Physics challenge!
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Original post by Keyhofi
There are 16 aliens on a house. The house is orange and weighs 10000kg. The aliens' height range is 120cm - 190cm as a Gaussian distribution.
Using the above numbers and physics, calculate how many strands of hair the aliens have on their head.
P.S. If you can't do this simple calculation you have no chance of getting into Oxbridge.
Using the above numbers and physics, calculate how many strands of hair the aliens have on their head.
P.S. If you can't do this simple calculation you have no chance of getting into Oxbridge.
How are we supposed to do this without telling us the aliens' names? Stupid question.
Original post by Plutonian
Thought this place was full of physics geniuses?
I think that OP is a 30 year old lonely Physics/Engineering phd student that uses TSR as a means of making himself feel good about himself.
No offence.
Original post by Rubgish
I didn't say I couldn't do it, I just said I couldn't do it *accurately*, there is a big difference.
As it is, I looked at your method. I even plugged some numbers into your final equation. Didn't match up with the correct answers. Not sure what you are doing tbh, especially with that whole angle thing in the middle that doesn't seem to have any impact on your final formula.
As for the method itself, it's pretty much exactly what anyone would do if that was all the information they had. You did f=ma and then put that into p=fv, not exactly ground breaking stuff.
As it is, I looked at your method. I even plugged some numbers into your final equation. Didn't match up with the correct answers. Not sure what you are doing tbh, especially with that whole angle thing in the middle that doesn't seem to have any impact on your final formula.
As for the method itself, it's pretty much exactly what anyone would do if that was all the information they had. You did f=ma and then put that into p=fv, not exactly ground breaking stuff.
Huh? Yes it does. You don't even know the answer?
I've never seen that equation anywhere regardless of it being easy to derive. And that isn't even the "new" part. the angle bit is. You are confused, there are 2 final equations, one to tell thrust in terms of power and the other to tell thrust in terms of angular velocity. Regardless of all this I'm not exactly looking for a nobel prize, I'm simply want to share my work with the hobbyist community (after making sure there isn't any other equally simple method by making this thread) because I found there is no simple theory for all of this and hobbyists are always asking how much lift their drone or model fighter will produce. My method has 2 super simple equations as opposed to integration and experimental constants and all that jazz in blade element theory or worse full blown FEM. As usual internet people are being hostile when I'm only trying to share my ideas. Don't know why I bother with forums.
Original post by Plutonian
Huh? Yes it does. You don't even know the answer?
I've never seen that equation anywhere regardless of it being easy to derive. And that isn't even the "new" part. the angle bit is. You are confused, there are 2 final equations, one to tell thrust in terms of power and the other to tell thrust in terms of angular velocity. Regardless of all this I'm not exactly looking for a nobel prize, I'm simply want to share my work with the hobbyist community (after making sure there isn't any other equally simple method by making this thread) because I found there is no simple theory for all of this and hobbyists are always asking how much lift their drone or model fighter will produce. My method has 2 super simple equations as opposed to integration and experimental constants and all that jazz in blade element theory or worse full blown FEM. As usual internet people are being hostile when I'm only trying to share my ideas. Don't know why I bother with forums.
I've never seen that equation anywhere regardless of it being easy to derive. And that isn't even the "new" part. the angle bit is. You are confused, there are 2 final equations, one to tell thrust in terms of power and the other to tell thrust in terms of angular velocity. Regardless of all this I'm not exactly looking for a nobel prize, I'm simply want to share my work with the hobbyist community (after making sure there isn't any other equally simple method by making this thread) because I found there is no simple theory for all of this and hobbyists are always asking how much lift their drone or model fighter will produce. My method has 2 super simple equations as opposed to integration and experimental constants and all that jazz in blade element theory or worse full blown FEM. As usual internet people are being hostile when I'm only trying to share my ideas. Don't know why I bother with forums.
Why don't you try The Physics Forum ?
Much more likely to elicit a response there? A lot of people there are quite senior in Physics positions and will be glad to help. Just don't be arrogant.
