Okay, I'm finally getting down to writing this. I've no idea how this will go, so right now I'll just be posting whatever I can think of, receive comments (especially on whether it is too technical or if there's any debatable points) and slowly clean it up and organize it, so that it'll have a more logical flow.

In essence, I'll be taking whatever's relevant to Beyblade from probably the first one and a half years or so of a uni physics course. So of course, you won't see a discussion on time dilation since Beyblades tend not to travel at the speed of light. This will be very heavily based on mechanics and kinetics, for obvious reasons.

I'll assume basic mathematics is known (e.g. addition, multiplication, and so on). Where more advanced concepts in geometry, algebra or calculus are involved, I'll take time to explain them.

Defining a particle
Firstly, I will discuss particles before moving on to more complex things like Beyblades. A particle is something that just occupies a point in space. More complex things are made up of many particles; the laws that describe particles can then be used to describe the complex things.

Mathematical interlude: Vector and scalar quantities
Quantities refer to how much of something exist. A scalar quantity is what we call "one-dimensional"; for example, weight of an attack ring or height of a beyblade. Vectors, on the other hand, have both "how much" and "in which direction". For example, if we say that something is "5 meters to the right", that is a vector quantity: the "how much" is 5, and the "direction" is to the right.

Defining the motion of a particle
We have a physical quantity known as displacement. Displacement is a vector, such as "5 meters to the right". Most vector quantities have an equivalent scalar quantity; the equivalent scalar for displacement is distance. For example, in the above case, we say that the distance away is 5 meters. This may seem trivial, but not in certain cases. For example, say the object does not move in a straight line, but takes a roundabout route to reach the point "5 meters to the right". Hence, the displacement is 5 meters to the right; but the distance is much larger since it took a long path to reach there.

Next, we have the vector "velocity", and corresponding scalar, "speed". This refers to how much the above values change with time. For example, a velocity of "5 meters per second, to the right" means that it moves 5 meters to the right every second. Again, while velocity is directional, speed is not. Thus, if an object moves in a circle, after it makes one round, it's velocity would be zero since it's back to where it started! However, its SPEED is non-zero, obviously.

Finally, we have "acceleration", which is how much the velocity changes with time. When something accelerates, it gets faster. I'll cover this in more detail later.

On the laws of motion
All objects in this world tend to obey Newton's laws when they're travelling at everyday speeds, namely that
1) A physical body will remain at rest, or at constant velocity, unless an external net force is exerted on it.
2) A force acting on an object will cause a change in momentum that is proportional to the strength of the force, and in the direction of the force.
3) Every action has an equal and opposite reaction.

The first two points are self-explanatory. Law 1 means things don't happen unless something makes it happen (i.e. the world makes sense). Law 2 is extremely commonsensical; if you push something, it's not going to come closer to you.

The third law, however, should be clarified. It means that everything comes in pairs.

One important example is the repulsion between the atoms in two surfaces in contact with each other. It results in what is known as the "normal force", not because it normally happens (although this is true). In this case, "normal" uses the mathematical definition "perpendicular to something".

So, say you push a wall. Although you're pushing, your hand does not go through the wall. This is because the wall exerts a force perpendicular to the wall, onto your hand. It would not make sense to be non-perpendicular: When you push straight on at a wall, your hand does not move sideways. At the same time, you may ask, although your hand does not go through the wall, the wall also does not go through your hand! This is because your hand is also exerting a normal force onto the wall, pushing the wall away from it at the same time that the wall is pushing your hand away from it.

Another example would be in gravity. Although many of us are aware that we're being pulled towards the Earth, it's very important to also note that the Earth is being pulled towards us.. This also applies for electric forces. However, these are not very much related to Beyblades as we usually play them, so I will not cover these.

Applications of the laws of motion
Now, we convert the above concepts into mathematical laws so they can be applied to calculations.

Firstly, the first two laws can be summarized to a simple equation,

Force = Mass x Acceleration

The more force you apply to an object, the more it will accelerate (or decelerate). The more massive an object is, the less likely it is to be affected by said force.

So let's say you're launching your Beyblade. You apply a force by pulling on the winder/ripcord. As a result, your Beyblade accelerates until the point where it falls off the launcher: this is the final speed of the Beyblade's spinning at the end of the acceleration.

How do we know this speed? Consider that acceleration is a measure of how much the speed or velocity changes with time. So, firstly, we can find the acceleration if we know the force and it's mass (actually, it's not as simple as this, but since spinning tops are so complex, it'll take a while to work up to it), and then we multiply the acceleration by time to find out how much the speed changed altogether.

