreadsheet of how good it is at shooting the ball.

and how do I relate a linear coefficient of restitution to the tangential movement of the wheel? it doesn't exactly bounce back and it doesn't solely rotate around its own axis. some examples I saw just use the wheel as a point mass at point of contact but I don't really know how accurate it is to use it that way.

so for example, if you are given a projectile which you know everything you want about, and are told to fire it at this and this velocity and fire this and this of them at this and this rate, how do you decide what wheel do you use? it doesn't have to mean "you will use this exact wheel with this exact variable statistics" but is mainly "if I want the ball to move at this velocity I need to either change this variable x or that variable y"

 

[cardboard_box]

Yes, you will be able.


Yes, that is the principle of a flywheel.
Your flywheel is solidly connected to a motor that can't be just pop out of existence.
The flywheel will rotate a little slower after shooting the ball, it will try to slowdown the motor, which will react applying a torque to speed up the flywheel until reaching the motor nominal rpm's.


The case is similar to a car trying to reach a high speed in first gear: the wheels receive lots of torque, but are unable to spin faster.
If we increase the mass of the wheel instead of velocity, our machine will be able to shoot heavier balls at the same velocity.


If no connected to a source of energy (the electrical motor in this case), the flywheel will have less momentum after giving some of it away to each launched ball.
As the mass of the flywheel can't be reduced, a reduction in rotational velocity will be the result of a free spinning flywheel launching several balls successively.


Yes.
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#c2

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#am

I already know some of these variables, but I am not fully sure I get every effect of them and I still don't understand the importance of RPM.

about the wheel imagine instead I spin it up and then disengage it so that it is spinning freely without any resistance but isn't accelerating and applying a force anymore.

since you use a car metaphor I'll try to use one too: imagine we lift a car up in the air and accelerate it, and a certain point we just disengage the engine so it isn't powering the car anymore, while the wheels still roll we let the car down on the ground. at this point the car should start moving, but the question is why, how does the RPM of the car determine how fast it moves (how much force it applies on the ground)

 

[Baluncore]

For an arbitrary set of launch conditions, the plot of force against time, is unpredictable, and will be different for every launch under those conditions. Yet, the integral of that plot, and so the ball final velocity, will be the same. You really don't need to go into the detail of the sliding friction, and chaotic capture of the ball. That may be fascinating, but it is not a tractable problem.

The KE of wheels before launch = KE of wheels after launch + KE of flying ball.
The velocity of the ball will be equal to the circumferential velocity of the wheels after the launch. You can compute that, it is a tractable problem.

I would compute the launched ball velocity in the easiest way.
1. Design your flywheel and evaluate its Moment of Inertia.
2. Evaluate the MoI of the mass of the ball, if spread thinly, on the periphery of the flywheel.
3. Work out the initial KE of the flywheel, when spinning at the specified RPM.
4. Add the MoI of the ball to the MoI of the wheel.
5. Back compute from initial energy, to the lower final RPM using the increased MoI. You now know the reduction in wheel RPM due to the launch.
6. The velocity of the launched ball will be the same as the final velocity of the flywheel periphery.

 

[cardboard_box]

For an arbitrary set of launch conditions, the plot of force against time, is unpredictable, and will be different for every launch under those conditions. Yet, the integral of that plot, and so the ball final velocity, will be the same. You really don't need to go into the detail of the sliding friction, and chaotic capture of the ball. That may be fascinating, but it is not a tractable problem.

The KE of wheels before launch = KE of wheels after launch + KE of flying ball.
The velocity of the ball will be equal to the circumferential velocity of the wheels after the launch. You can compute that, it is a tractable problem.

I would compute the launched ball velocity in the easiest way.
1. Design your flywheel and evaluate its Moment of Inertia.
2. Evaluate the MoI of the mass of the ball, if spread thinly, on the periphery of the flywheel.
3. Work out the initial KE of the flywheel, when spinning at the specified RPM.
4. Add the MoI of the ball to the MoI of the wheel.
5. Back compute from initial energy, to the lower final RPM using the increased MoI. You now know the reduction in wheel RPM due to the launch.
6. The velocity of the launched ball will be the same as the final velocity of the flywheel periphery.

the problem is that I am trying to understand how to determine how I design the flywheel. So get the data on what how I need to projectiles to behave and work back from that what mass and velocity and so on the flywheel should be.

I think I am just not wording this well enough and not really explaining what specific problems I am having to understand. Thank you for trying to help and I maybe can just use your calculations, but for now I think I'll just open a different thread with a more coherent and generalized set of questions then "how to calculate flywheel".

take care.

 

[Baluncore]

"how to calculate flywheel".

Why ask when you can read all about it.
https://en.wikipedia.org/wiki/Flywheel#Design
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

 

[cardboard_box]

sadly this is more about flywheels as a tool for conserving energy, while I'm mainly interested in using them to shoot out a projectile.

 

[renormalize]

the problem is that I am trying to understand how to determine how I design the flywheel. So get the data on what how I need to projectiles to behave and work back from that what mass and velocity and so on the flywheel should be.

Why are you trying to reinvent the (fly)wheel? Just use an existing calculator like this one: https://calculator.frc4322.com/shooting-mechanism.

 

[cardboard_box]

Why are you trying to reinvent the (fly)wheel? Just use an existing calculator like this one: https://calculator.frc4322.com/shooting-mechanism.

somehow I did not stumble upon this calculator, very useful indeed. I still want to learn how it works though, since without fully understanding it feels kinda empty.

I can not stretch this enough though but this calculator seems incredible, and now I got more of a focus on variables to study.

is a bit sub optimal to use this calculator since you input the variables for the result instead of the other way around, as well as variables that I (at least at the moment) do not understand how you could possibly know.
but again, a great help so thanks. any chance you know if I can view the way the calculator itself works by any chance?

 

[renormalize]

but again, a great help so thanks. any chance you know if I can view the way the calculator itself works by any chance?

This calculator is provided by a California robotics team. You should contact them for more info: https://frc4322.com/contact/.

 

[cardboard_box]

This calculator is provided by a California robotics team. You should contact them for more info: https://frc4322.com/contact/.

thank you.

 

[A.T.]

any chance you know if I can view the way the calculator itself works by any chance?

You cloud look at its source code:
https://calculator.frc4322.com/main.582e45e1ffceb5fd.js
But first use some tool to reformat it by inserting line breaks and indentation again.