Original post by JD1lla
Is this for your prototype? You want somebody here to improve your formula so that you can take their answer (well, pay them for it) and improve your drone?
It is actually. No I know it works, as I said I just wanted to tell others because I too struggled to find a simple answer before I sat down and worked it out. The "competition" was just to see if it was the simplest way/already in mass usage.
Original post by JD1lla
Why don't you try The Physics Forum ?
Much more likely to elicit a response there? A lot of people there are quite senior in Physics positions and will be glad to help. Just don't be arrogant.
Much more likely to elicit a response there? A lot of people there are quite senior in Physics positions and will be glad to help. Just don't be arrogant.
I have to join that, it's long. I might try physics stack exchange though. I already have an account with mathematics stack exchange.
Original post by Plutonian
I have to join that, it's long. I might try physics stack exchange though. I already have an account with mathematics stack exchange.
What do you mean, "its long" ?
takes a few mins to sign up?
Original post by JD1lla
What do you mean, "its long" ?
takes a few mins to sign up?
takes a few mins to sign up?
I'm not joining a forum just to ask one question.
Original post by Plutonian
Huh? Yes it does. You don't even know the answer?
I've never seen that equation anywhere regardless of it being easy to derive. And that isn't even the "new" part. the angle bit is. You are confused, there are 2 final equations, one to tell thrust in terms of power and the other to tell thrust in terms of angular velocity. Regardless of all this I'm not exactly looking for a nobel prize, I'm simply want to share my work with the hobbyist community (after making sure there isn't any other equally simple method by making this thread) because I found there is no simple theory for all of this and hobbyists are always asking how much lift their drone or model fighter will produce. My method has 2 super simple equations as opposed to integration and experimental constants and all that jazz in blade element theory or worse full blown FEM. As usual internet people are being hostile when I'm only trying to share my ideas. Don't know why I bother with forums.
I've never seen that equation anywhere regardless of it being easy to derive. And that isn't even the "new" part. the angle bit is. You are confused, there are 2 final equations, one to tell thrust in terms of power and the other to tell thrust in terms of angular velocity. Regardless of all this I'm not exactly looking for a nobel prize, I'm simply want to share my work with the hobbyist community (after making sure there isn't any other equally simple method by making this thread) because I found there is no simple theory for all of this and hobbyists are always asking how much lift their drone or model fighter will produce. My method has 2 super simple equations as opposed to integration and experimental constants and all that jazz in blade element theory or worse full blown FEM. As usual internet people are being hostile when I'm only trying to share my ideas. Don't know why I bother with forums.
It's terribly easy to just google "ducted fan 90mm 11 blade" and click on the result (second result technically). That gave me the actual answers to compare against.
As for the equation part, I looked up "thrust" on wikipedia and under the 'Thrust to Power' subheading, it has your equation (albeit with an extra factor of 4 which explains why I was getting incorrect values using your equation).
So yeah, that equation isn't new. I'm also a little suspect of your angles work, but I've yet to actually go through it and check to see what the mistakes with it are.
Original post by Plutonian
I'm not joining a forum just to ask one question.
you joined this one to ask the same question and potentially pay somebody money? your interests such as drones, 3d printing, physics will likely be more warmly received in a physics/engineering forum. Just saying.
Original post by Rubgish
It's terribly easy to just google "ducted fan 90mm 11 blade" and click on the result (second result technically). That gave me the actual answers to compare against.
As for the equation part, I looked up "thrust" on wikipedia and under the 'Thrust to Power' subheading, it has your equation (albeit with an extra factor of 4 which explains why I was getting incorrect values using your equation).
So yeah, that equation isn't new. I'm also a little suspect of your angles work, but I've yet to actually go through it and check to see what the mistakes with it are.
As for the equation part, I looked up "thrust" on wikipedia and under the 'Thrust to Power' subheading, it has your equation (albeit with an extra factor of 4 which explains why I was getting incorrect values using your equation).
So yeah, that equation isn't new. I'm also a little suspect of your angles work, but I've yet to actually go through it and check to see what the mistakes with it are.