Therefore, this brings us to the next commonsensical point! The more force you apply, the faster it'll become! Therefore, the harder you pull your ripcord, the faster your Beyblade will be!

We can apply this to see why long ripcords work better. Let's say you're pulling at the same force throughout, so there's a constant acceleration. But! With a longer ripcord, you'll be pulling it, and hence accelerating it, for a longer period of time! As a result, the final speed will be greater than with a short ripcord. However, it's really important to take note that with good technique, you can attain really fast speeds even with a short ripcord: it's about how you pull it.

I understand all that is rather... daunting to read. But! I'm afraid if I jump straight into serious discussion, many people without appropriate background knowledge will be lost! So, I'll need to know how much to put in, how much to give a miss! The next time I update, I'll be covering on analyzing forces, and what forces are likely to act on a Beyblade, then on momentum and how to do analysis related to momentum. Up to here is mainly background information to lay the groundwork for the discussion on Beyblade.

I'll then proceed to discuss on rotational motion (torque, angular speed, angular inertia, etc.), which, together with the discussion above, will allow for detailed analysis of a Beyblade. I'll also be able to give a lot more examples from here on.

Finally, I'll cover precession, which explains why the Beyblade doesn't just topple over. It's a very horrible topic to explain, but I'll try xD






(I feel like I'm writing a physics textbook.)

---

I think the first two or three paragraphs are unnecessary, but perhaps it will be useful for what you will write next ...

Will you have pictures/diagrams to support this ?

tbh, I think its too long and explaining on a particle level to familiarize newton's laws of motion is i think to distant from bey blade when you could just use beyblade.

you typed all that, though? wow kinda intense

my ghetto homies wud be liek, "u krazee fam"

u krazee fam

and I think that half of the people who use these forums wouldn't care about physics, they'd rather talk about monsters cuming (ew) out of the bit-chips.

I really enjoyed reading it though

---

balls

my ghetto homies wud be liek, "u krazee fam"

u krazee fam

and I think that half of the people who use these forums wouldn't care about physics, they'd rather talk about monsters cuming (ew) out of the bit-chips.

I really enjoyed reading it though

what the hell is wrong with you

LOL Elmo. I know what you mean.
hahhahaha

Rewriting of the opening of the first section, to make it more accessible to laymen. I also realize that most readers will not want a complete understanding on the physics behind Beyblade, just a general idea; this is certainly a relief in writing.

Also, I will probably categorize this within "Background knowledge", so that readers can skip it if they're confident of understanding.

Defining the motion of a particle
To know how something moves, we need to know where it is, and how fast it is moving.

We can do this by measuring "displacement". This indicates position, for example, we can say that the Beyblade is "five meters to the right". Besides this, we also need to know how this changes with time. To know this, we have another measurement known as "velocity", which is essentially the displacement over time: For example, "three meters per second towards the North" means that every second, it would move three meters Northwards.

Finally, we also need to know how much its velocity or speed changes with time, which is measured by another quantity known as "acceleration". When an object accelerates, it changes speed.

On the laws of motion
After we know how to describe how an object moves, we'll need to know what causes it to move that way.

All objects in this world tend to obey Newton's laws when they're travelling at everyday speeds, namely that
1) A physical body will remain at rest, or at constant velocity, unless an external net force is exerted on it.
2) A force acting on an object...

Addition of a discussion on forces, momentum and energy, as promised.

On Forces
The effect of forces were summed up previously; we can easily say all motion in this world is caused by forces.

Generally, when we analyze most moving objects, we will ignore annoying things like air resistance, which has very complicated calculations and negligible effects in Beyblades.

In fact, when considering Beyblades, there's only three main forces that you need to be concerned about; friction, gravity, and impacts between Beyblades.

Gravity, obviously, pulls the Beyblade downwards (and also pulls the Earth and everything else up towards your Beyblade, albeit negligibly). This has very serious consequences on your game. It causes the Beyblade to tend to stick to the stadium most of the time except when your Beyblade receives a strong upper attack.

Gravity also has another effect. By pushing the Beyblade downwards at the stadium, the resulting normal force between the Beyblade and the stadium causes friction. Friction can be said to be a resistive force; common sense tells you that friction makes things stop. Push a book across a smooth table, and it will slide further than pushing it against a rough table. If you use a heavier book, it probably won't slide as far.

