This right here my friend is exactly what I was looking for! I love that you put in real numbers and not just the writing out the equation. Thanks for helping me grasp this concept!
Thank you. I'm glad it helped. Might I add that I greatly admire your work. One of my favorite being your interview with Elon Musk. Seeing his eyes light up as you began asking about closed cycle combustion was just delightful. Cheers.
Thanks ... I think. If I could have seen away around the calculus, I would have. I don't think it would have been reasonable to assume an audience that knows integral calculus. Even if you've taken it, math algorithms vanish pretty quickly once you stop using them. That said, I don't think there's anything wrong with aiming high. I watch a fair amount of recreational math videos, and my favourites aren't afraid of tackling tougher ideas. I like to make math as simple as I can, but no simpler. I hope I've done that.
you can split -ln(m0/mf) into -ln(m0)+ln(mf) and find the derivative in a simpler way. The derivative of -ln(m0), as we assumed it a constant will just be 0. skipping some steps. i know i'm quite late in but thank you so so much for these videos! i'm enjoying them immensely and they've been helpful and informative to many others like me. you've made these topics really fun and simple to understand :)
Great video this will actually help me with the real rocket I’m building . Have to brush up on the calculus parts but that’s the fun of learning keep up on the math videos
Nice vid. Bit of feedback from a non-mathematician/layman - a box in one of the corners perpetually showing what each symbol signifies would be very useful.
Good observation, but this would be dealt with by attaching the signs to the velocities. As they are in opposite directions, the velocities would have opposite signs. With v_e negative, we would get the formula you're intuitively seeing. This is kind of a hidden subtlety in this part of the video, as I never put in any numbers. What's neat is that the later calculus takes care of the negatives anyway. Ain't math great!
Bravo mike, though be careful with some of those calculations, as go in this case is just gravity, but kerbal has a similar enough gravity for this to slide. Doing similar calculations here so yeah
Gonna look seriously stupid here but... Does the gravitational constant change depending on which moon or planet you're going to be leaving? For example, would 9.81 be used for Duna, the Mun and Minmus?
The gravitational parameter μ = GM changes for each celestial body. It depends on how much mass that body has. en.wikipedia.org/wiki/Standard_gravitational_parameter
Great vid, but I'm baffled as to why we need the ln? I get that it's the natural log...but what does it provide here? What does taking the natural log of the intial and final mass?
From calculus, the natural log appears because we are taking an integral of a fraction with the variable in the denominator. This is because the derivative of y = ln(x) (that is the rate at which that function is changing for any particular value of x) is dy/dx = 1/x. The why of this can't help but get into calculus as far as I can see. Maybe I'll do some googling and see if there is a more intuitive way to see this. That said, it's not surprising that logs have to show up. If you imagine the graph of the velocity of a rocket compared to the mass of fuel consumed, it can't be linear. This is because as you consume fuel, mass decreases, which means the rate at which velocity increases (ie. acceleration) won't be constant. Acceleration will be constant increase as well. Behaviours in which the rate of change is itself changing at a constant rate will always involve an exponential function (y = b^x) or its inverse, the logarithmic function. Why the natural logarithm (ie. log base e rather than another base)? Well, it ain't called the natural log for nothing. ln(x) and e^x pop up whenever you are looking at real world examples of exponential change.
@@MikeAben Thanks so (so very) much for your time to reply. Yea I haven't hit integrals yet in calc, i'm still trying to get my head around deriviatives and the different product, quotient and chain rules... And I"ve touched on acceleration and velocity in physics. So yea that part I get. I suppose I'm putting too much thought into questiong, but the whole idea of just accepting 'e' as the answer bothers me! lol thanks though!
@@horizonbrave1533 You can never put too much thought into questioning. Question away. :) Here's a 3-blue 1-brown video on the importance of e^x. A great math follow, by the way. ruclips.net/video/m2MIpDrF7Es/видео.html Un truth, you could use any base for the logarithm in the rocket equation, but it would introduce an extra constant that disappears when you use log base e, which is ln. It all stems from the property of e^x being its own derivative. There is something special about the number e. Personally, I find that the derivative of ln(x) is 1/x to be one of the most surprising things in introductory calculus.