Lets go through the calculation:
A= pi*(45/1000)^2 = 6.3617*10^-3 m^2
P=22.2*76 =1687.2 W
F = (1687.2^2 * 1.2 * 6.3617*10^-3)^1/3 =27.9051 N
27.9051/9.81 = 2.8446 Kg
2.8446*1000= 2845 g
Would you look at that! Wikipedia (and you) are wrong!
Original post by Plutonian
Lets go through the calculation:
A= pi*(45/1000)^2 = 6.3617*10^-3 m^2
P=22.2*76 =1687.2 W
F = (1687.2^2 * 1.2 * 6.3617*10^-3)^1/3 =27.9051 N
27.9051/9.81 = 2.8446 Kg
2.8446*1000= 2845 g
Would you look at that! Wikipedia (and you) are wrong!
A= pi*(45/1000)^2 = 6.3617*10^-3 m^2
P=22.2*76 =1687.2 W
F = (1687.2^2 * 1.2 * 6.3617*10^-3)^1/3 =27.9051 N
27.9051/9.81 = 2.8446 Kg
2.8446*1000= 2845 g
Would you look at that! Wikipedia (and you) are wrong!
For the second one, 22.2V @ 98A, correct answer is 3200g, your formula gives 3370g. For the 29.6v @ 105A, correct answer is 4100g, your formula gives 4275g.
It just so happens that you get the correct answer for the first one. That isn't to say your formula is the correct one.
Original post by Rubgish
For the second one, 22.2V @ 98A, correct answer is 3200g, your formula gives 3370g. For the 29.6v @ 105A, correct answer is 4100g, your formula gives 4275g.
It just so happens that you get the correct answer for the first one. That isn't to say your formula is the correct one.
It just so happens that you get the correct answer for the first one. That isn't to say your formula is the correct one.
Haha it's a bit more correct than if you were to put in that "4" you were talking about though eh?
Let's do another calculation
, this time of an esoteric little piece of US military hardware that I was reading about yesterdayhttps://en.wikipedia.org/wiki/Hiller_VZ-1_Pawnee
2 x 40 hp engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with my equation
Power = 2*40*746= 59680W
Area= pi * 1.065^2 = 3.563m^2
(1/9.81) * (59680^2 * 1.2 * 3.563)^(1/3) = 253 kg
But my formula isn't correct right? 253 is nowhere near 252, absolute trash result that can only be rectified by the introduction of a "4"!
(edited 8 years ago)
Original post by Plutonian
Haha it's a bit more correct than if you were to put in that "4" you were talking about though eh?
Let's do another calculation , this time of an esoteric little piece of US military hardware that I was reading about yesterday
https://en.wikipedia.org/wiki/Hiller_VZ-1_Pawnee
2 x 30kW engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with my equation
(1/9.81) * (60000^2 * 1.2 * (pi * 1.065^2))^(1/3) = 254 kg
But my formula isn't correct right? 254 is nowhere near 252, absolute trash result that can only be rectified by the introduction of a "4"!
Let's do another calculation
, this time of an esoteric little piece of US military hardware that I was reading about yesterdayhttps://en.wikipedia.org/wiki/Hiller_VZ-1_Pawnee
2 x 30kW engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with my equation
(1/9.81) * (60000^2 * 1.2 * (pi * 1.065^2))^(1/3) = 254 kg
But my formula isn't correct right? 254 is nowhere near 252, absolute trash result that can only be rectified by the introduction of a "4"!
It says "max takeoff weight", which isn't the same thing. It'll have a nice bunch of extra power beyond that to actually accelerate (I mean it has a top speed of 16mph so the power to achieve that has to come from somewhere). Combine this fact with the fact that you are assuming 100% efficiency, where as it'll likely be lower, maybe 80% if we are generous.
Basically just because your number happens to fit for a couple (or even 100's) of examples, does not mean it is correct. I've already shown it's pretty far off for two known examples that only differ in power input. The simplified nature of the model means it just doesn't really apply to real world applications, hence why (as you said earlier), most of the time people use more complicated models and mathematics to work out this kind of thing.