From these observations, we can figure out that friction is dependent on the surface in contact, and the weight involved. This is why the Grip Sharp tip of Seaborg tends to have really good defence, as compared to a Metal Sharp, from Master Dranzer; yet, Master Dranzer can spin longer, because there's less friction involved to slow it down- the same friction that stops Seaborg from being knocked out also causes it to slow down.

Finally, we come to the really exciting part, the impacts between Beyblades! When two Beyblades hit together, both of them will be exerting forces on each other, so you would think that it's really easy to figure out how much force is involved, and using the analysis from above, we find the resultant acceleration and thus the change in speed.

However, it's actually really difficult to find out how much force is involved when two Beyblades collide, and furthermore, the time which they spend hitting each other and exerting said forces is infinitesimally small. As a result, we'll have to use a different analysis when it comes to collisions.

On momentum
To solve this problem without advanced measurement techniques and application of calculus, we can use "momentum".

Firstly, what is momentum? It's a measure of how much impact an object has.

In mathematical terms, Momentum = Velocity x Mass

So for example, if your Beyblade is moving at 0.2m/s (meters per second), and it's 40g (grams) in weight, then it will have a momentum of 8mg/s (gram-meters per second).

How does this reflect on our analysis? Well, in all instances, unless an external force is applied, momentum must be conserved. An external force would be, say, when the two Beyblades collide, you pick one up: you're interfering with the collision, so we can't apply this rule.

However, if you just let them collide without disturbing them with your itchy fingers, then the total momentum of all Beyblades involved will be the same before and after the collision.

For example, if the Beyblade mentioned above is moving to the right with 8mg/s, and it hits a Beyblade weighing 20g that's moving to the left with 0.3m/s, what we can say is that

Initial momentum of Beyblades 1 and 2 = Final momentum of Beyblades 1 and 2

thus,

Mass of Beyblade 1 x Original velocity of Beyblade 1 + Mass of Beyblade 2 x Original velocity of Beyblade 2 = Mass of Beyblade 1 x Final velocity of Beyblade 1 + Mass of Beyblade 2 x Final velocity of Beyblade 2

Using this analysis, we can find out what happens to the two Beyblades after the collision.

We'll still need another component, on energy, to calculate what speeds will occur, but from this understanding, we can already roughly know that generally, the heavier and faster your Beyblade is, the more easily it can knock out the opponent.[quote]

I'll cover energy next. So far, is mainly basic information for beginners that is mostly intuitive.

After that, I'll go through some stuff on rotational motion. From observation on this forum, I realize that many beginners don't understand much about Beyblade rotation; once this is covered, I'll be able to explain why Beyblades move around the stadium like they do.

Following which, I'll cover topics for more advanced bladers, like the tradeoff between weight and speed, between using 10-wide and 10-heavy, and so on. Please give me suggestions on what other issues to write about for these.

I think the first two or three paragraphs are unnecessary, but perhaps it will be useful for what you will write next ...

Will you have pictures/diagrams to support this ?

Yes.

[quote=Blue]
tbh, I think its too long and explaining on a particle level to familiarize newton's laws of motion is i think to distant from bey blade when you could just use beyblade.

you typed all that, though? wow kinda intense

It's just that particles will make it much easier to explain rotational motion later. I suppose I'll just try to explain it without that.

---

Don't really have any suggestions, just keep up the good work. I like what I've read so far, it eases the reader through the different concepts well.

There was only one thing I noticed in the last part that I thought was kind of redundant,

Mass of Beyblade 1 x Original velocity of Beyblade 1 + Mass of Beyblade 2 x Original velocity of Beyblade 2 = Mass of Beyblade 1 x Final velocity of Beyblade 1 + Mass of Beyblade 2 x Final velocity of Beyblade 2

Does that really need to be there considering you basically stated the exact same thing two lines before? It does clarify the original statement, but just putting that the initial momentum of Beyblades 1 and 2 equal their final momentums was enough I think.

---

Don't really have any suggestions, just keep up the good work. I like what I've read so far, it eases the reader through the different concepts well.

There was only one thing I noticed in the last part that I thought was kind of redundant,

Mass of Beyblade 1 x Original velocity of Beyblade 1 + Mass of Beyblade 2 x Original velocity of Beyblade 2 = Mass of Beyblade 1 x Final velocity of Beyblade 1 + Mass of Beyblade 2 x Final velocity of Beyblade 2

Does that really need to be there considering you basically stated the exact same thing two lines before? It does clarify the original statement, but just putting that the initial momentum of Beyblades 1 and 2 equal their final momentums was enough I think.