I agree there probably isn't a non-calculus solution to this, and picking the audience is hard. I'm just not sure if you can give an explanation involving calculus to a non-calculus audience. I guess the only other way is to say "This is a calculus proof, trust me," but that's not exactly satisfying, as you pointed out.
No one should ever trust me, at least that's what I tell me students. In my opinion, everything should be justified with the rigour of the justification appropriate to the audience. Even things that are defined as true should at least be demonstrated as sensible. The rocket equation is such a biggy, though. It feels right dealing with it early. With that done though, the rest of this series should be calculus free.
yeah, that's a tough one. Imagining adding up all of the little dvs that are the result of chucking all the dms out the back of the rocket, plus knowing about conservation of linear momentum and not having any calculus to hang it on, would be daunting. But if you have the basics, it was a fun into to the rocket equation that makes plenty of sense.
If you're talking about the calculus, I jumped pretty much into the middle of an introductory calculus course (or an advanced Newtonian Mechanics course). There is a lot of build up to get to the point in this video that I tried to more-or-less smooth over, but knew that it was a tough ask of anyone who had never taken a calculus course. There is a lot of content on RUclips. I've not surveyed it in any depth, but if you do a search for "introduction to calculus" you'll find playlists that should build to the point where you will begin to understand differentiation and integration.
I tried to make sense of your vis viva equation video and my god you must be some kind of maths god or something cuz i have no clue about any of it haha you said you only needed some high school physics to be able to understand but unfortunately I failed GCSE science at school so I don't have the necessary knowledge to understand vis viva equation, so I was wondering if you could point me in the direction of high school physics that I could learn?? Thanks for all these tutorials though, it's super kind of you
It depends on your calculator. For many (like the typical ones on phones) you have to do the order right for it. That's inside parentheses, then functions like ln, then multiplying, then adding. To do the calculation at 3:00 with such a calculator it's first 7.65 / 6.85. Get that answer and push ln, then multiply by 345 and 9.8. Again, this can be different on different calculators. They're not all the same. Play around until you get my answer to figure out how your calculator works. I hope this helps.
@@MikeAben oh wait, i am using a scientific calculator and wait i think i got that wrong, can you elaborate just a bit more with the log part please, many thanks your wesome!!!!!
@@rumcoke8909 On a scientific calculator, you should find a button labeled 'ln', that's a lower case 'L'. This may be written above the button in which case you'll have to push '2nd' or 'shift' first. The button works differently on different calculators. On some you can type pretty much what you see, ln(7.65/6.85). It'll know to do the dividing in the brackets first, but some calculators don't do this. You have to do the dividing first, which gets 1.11678..., then take the log. Good job on sticking this out. It's tough when you haven't done this is school. If your still having trouble, my advice would be to google the make and model of your calculator. You should be able to find a manual with specific instructions.
Nah, you really need some (recent) calculus to get what's going on in the end. It's not really fair and if I could have thought of a way without calc, I would have done it that way. It's just such an important equation latter.
@@MikeAben Yea, to take calculus way back when, when I was in high-school, you had to be on the advanced diploma. I was removed from it because it also required spanish; which I failed... I appreciate the kind words, though. I like math and usually don't even break a sweat with these kind of things but I get lost with how many times the formula changes and end up losing focus.
This right here my friend is exactly what I was looking for! I love that you put in real numbers and not just the writing out the equation. Thanks for helping me grasp this concept!
Thank you. I'm glad it helped. Might I add that I greatly admire your work. One of my favorite being your interview with Elon Musk. Seeing his eyes light up as you began asking about closed cycle combustion was just delightful. Cheers.
OMG
I absolutely love how you respect the audience's ignorance. :) This was very interesting, thank you so much!
Thanks ... I think. If I could have seen away around the calculus, I would have. I don't think it would have been reasonable to assume an audience that knows integral calculus. Even if you've taken it, math algorithms vanish pretty quickly once you stop using them. That said, I don't think there's anything wrong with aiming high. I watch a fair amount of recreational math videos, and my favourites aren't afraid of tackling tougher ideas. I like to make math as simple as I can, but no simpler. I hope I've done that.