I really don't want to discourage you from trying to work out things like this, but please don't be so egotistical about it. Next time just post something along the lines of "I've been thinking about power and thrust and have come up with these equations that I think are pretty useful. Can anyone spot anything obviously wrong with this, or is this correct and if so is it useful?"
Original post by Rubgish
It says "max takeoff weight", which isn't the same thing. It'll have a nice bunch of extra power beyond that to actually accelerate (I mean it has a top speed of 16mph so the power to achieve that has to come from somewhere).
That's why my equation got higher than the max take off weight. Max speed does not have to be at the max takeoff weight.
Original post by Rubgish
Combine this fact with the fact that you are assuming 100% efficiency, where as it'll likely be lower, maybe 80% if we are generous.
Combine this fact with the fact that you are assuming 100% efficiency, where as it'll likely be lower, maybe 80% if we are generous.
Contra-rotating props are over 90% efficient,
Original post by Rubgish
Basically just because your number happens to fit for a couple (or even 100's) of examples, does not mean it is correct. I've already shown it's pretty far off for two known examples that only differ in power input. The simplified nature of the model means it just doesn't really apply to real world applications, hence why (as you said earlier), most of the time people use more complicated models and mathematics to work out this kind of thing.
Basically just because your number happens to fit for a couple (or even 100's) of examples, does not mean it is correct. I've already shown it's pretty far off for two known examples that only differ in power input. The simplified nature of the model means it just doesn't really apply to real world applications, hence why (as you said earlier), most of the time people use more complicated models and mathematics to work out this kind of thing.
"Just because you prove your equation to work countless times doesn't mean it works"
Why don't you just accept that you got completely BTFO instead of coming up with these laughable attacks on my equation?
Who made you arbiter of what model is good enough for real world applications? If you're casually making a model plane it's pretty damn useful. The fact that it's not good enough for NASA is of no relevance. Again this is one of the weakest counter arguments I have ever seen. Bottom line is that you said that my equation did not work, and that the method to calculate thrust was already clearly posted on wikipedia. I both showed that my equation does work AND that the equation on wikipedia does not, yet you continue to trash my equation and defend one that is completely wrong! If that's not stubbornness I just don't know.
Let's do the calculation with the wikipedia equation you claim is the correct one
2 x 40 hp engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with YOUR equation
Power = 2*40*746= 59680W
Area= pi * 1.065^2 = 3.563m^2
(1/9.81) * (4 * 59680^2 * 1.2 * 3.563)^(1/3) = 401 kg
Oh my god! It's completely flipping wrong! OH BUT WAIT! that extra 150 kg of thrust is what you need to accelerate that little hovercraft to 16 mph am I right? You want to bring up efficiency? Even if it was 70% efficient the formula is still way off! Let's not mention that the page for the ducted fan stated "thrust" and nothing more
Original post by Rubgish
I really don't want to discourage you from trying to work out things like this, but please don't be so egotistical about it. Next time just post something along the lines of "I've been thinking about power and thrust and have come up with these equations that I think are pretty useful. Can anyone spot anything obviously wrong with this, or is this correct and if so is it useful?"
I really don't want to discourage you from trying to work out things like this, but please don't be so egotistical about it. Next time just post something along the lines of "I've been thinking about power and thrust and have come up with these equations that I think are pretty useful. Can anyone spot anything obviously wrong with this, or is this correct and if so is it useful?"
Don't be so egotistical? I wasn't being egotistical then, you're the one who felt threatened because someone posted a problem that you couldn't solve so you decide to trash them and ended up getting completely destroyed. NOW I'm being egotistical at feeling the satisfaction of knocking down a peg some ******* who completely trashed my methods and now is seriously scraping the barrel to avoid just admitting that he was wrong.
(edited 8 years ago)
Original post by Plutonian
That's why my equation got higher than the max take off weight. Max speed does not have to be at the max takeoff weight.
Contra-rotating props are over 90% efficient,
"Just because you prove your equation to work countless times doesn't mean it works"
Why don't you just accept that you got completely BTFO instead of coming up with these laughable attacks on my equation?