It's required for calculations =/

---

It's required for calculations =/

Hmm, yeah. Probably better to show it than not since some people might not be able to figure it out themselves.

---

you know, i actually get this...although im more with elmo's statement since everything around me is like that and i grew up like that but i can use proper englis and understand things sucha as this with ease...i would like more, could u continue?

you know, i actually get this...although im more with elmo's statement since everything around me is like that and i grew up like that but i can use proper englis and understand things sucha as this with ease...i would like more, could u continue?

lol my mistake..i type too fast

Good stuff..I took a half year of physics and this makes a lot of sense. Great refresher...although I'm not sure how a noob might take to this. But it seems as if this is as simple as it gets, without throwing out base formulas that anyone can understand (which actually just may be easier, typing formulas, explaining how this formula relates to beyblade, translate it to beyblade terms, and viola, physics for noobs.

---

Good stuff..I took a half year of physics and this makes a lot of sense. Great refresher...although I'm not sure how a noob might take to this. But it seems as if this is as simple as it gets, without throwing out base formulas that anyone can understand (which actually just may be easier, typing formulas, explaining how this formula relates to beyblade, translate it to beyblade terms, and viola, physics for noobs.

Explaining things using formulas just doesn't feel right to me. My approach to physics is starting with concepts and deriving equations from it; it has to tie into real life in a way that makes sense.

Anyway, I've been very lazy on this. I'll continue tonight if I don't have a date later.

---

Double post at 9PM =O looks like I don't have a date tonight!

Anyway, I'm putting this list of contents here to remind myself what I need to finish up.

Basics:
Energy and momentum
Rotational motion

More advanced:
General:
Precession
Tradeoff between various WD (in relation to rotational inertia)
Engine Gears

Attack:
Upper attack
Tradeoff between weight and speed: Why lighter Beyblades make better smash attackers
Selection of base type and tip width

Defence:
Deflection of opponent attacks
Spin rate vs grip

Survival:
Bearing tips
Sharp tips

Now for today's little update

On energy and momentum to solve kinematics
We consider from the previous discussion on momentum, where we know that: Say we have objects at speed V1 and V2 respectively, and masses M1 and M2 respectively,

Initial M1 x V1 + M2 x V2 = Final M1 x V1 + M2 x V2

The problem with this equation is that there are an infinite number of possibilities that can occur. However, we can broadly divide collisions into two varieties, "Elastic Collisions" and "Inelastic collisions".

Both of these are ideal examples. You must realize that in theoretical physics, we get "perfect" situations, but in the real world, we'll only get very close approximations.

Anyway, in elastic collisions, all energy is conserved. You may have heard of the term "conservation of energy" before, that energy cannot be created or destroyed. We can describe the movement of objects using "kinetic energy"; In these elastic collisions, all the kinetic energy is conserved.

The kinetic energy of an object can be defined as

1/2 x mass x (velocity)^2, or (m v^2)/2 for short.

Since we know that the energy is conserved, thus, if we continue using m1, v1, etc. to denote the masses and speeds like above,

Initial (m1 v1^2)/2 + (m2 v2^2)/2 = Final (m1 v1^2)/2 + (m2 v2^2)/2

Used together with the above equation,
Initial M1 x V1 + M2 x V2 = Final M1 x V1 + M2 x V2

We can solve to find out what happens to both Beyblades after the collision.

Generally, there are two main scenarios that can happen.

Firstly, if both are almost equal in weight, you'll find that they seem to "exchange speeds". Say both Beyblades are 35 grams, but one is moving at 5cm per second while the other is moving at 1cm per second. After colliding, it's very likely that you'll find the first one recoiling at 1cm per second, while the second flying off at 5cm per second. This is rather intuitive- when something fast hits something slow, it'll slow down while the target will get thrown off quite fast.

The other possibility is that one is much heavier than the other. In this case, you'll find that the heavier one won't experience much change in it's velocity, but the lighter one will be more affected by the hit. This, again, is intuitive: if you throw a baseball at a car, you're not likely to stop the car (unless you're really really good at throwing baseballs) Of course, practically, you'll find that the first case is more advantageous for attacker types, but that'll be covered more in-depth later.

As mentioned earlier, this is an ideal situation. What happens in most real life situations is that some energy is lost. There may be friction between the two when they collide, slowing down both Beyblades and converting some kinetic energy to heat instead. Also, you might notice that Beyblades get dented in battle- and energy is needed to compress them to dent it. In fact, energy loss through compression doesn't need to be as drastic as denting- if your parts are lose, hitting the attack ring will put some energy into moving it slightly- so not all the energy is put into attacking, but some of it is wasted. It's very much like a bike helmet: they're designed to crumple up on collision, so that energy is lost crumpling up the helmet instead of crumpling up your head.