I know its been 5 years but Im using this for my Math IA on the Rocket Equation and Ive never played KSP. Thanks King
Man, I miss the brain i had that could do calculus without thinking 😂 Great video!
you can split -ln(m0/mf) into -ln(m0)+ln(mf) and find the derivative in a simpler way. The derivative of -ln(m0), as we assumed it a constant will just be 0. skipping some steps.
i know i'm quite late in but thank you so so much for these videos! i'm enjoying them immensely and they've been helpful and informative to many others like me. you've made these topics really fun and simple to understand :)
Great video this will actually help me with the real rocket I’m building . Have to brush up on the calculus parts but that’s the fun of learning keep up on the math videos
Awesome. Thanks. Good luck with your rocket.
Nice vid. Bit of feedback from a non-mathematician/layman - a box in one of the corners perpetually showing what each symbol signifies would be very useful.
Good idea, especially in this one which is very symbol heavy.
I i havent yet learned a single calculus and i now feel my brain is flotaing away. Jesus. I srsly dont know what those symbol on calculus part is.
That's why I left it to the end.
Life saving video. Thumbs up
Surely when calculating momentum at the start we minus the delta m times exhaist velocity since it is travelling opposite to the rocket.
Good observation, but this would be dealt with by attaching the signs to the velocities. As they are in opposite directions, the velocities would have opposite signs. With v_e negative, we would get the formula you're intuitively seeing.
This is kind of a hidden subtlety in this part of the video, as I never put in any numbers. What's neat is that the later calculus takes care of the negatives anyway. Ain't math great!
Bravo mike, though be careful with some of those calculations, as go in this case is just gravity, but kerbal has a similar enough gravity for this to slide. Doing similar calculations here so yeah
Watched till the end :)
Gonna look seriously stupid here but...
Does the gravitational constant change depending on which moon or planet you're going to be leaving? For example, would 9.81 be used for Duna, the Mun and Minmus?
Not stupid at all. The answer is no. It's always 9.81. The definition of ISP is very Earth centric.
The gravitational parameter μ = GM changes for each celestial body. It depends on how much mass that body has.
en.wikipedia.org/wiki/Standard_gravitational_parameter
Next - #4: Eliptical Insertions - ruclips.net/video/ZZvvj3i2eZk/видео.html
Hey dude can you make one about Hover-slam (Suicide burn)
Great job!!!
Ngl, I did have calculus in middle and highschool, but it has been 3 years since that time... 99% of it is gone.
Great vid, but I'm baffled as to why we need the ln? I get that it's the natural log...but what does it provide here? What does taking the natural log of the intial and final mass?
From calculus, the natural log appears because we are taking an integral of a fraction with the variable in the denominator. This is because the derivative of y = ln(x) (that is the rate at which that function is changing for any particular value of x) is dy/dx = 1/x. The why of this can't help but get into calculus as far as I can see. Maybe I'll do some googling and see if there is a more intuitive way to see this. That said, it's not surprising that logs have to show up. If you imagine the graph of the velocity of a rocket compared to the mass of fuel consumed, it can't be linear. This is because as you consume fuel, mass decreases, which means the rate at which velocity increases (ie. acceleration) won't be constant. Acceleration will be constant increase as well. Behaviours in which the rate of change is itself changing at a constant rate will always involve an exponential function (y = b^x) or its inverse, the logarithmic function. Why the natural logarithm (ie. log base e rather than another base)? Well, it ain't called the natural log for nothing. ln(x) and e^x pop up whenever you are looking at real world examples of exponential change.
@@MikeAben Thanks so (so very) much for your time to reply. Yea I haven't hit integrals yet in calc, i'm still trying to get my head around deriviatives and the different product, quotient and chain rules... And I"ve touched on acceleration and velocity in physics. So yea that part I get. I suppose I'm putting too much thought into questiong, but the whole idea of just accepting 'e' as the answer bothers me! lol thanks though!