Who made you arbiter of what model is good enough for real world applications? If you're casually making a model plane it's pretty damn useful. The fact that it's not good enough for NASA is of no relevance. Again this is one of the weakest counter arguments I have ever seen. Bottom line is that you said that my equation did not work, and that the method to calculate thrust was already clearly posted on wikipedia. I both showed that my equation does work AND that the equation on wikipedia does not, yet you continue to trash my equation and defend one that is completely wrong! If that's not stubbornness I just don't know.
Let's do the calculation with the wikipedia equation you claim is the correct one
2 x 40 hp engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with YOUR equation
Power = 2*40*746= 59680W
Area= pi * 1.065^2 = 3.563m^2
(1/9.81) * (4 * 59680^2 * 1.2 * 3.563)^(1/3) = 401 kg
Oh my god! It's completely flipping wrong! OH BUT WAIT! that extra 150 kg of thrust is what you need to accelerate that little hovercraft to 16 mph am I right? You want to bring up efficiency? Even if it was 70% efficient the formula is still way off!
Don't be so egotistical? I wasn't being egotistical then, you're the one who felt threatened because someone posted a problem that you couldn't solve so you decide to trash them and ended up getting completely destroyed. NOW I'm being egotistical at feeling the satisfaction of knocking down a peg some ******* who completely trashed my methods and now is seriously scraping the barrel to avoid just admitting that he was wrong.
Contra-rotating props are over 90% efficient,
"Just because you prove your equation to work countless times doesn't mean it works"
Why don't you just accept that you got completely BTFO instead of coming up with these laughable attacks on my equation?
Who made you arbiter of what model is good enough for real world applications? If you're casually making a model plane it's pretty damn useful. The fact that it's not good enough for NASA is of no relevance. Again this is one of the weakest counter arguments I have ever seen. Bottom line is that you said that my equation did not work, and that the method to calculate thrust was already clearly posted on wikipedia. I both showed that my equation does work AND that the equation on wikipedia does not, yet you continue to trash my equation and defend one that is completely wrong! If that's not stubbornness I just don't know.
Let's do the calculation with the wikipedia equation you claim is the correct one
2 x 40 hp engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with YOUR equation
Power = 2*40*746= 59680W
Area= pi * 1.065^2 = 3.563m^2
(1/9.81) * (4 * 59680^2 * 1.2 * 3.563)^(1/3) = 401 kg
Oh my god! It's completely flipping wrong! OH BUT WAIT! that extra 150 kg of thrust is what you need to accelerate that little hovercraft to 16 mph am I right? You want to bring up efficiency? Even if it was 70% efficient the formula is still way off!
Don't be so egotistical? I wasn't being egotistical then, you're the one who felt threatened because someone posted a problem that you couldn't solve so you decide to trash them and ended up getting completely destroyed. NOW I'm being egotistical at feeling the satisfaction of knocking down a peg some ******* who completely trashed my methods and now is seriously scraping the barrel to avoid just admitting that he was wrong.
You're pretty confident for a guy who suffers from anxiety.
Original post by JD1lla
You're pretty confident for a guy who suffers from anxiety.
Ad hominem attacks, nice.
Original post by mrno1324
You are one of the biggest ***** I've ever seen.
Please explain to me why I'm the bad person when this guy said my calculation was wrong, I showed him it was right, and he still continued to claim it was wrong? I have a right to get annoyed. Before he just says "my bad" and we continue a positive discourse, he has to pull out everything under the sun to not admit he was wrong.
Original post by Plutonian
That's why my equation got higher than the max take off weight. Max speed does not have to be at the max takeoff weight.
Contra-rotating props are over 90% efficient,
"Just because you prove your equation to work countless times doesn't mean it works"
Why don't you just accept that you got completely BTFO instead of coming up with these laughable attacks on my equation?
