There is virtually no loco that use diesel as the traction motor. Yes, that would also include diesel locos. Most diesel locos have a hydrolic or electric traction motor. Very rare examples does have gearboxes. Also the most common restriction for old locos is line frequency. Newer locos that run on Ac-dc-ac they are pretty much just restricted for regulation or power. That is, a modern loco with no load on it would accelerate untill something else breakes.
I use some ()-s so it is always clear what I mean. P is power, F is force, t is time, v is velocity, d is distance, m is mass, W is work A) If you meant P=F/(tv), we would get P=F/t(d/t), which simplified with t is P=F/d. F/d would be something like m/t^2" which doesn't make much sense, so this cannot work. B) If you meant P=(F/t)v, we can rewrite it like this: P=(F/t)(d/t), which is (Fd/t^2). W=Fd, so we get P=W/t^2 which is also not true, because P=W/t. So here you would divide by an extra t. Our friend in the video is 100% accurate.
Very informative. all new to me. Best thing I learned? fuel burn time limits max engine rpm
Very nice explanation bro. Thanks
very informative, pls make more videos! thanks
Thank you for this accurate video.
Awesome!
Thanks!
How to consider/show starting resistance in this tractive effort curve.
What do you think was the best thing you learned from this video?
adhesive limit
When in doubt, more EMD🚂🇨🇦🙋
Maybe a little bit about the power point will complete it in totality..what do you think?
Could you please elaborate a bit more? I will be sure to expand on it.
There is virtually no loco that use diesel as the traction motor. Yes, that would also include diesel locos. Most diesel locos have a hydrolic or electric traction motor. Very rare examples does have gearboxes.
Also the most common restriction for old locos is line frequency. Newer locos that run on Ac-dc-ac they are pretty much just restricted for regulation or power. That is, a modern loco with no load on it would accelerate untill something else breakes.
Wouldn't the peak power calculation be: Peak Power = Force/Time × Velocity?
I use some ()-s so it is always clear what I mean. P is power, F is force, t is time, v is velocity, d is distance, m is mass, W is work
A) If you meant P=F/(tv), we would get P=F/t(d/t), which simplified with t is P=F/d. F/d would be something like m/t^2" which doesn't make much sense, so this cannot work.
B) If you meant P=(F/t)v, we can rewrite it like this: P=(F/t)(d/t), which is (Fd/t^2). W=Fd, so we get P=W/t^2 which is also not true, because P=W/t. So here you would divide by an extra t.
Our friend in the video is 100% accurate.