Consider the equation used above; we add another component like shown-

Initial (m1 v1^2)/2 + (m2 v2^2)/2 = Final (m1 v1^2)/2 + (m2 v2^2)/2 + WASTED ENERGY

What are the implications of these? The two scenarios shown above will largely hold true even with this wasted energy, the main implication of this is that both Beyblades will be slightly slower than expected after the collision- which is also perfectly intuitive! Beyblades slow down after launching and during the battle, obviously.

Finally, we have the extreme case, a completely inelastic collision. This happens when both things collide, and they stick together. We modify the equation slightly;

Initial (m1 v1^2)/2 + (m2 v2^2)/2 = [(m1 + m2) v^2]/2

Since both masses stick together, we can consider them to become one big object after colliding, and solve the equations accordingly.

Also, we have another case where one Beyblade breaks apart. This is the opposite, and we have:

[(m1 + m2) v^2]/2 = Final (m1 v1^2)/2 + (m2 v2^2)/2

We can generally say that ALL collisions can be solved in this manner. Keep track of everything that can happen to it during collision, and you'll have a rough idea of what's going on.

That's it for today, in any case, my dough has almost finished rising and I need to bake it.

---

I really love reading these articles.

Plus 2 points for baking your own bread.

---

Credit for the avatar and sig go to Nic. THANK YOU!

START LIVING BEFORE YOU START DYING

i do not care about physics

death gargoyle ms better than draciel ms rite?

i do not care about physics

death gargoyle ms better than draciel ms rite?

correct!

My only problem with this is that calculating the equations doesn't seem so necessary.

this is good. keep doing this. im interested especially since i havent taken physics yet.

---

Thanks, Kai-V, for my banner

i think you second article was more straight forward than the first.
collisions are spot on

Good, but at points the article seems to lose focus, as we don't see practical applications, instead, we just see pure theory.

---

I really love reading these articles.

Plus 2 points for baking your own bread.

[Image: P4100775.jpg]

i think you second article was more straight forward than the first.
collisions are spot on

Getting the hang of it, thankfully.

i do not care about physics

death gargoyle ms better than draciel ms rite?

driger ms beats death gargoyle ms anytime!!

My only problem with this is that calculating the equations doesn't seem so necessary.

I'm leaving those in for those who want to work with specifics; also, I'm generally not covering on how to solve the equations, only putting down what equations are useful. Those who're more interested in mathematics can probably solve them by themselves.

I'm very much against education being a "I say it is this way, therefore it is this way" thing- I'll very much prefer to build up the evidence for it and demonstrate it.

---

I am generally in the middle of studying this subject, and my understanding is mostly due to how you explain things.

The part about elastic and inelastic collisions could be separated in a better way, though. Just adding "Elastic collisions: [...] Inelastic collisions: [...]" could help. More subtitles in general would be a good thing ...

This one is going to be a pain in the carp to write about. Bleh.

If you're already familiar with linear motion, it'll make things much easier when studying angular motion. You can apply the same concepts, just that things keep going round in circles when discussing angular motion. Honestly, linear motion is straightforward carp.

On rotational motion

Rotational motion, also commonly known as spinning, is a very easily misunderstood topic. To start off, I should first begin to define it.

In normal kinematics, as covered earlier, we had a few measurements; we had distance/displacement, speed/velocity and acceleration.

For rotational motion, we have an angular equivalent for most of these values, named because most measurements are in terms of angles. For example, we could say that the object has rotated 20 degrees. In comparison, non-rotational motion is known as "linear" instead of angular because it tends to move in lines instead of angles.

[Image: angular.jpg]

Here's an example. Usually, when we measure angles, we go anti-clockwise by convention.

We usually use the symbol θ (theta) to represent angle. It's measured in "radians": one complete circle is 2π radians. This makes it very convenient to convert between angle and distance: geometrically, one circle has a circumference of 2πr.

In this measurement system, since one circle has an angle of 2π, any arc that is bigger or smaller than one complete circle will have a distance of 2πr x (θ/2π) = θr, where r is the radius. That links distance and angle together.

If two objects move through the same angle, they don't necessarily cover
the same distance.

