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.
What do you think was the best thing you learned from this video?
adhesive limit
Awesome!
Thanks!
When in doubt, more EMD🚂🇨🇦🙋
How to consider/show starting resistance in this tractive effort curve.
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.