@@horizonbrave1533 You can never put too much thought into questioning. Question away. :)
Here's a 3-blue 1-brown video on the importance of e^x. A great math follow, by the way.
ruclips.net/video/m2MIpDrF7Es/видео.html
Un truth, you could use any base for the logarithm in the rocket equation, but it would introduce an extra constant that disappears when you use log base e, which is ln. It all stems from the property of e^x being its own derivative. There is something special about the number e. Personally, I find that the derivative of ln(x) is 1/x to be one of the most surprising things in introductory calculus.
An introduction to calculus? Well, hopefully someone seeing this will choose to take that class.
I couldn't figure out, nor find, a non-calculus way of doing this one. That's why I split it up the way I did putting the calculus off until the end.
I agree there probably isn't a non-calculus solution to this, and picking the audience is hard. I'm just not sure if you can give an explanation involving calculus to a non-calculus audience.
I guess the only other way is to say "This is a calculus proof, trust me," but that's not exactly satisfying, as you pointed out.
No one should ever trust me, at least that's what I tell me students. In my opinion, everything should be justified with the rigour of the justification appropriate to the audience. Even things that are defined as true should at least be demonstrated as sensible.
The rocket equation is such a biggy, though. It feels right dealing with it early. With that done though, the rest of this series should be calculus free.
True enough.
yeah, that's a tough one. Imagining adding up all of the little dvs that are the result of chucking all the dms out the back of the rocket, plus knowing about conservation of linear momentum and not having any calculus to hang it on, would be daunting. But if you have the basics, it was a fun into to the rocket equation that makes plenty of sense.
I enjoy this, but know very little about any of this I am taking pre-cal next semester do you know the best way I can learn this?
If you're talking about the calculus, I jumped pretty much into the middle of an introductory calculus course (or an advanced Newtonian Mechanics course). There is a lot of build up to get to the point in this video that I tried to more-or-less smooth over, but knew that it was a tough ask of anyone who had never taken a calculus course.
There is a lot of content on RUclips. I've not surveyed it in any depth, but if you do a search for "introduction to calculus" you'll find playlists that should build to the point where you will begin to understand differentiation and integration.
Thanks, will do.
I tried to make sense of your vis viva equation video and my god you must be some kind of maths god or something cuz i have no clue about any of it haha you said you only needed some high school physics to be able to understand but unfortunately I failed GCSE science at school so I don't have the necessary knowledge to understand vis viva equation, so I was wondering if you could point me in the direction of high school physics that I could learn?? Thanks for all these tutorials though, it's super kind of you
wait, , what order and how do you calculate the numbers when using the calculator, please respond i need help :(
It depends on your calculator. For many (like the typical ones on phones) you have to do the order right for it. That's inside parentheses, then functions like ln, then multiplying, then adding. To do the calculation at 3:00 with such a calculator it's first 7.65 / 6.85. Get that answer and push ln, then multiply by 345 and 9.8. Again, this can be different on different calculators. They're not all the same. Play around until you get my answer to figure out how your calculator works.
I hope this helps.
@@MikeAben thank you so much, you are a great teacher, your awesome, i also subscribed to you, THANKS YOU!!!!!
@@MikeAben oh wait, i am using a scientific calculator and wait i think i got that wrong, can you elaborate just a bit more with the log part please, many thanks your wesome!!!!!
@@rumcoke8909 On a scientific calculator, you should find a button labeled 'ln', that's a lower case 'L'. This may be written above the button in which case you'll have to push '2nd' or 'shift' first. The button works differently on different calculators. On some you can type pretty much what you see, ln(7.65/6.85). It'll know to do the dividing in the brackets first, but some calculators don't do this. You have to do the dividing first, which gets 1.11678..., then take the log.
Good job on sticking this out. It's tough when you haven't done this is school. If your still having trouble, my advice would be to google the make and model of your calculator. You should be able to find a manual with specific instructions.
Figured 3rd time's the charm. Conclusion: I'm just too dumb for this.
Nah, you really need some (recent) calculus to get what's going on in the end. It's not really fair and if I could have thought of a way without calc, I would have done it that way. It's just such an important equation latter.
@@MikeAben Yea, to take calculus way back when, when I was in high-school, you had to be on the advanced diploma. I was removed from it because it also required spanish; which I failed...
I appreciate the kind words, though. I like math and usually don't even break a sweat with these kind of things but I get lost with how many times the formula changes and end up losing focus.