Who made you arbiter of what model is good enough for real world applications? If you're casually making a model plane it's pretty damn useful. The fact that it's not good enough for NASA is of no relevance. Again this is one of the weakest counter arguments I have ever seen. Bottom line is that you said that my equation did not work, and that the method to calculate thrust was already clearly posted on wikipedia. I both showed that my equation does work AND that the equation on wikipedia does not, yet you continue to trash my equation and defend one that is completely wrong! If that's not stubbornness I just don't know.
Let's do the calculation with the wikipedia equation you claim is the correct one
2 x 40 hp engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with YOUR equation
Power = 2*40*746= 59680W
Area= pi * 1.065^2 = 3.563m^2
(1/9.81) * (4 * 59680^2 * 1.2 * 3.563)^(1/3) = 401 kg
Oh my god! It's completely flipping wrong! OH BUT WAIT! that extra 150 kg of thrust is what you need to accelerate that little hovercraft to 16 mph am I right? You want to bring up efficiency? Even if it was 70% efficient the formula is still way off! Let's not mention that the page for the ducted fan stated "thrust" and nothing more
Don't be so egotistical? I wasn't being egotistical then, you're the one who felt threatened because someone posted a problem that you couldn't solve so you decide to trash them and ended up getting completely destroyed. NOW I'm being egotistical at feeling the satisfaction of knocking down a peg some ******* who completely trashed my methods and now is seriously scraping the barrel to avoid just admitting that he was wrong.
Contra-rotating props are over 90% efficient,
"Just because you prove your equation to work countless times doesn't mean it works"
Why don't you just accept that you got completely BTFO instead of coming up with these laughable attacks on my equation?
Who made you arbiter of what model is good enough for real world applications? If you're casually making a model plane it's pretty damn useful. The fact that it's not good enough for NASA is of no relevance. Again this is one of the weakest counter arguments I have ever seen. Bottom line is that you said that my equation did not work, and that the method to calculate thrust was already clearly posted on wikipedia. I both showed that my equation does work AND that the equation on wikipedia does not, yet you continue to trash my equation and defend one that is completely wrong! If that's not stubbornness I just don't know.
Let's do the calculation with the wikipedia equation you claim is the correct one
2 x 40 hp engines
2.13 m rotor diameter
Wiki page says 252 kg of thrust
Let's see what we get with YOUR equation
Power = 2*40*746= 59680W
Area= pi * 1.065^2 = 3.563m^2
(1/9.81) * (4 * 59680^2 * 1.2 * 3.563)^(1/3) = 401 kg
Oh my god! It's completely flipping wrong! OH BUT WAIT! that extra 150 kg of thrust is what you need to accelerate that little hovercraft to 16 mph am I right? You want to bring up efficiency? Even if it was 70% efficient the formula is still way off! Let's not mention that the page for the ducted fan stated "thrust" and nothing more
Don't be so egotistical? I wasn't being egotistical then, you're the one who felt threatened because someone posted a problem that you couldn't solve so you decide to trash them and ended up getting completely destroyed. NOW I'm being egotistical at feeling the satisfaction of knocking down a peg some ******* who completely trashed my methods and now is seriously scraping the barrel to avoid just admitting that he was wrong.
You have shown nothing. The fact that your model gives 253kg vs 252 max takeoff weight suggests you have 1kg extra for acceleration, otherwise known as this thing accelerates at 1/253 m/s^2 during takeoff. Then if your efficiency is below 99.5% or so, then you have absolutely no lift at all and it wouldn't leave the ground.
What's more realistic, 400kg of thrust, at 80% efficiency leaving you with 320kg effective thrust on a 250kg platform, giving ~70kg worth to accelerate you at approx 0.3 m/s^2. Or 253kg of thrust at 100% efficiency giving you 1kg effective thrust accelerating you at approx 0.003m/s^2?
Essentially, you have worked out a simple model, the model underestimates the thrust by a factor of (4)^(1/3) compared to the more accurate simple model. It just so happens that when you factor in how efficient the real system is and various other real world quantities, that your method *looks* more accurate for /some/ thrust values. Once you begin to factor in these relevant real world factors, the model on wikipedia is more accurate, and yours underestimates the thrust.
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