Endercreeper01

I'll play. I just registered at chess.com

I do

The difference is that even though 1/ is infinite, 1/- is negative infinity. You can have a positive and a negative . cannot be 0 because 0 is neither negative or positive.

x/0 is undefined because if you take the limit as n approaches 0 of x/n, then from the positive side, it approaches infinity and it approaches negative infinity from the negative side, so therefore, x/0 is undefined

d would be for the derivative. If you had a function f(x) which is the same as y, then you could write the derivative 2 ways. One is f'(x). The second is dy/dx. With the second derivative, you could also write it 2 ways. One is f''(x). The other is d2y/dx2

The acceleration would be the second derivative of the distance with respect to time, which is written as d2s/dt2 Where s is distance and t is time

D is the drag coefficient it would have if it was a rectangular prism divided by l! What part don't you understand?! When you have Dl, it becomes the drag coefficient it would have if it was a rectangular prism with that same length! What is so hard to understand about that?! This is just pissing me off! D Now has units of 1/l because of what I said! It isn't just all of a sudden! I told you why it was in units of 1/l ! Seriously, listen to what I am saying. And now, you will just consider this "trolling".

The thing is, D is not dimensionless and has units of 1/l. Therefore, when you have l in the equation, then the 1/l unit and the l unit cancels out. This is because the definition of D is that it is the form drag coefficient it would have if it was a rectangular prism divided by the length, which makes it the form drag coefficient it would have if it was a cube, but with units of 1/l. When you multiply by l, the 1/l and l units cancel out, giving you no dimensions. So basically, Dl is the form drag coefficient it would have if it was a rectangular prism with the same length.

You have not answered my question about how I am "trolling" you. How am I trolling you?! And how can you criticize my theory if you don't even know what it means?!

It would not be wmr and it would instead be wmr2 because w is the angular velocity, and the velocity is v=wr. Since L=mvr, then it is also L=wmr2 because v=wr, and if we plug that into the equation, we get L=(wr)mr, and then we get wmr2.

Yes, you can do that. That's true, because you have the arc length further away. You can imagine it as if the object was moving along the circumference of a circle. If it is further away from the circle (a bigger radius), it is traveling along a bigger length, since circumference is 2 pi * r, where r is radius. Because angular velocity is how much the angle change per unit time, this means that if you are further away and you have the same angular velocity as something else that is closer, you have a bigger velocity. This is because you are traveling across a bigger arc length in the same period of time. This means that velocity is the angular velocity times the radius v=wr

v=r*w means that the velocity is the radius times the angular velocity. The angular velocity is a measure of how much the angle is changing (in radians) per unit time, and it is 2pi*f, where f is the frequency. It can also be written as the derivative of the angle with respect to time, or da/dt, where a is the angle and t is time And yes, the tangental velocity would be s/t, where s is distance, but when you have accelerations, you would use the derivative of distance with respect to time, or ds/dt for velocity.

You could still divide, but you would have to convert the units. How would it have units if it is using the length without units?

You would only divide by the number if they both used the same units for measuring length.

I know there is already a length in Re already, that's why you would divide by length without using units if both Reynolds numbers were using the same units

Then, how would you explain the fact that the Re for a unit cube and the Re for a rectangular prism with length l differ by a factor of l ?! You have even shown signs of not reading my posts on previous pages. And now, you are calling me a troll! How am I trolling you?!

I clearly said we use the length, but do not include the units! Where do you not see this?! Did you even read the post?! This is starting to piss me off. Why don't you pay attention to what I say?!

This is the drag coefficient In this case, we have the Re for a unit cube. If you compare it to the Re for a rectangular prism with length l, the Re would be Re=Re0/l, where Re0 is the Re for the rectangular prism. With the thing about units, this makes us use the length value without units You don't even seem to know what I mean by D. You think that's the drag itself. No it is the drag coefficient for a unit cube with the same ratio of Re / l

l is length in my equation. Dl is basically the form drag coefficient for a rectangular prism with the same Re, and D is the form drag coefficient for the object with the same Re / l

1. I assumed there was no skin friction. 2. My theory is just a general equation. It cannot solve things like that unless we have a value for D. You keep talking about the Navier Stokes equations. Do you want a solution of D from the Navier Stokes equation?

The average angle is 45, and I assumed a was for the midpoint on the circle, with that making a=45. However, da/dx I snot 0 since the angle changes over the sphere 24/Re can be used for Re<<1 for a sphere, but my theory uses the other variables in order to determine the coefficient. Using my theory, we can also determine the value of D for Re<<1 We will assume there is little skin friction Since both average angles are the same and the average angle is 45, we can make an equation: 24/Re=Dlcos(45)=Dl21/2/2 Divide both sides by 21/2/2, and you get Dl=33.94/Re We now can have the following equation: D=33.94/(Re/l)=33.94l/Re Now, we have a value for D at Re<<1, D=33.94l/Re

I was just assuming random values, such as D being 1.05 at v=0.1 m/s. Do you want me to make actual calculations? If so, you must tell me all of the information needed in my equation, or else I cannot make accurate predictions., such as viscosity of the air at that temperature, exact velocity, etc. Also, I am unable to find a graph that relates the Re to Cd for a cube. Don't believe me (since you never do) ? If so, you try finding a cube Cd vs Re graph. You try assuming random values and comparing to experimental results. You try putting in all of these numbers in only half an hour, something I had to do. And, you have to compare your results that had random values inserted into the equation with experimental data You try doing all this. Also, don't you remember me saying something about a d2a/dx2? That means it was never complete. Didn't you notice? I don't even think you understand my theory. Also, how can I make predictions if I don't know the value of D?

According to my theory, I need certain things, but since you will just not give them to me and you just think I am severely lacking in knowledge, I will have to make up my own values. I need length for the form drag, but I will assume that the length is 1. I needed temperature for the viscosity, so I will just assume it is at 500 degrees kelvin, making the viscosity about 2.7. I will assume the density is about 1.27. Notice how all I can do is assume? Now to the calculations, To calculate the form drag, we know some things first. We know Re is less then 1, and I will assume it is 0.1, making the velocity abut .21 m/s. The average angles for both I sides of the sphere are both 45, and it is length 1, so we can reduce it to Dcos(45), which makes 21/2D / 2 Because I could not find a graph relating the Re to Cd for a cube, then I will just assume it is 1.05, making the form drag coefficient about .74 Now, I will calculate the skin friction drag coefficient. In this case, q=.028, and the viscosity divided by q will be about 96.43 Now, I will assume a=45, and we first need to find ucos(45), which will be .21cos(45), which is about .11 Now, we will need to find dy/dx, which we will assume is 1, so we will be dividing by 2, giving us .055 We now need to find d2y/dx2, which for a sphere, is going to be d2/dx2 r*sin(a) We will get -rsin(a)d2a/dx2, and a=45, so we get about -0.7d2a/dx2 Now, I will try to figure our d2a/dx2, and I will tell you when I do, but it must be small We now multiply that by 96.43 to get about -6.75d2a/dx2.

Is it -| x * 31/2 * | x |1/2 * ( 1 / x2)1/2 | / 2 + 1.251/2/2, where |x| represents the absolute value, and assuming x0 = - x2?

What exactly is the length of this sphere? The temperature of the air? Is the flow laminar or turbulent? I need to know all of the variables and factors.