Recent content by Roboto

My Lagrangian Dynamics textbooks from 30+ years ago had the comment that the derivation of the classical Kinetic Energy equation assumes acceleration is a constant. I just don't understand why multiple authors would make comments that the derivation assumes acceleration is a constant.

Yeah, I understand the conservation of energy, and the derivations. Easy peasy. What I haven't figured out is why would my textbook authors, along with searches on the internet, have this comment that the derivation of the kinetic energy equation is based on the assumption that acceleration is...

The classical definition to the Kinetic Energy equation is KE=integral of F*dx where F=d(m*v)/dt. When mass is constant, KE=(1/2)m*v^2. I am working on a vibration problem at work and having to review my Lagrangian Dynamics books from 30 years ago. So my question is about all of the authors...

Thanks, I like the way you got the velocity component worked into this. Awesome, thanks.

Ultimately I am working on the change in temperature within the tank as the high velocity waterjet is dissipated in the tank. m*Cp*dT/dt = Qin - mdot*Cp*Tout - Q_sides - Q_surface where m = mass of the water in the tank Cp = specific heat of the water dT/dt = rate of Temperature change in the...

Here is a fun question for you heat transfer experts. I have a high pressure waterjet firing into a open tank of water. The hydraulic power of the jet entering the tank is known, along with the flow rate of the jet entering into the tank. All of the energy of the jet is dissipated into the...

In this example, it happens to be that way. The function can have numerous parameters passed to it. In this example, time is not a variable that is being passed. if we want to get really formal, dy/dt=f(x, dy/dt, dy2/d2t,t). since the original equation was mass*d2y/dt2=K*(dy/dy)2 For Runge...

if k1=u2 then k2=u2+h/2*k1=u2+h*u2/2 Then k3=u2+h/2*k2=u2+(h/2)*(u2+h*u2/2)=u2+h*u2/2+h2*u2/4 then k4=u2+h*k3=u2+h(u2+h*u2/2+h2*u2/4)=u2+h*u2+h2u2/2+h3u2/4 thus yn+1=yn+(h/6)*(k1+2k2+2k3+k4) = yn+1=yn+(u2*h/6)*(6+3h+h2+h3/4) See how this happens? Crazy right? Units are not preserved, and yet...

if we take a simple particle velocity decay in the air, mass*acceleration = K*Velocity2 where K is drag and geometry coefficents of the particle. If we let u1=position, y, and du1/dt=velocity, and then we let u2=velocity and du2/dt=acceleration With Runge Kutta, dy/dt=f(u(t),t)...

Exactly, that was my thoughts. The first sanity check of something is to check the units. But does that have to be maintained when using numerical methods to solve something as opposed to algebraic. I will get my document cleaned up and figure out how to post it here. It several pages.

When conducting numerical methods using 4th Order Runge-Kutta do the physical units have to be maintained? This never occurred to me until I was writing out all the steps in detail when showing someone I work with the method using a simple projectile motion with drag. It had 4th Order time...

Thanks for the reply. My interest is in the power stroke to generate that initial velocity and launch angle. This is the part of archery physics that is utterly ignored, and yet it is the most critical part of archery. The reason I was looking at the pitch from the tail end of the arrow, nock...

I am working on modeling the acceleration and initial launch angle of an arrow shot from a bow during the power stroke. From an aiming point of view, the launch angle of the arrow doesn't change with respect to the target. But changing the nock position where the arrow nock attaches to the...