[Image: angular2.jpg]

As you can see, the two distances are obviously not equal. You might notice that usually attack types make large sweeping arcs while launched normally, while a sliding shoot covers many small circles. Both are traveling at the same speed, but because the sliding shoot makes arcs with a smaller radius, it can cover more circles in the same time.

This brings us to the next definition, "angular velocity". It's usually represented by ω (omega). It's how much angle is covered over time: if the angular velocity is two radians per second, that means that every second, it covers two radians.

Finally, we have angular acceleration, usually denoted by α (alpha). Angular acceleration is a measurement of how fast angular velocity changes, just like acceleration is a measurement of how fast velocity changes.

Angle, angular velocity and angular acceleration are akin to distance, speed and acceleration. You can use the same concepts for both; just bear in mind that angular motion makes it go in circles while linear motion makes it move in lines.

On torque and moment of inertia
Torque is the angular equivalent of force, and the moment of inertia is the angular equivalent of mass.

Torque is simple the cross-product of force and radius. That is to say, it depends on both how strongly you push, AND, how far from the center you push. But bear in mind also that the further away from the center you apply your force, although you'll have more torque, you'll also have to push a further distance: this is just like bicycle gears. On low gear settings, it's really easy to pedal, because your pedaling is connected to the larger gears so you'll be able to push with greater torque for less force: but, you'll have to pedal more to cover the same distance. However, on higher gear settings, although you feel more resistance while pushing, every one round you make covers a lot more distance on the ground.

Also, since it's a cross-product, you'll have to make sure you're pushing in the right direction. I won't go too much into vector analysis here, so I'll put this simply; when you want to make something spin, you would probably push it in the direction you want it to spin:

[Image: angular_3.jpg]

This is fairly obvious. Now, let's consider what happens when we push at an angle.

[Image: angular4.jpg]

You can divide it into two components, as shown. Only part of the force makes it spin. Actually, the other part isn't really "wasted": just that it doesn't apply to the spinning, so it isn't useful in this case.

The force marked in red is applied linearly, instead of angularly. In a collision between two Beyblades, you don't just have to consider their linear motion, because they're spinning. So this contributes to why faster-spinning Beyblades usually can perform smash attack better: you have both the linear motion of the Beyblade slamming the opponent, and also, the spinning will exert another force on the opponent as well.

This also explains why same-spin direction is better for smash attack: when they're spinning in the same direction, you have both the linear and angular component to your attack, while if they're spinning in opposite directions, usually, the angular component is lost.

Now, we cover moment of inertia. In linear motion, we just need to consider the mass: whatever shape it is, two objects of the same mass will travel as fast after being pushed by the same force.

However, you might wonder why 10-wide is supposedly better for survival than 10-heavy, even though it's lighter, since it's always easier to stop a lighter object than a heavier object in linear kinematics.

This is because in angular motion, the "moment of inertia" is dependent both on mass, and the square of the radius: the radius has a much greater effect on the moment of inertia, as compared to torque or angular acceleration.

Since 10-wide has a much greater radius, you can easily tell at a glance that it has a much greater moment of inertia than 10-heavy, even though it's lighter.

So, it's harder to stop 10-wide from spinning. However, because 10-heavy has a smaller moment of inertia, it's easier to make it spin faster when you launch it; and because of the simple fact that it's heavier and spinning faster, it makes it difficult to knock out and rather suitable for compacts or other combos that make use of high spin rate.

That's also the idea behind the HMC; a centralized weight so it can spin faster.

Remember that the torque applied is dependent on both force and radius. However, whichever Beyblade you're using, when you attach it to your shooter, the prongs of the shooter still stick into it at the same places: so, the place where you apply the force is the same, no matter what parts you're using. What makes the difference in spin rate would thus be the moment of inertia associated with your Beyblade.

Actually, it's not so simple as just considering the radius. You also have to consider the shape itself. 10-wide is an easy case, since almost all the weight is at the sides. However, the magnet WD, which has nearly the same radius as the 10-wide, has its weight evenly distributed. Although it is slightly heavier, I would think it has a lower moment of inertia, because most of its weight is more towards the center as compared to the 10-wide, which has everything concentrated at the sides.

However, since I doubt most people who will be reading this will have covered integration, I'll just leave it at that. In any case, after Beyblading, you'll probably have the intuition to figure this out roughly.

I hated explaining that. I'll write on angular momentum, energy analysis for rotational motion and finally, be able to explain why Beyblades move the way they do in the stadium, and that'll conclude the beginner's guide. Advanced guide will come later after my exams. Coincidentally, does anyone here take AP or SAT exams?

---