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Can Toothless Fly?
Topic Started: 06 Jun 2014, 13:04 (2983 Views)
TheCube42
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Start of good physics work.

Though, two questions come to me:

1. What does lowercase "a" and uppercase "B" signify in this context?
2. I realize that v(t) should approach v-terminal as t approaches infinity. However, wouldn't the limit of (e^β-1)/(e^β+1) at t = infinity be 1, since as t approaches infinity, β approaches infinity? Or is there something beyond the simple limit that causes the limit to be sqrt(2)?
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Zer0x
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[b]Banhammerdragon
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MC_Oceans
06 Jun 2014, 13:30
okay so solid bones=more weight. That makes thing more difficult for dragons in terms of flying right?The force needed to propel Toothless increases and makes it hard for Toothless to stay in the air. But how much do the bones weigh? Is there a limit to the weight Toothless can have before he can fly?
Hollow cylinders are more resistant than solid ones, especially when they are strutted inside. They could also be more flexible than the bones from normal tetrapods (maybe with some sort of natural aramid-like aramids [No, I did not steal this from Avatar!]).


TheCube42
06 Jun 2014, 17:04
Start of good physics work.

Though, two questions come to me:

1. What does lowercase "a" and uppercase "B" signify in this context?
2. I realize that v(t) should approach v-terminal as t approaches infinity. However, wouldn't the limit of (e^β-1)/(e^β+1) at t = infinity be 1, since as t approaches infinity, β approaches infinity? Or is there something beyond the simple limit that causes the limit to be sqrt(2)?

1.) a, B and β are just variables I've used for substitutions, they were necessary to solve the integral. It also makes the formula clearer.

2.) When t approaches infinity the whole second term with the euleric function becomes 1, that means v(t)=a and then you are at

v=sqrt((2xmxg)/(pxAxc))

which is the formula for the maximum speed you coud reach while falling.

t is the ony variable, all others are constant. Maybe this explains it a bit better. (Everything after the integral is not correct.)
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TheCube42
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Zer0x
07 Jun 2014, 12:39
2.) When t approaches infinity the whole second term with the euleric function becomes 1, that means v(t)=a and then you are at

v=sqrt((2xmxg)/(pxAxc))

which is the formula for the maximum speed you coud reach while falling.

t is the ony variable, all others are constant. Maybe this explains it a bit better. (Everything after the integral is not correct.)
But a = sqrt(mg/ρAc) as you've written at the end of the sheet. Or was that supposed to be sqrt(2mg/ρAc), in which case a=v_terminal?

EDIT: Yep, from your link, right before you define/substitute B near the beginning, the 1/2 in drag force disappears. If you keep it, B = ρAc/2mg, so a = sqrt(2mg/ρAc) = v_terminal, and everything makes sense.
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Zer0x
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Yes, sorry. That was a mistake. -_-
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mr.k.c.w.
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Zer0x
01 Jun 2014, 18:58
In the next days I will make a detailed sketch with all affecting forces.


I mentioned before that we have to know how fast Toothless can go. For that I use the maximum speed he could reach after a dive.

Here is the formula for the max. speed:

v=sqrt((2xmxg)/(pxAxc))


And here is the formula for the speed at any time while falling:



Don't ask me where it comes from. It needs more than 3 sheets of paper, 3 substitutions, separation of variables and a heavy integral.


m - mass (I used 130kg for it, if his is higher he will be faster)
g - acceleration (9.81 m/s^2)
p - density of the medium (air at level 0: 1.2041 kg/m^3)
A - area of the falling axis (according from the size of a 3D model it is around 0.8 m^2)
c - drag coefficient (without his wings he have a nearly ideal streamline shape (0.02), I used 0.06, that's a bit lower than an airplane)
v0 - speed at the start (mostly 0)
t - time
Your statement about the falling formula might not be correct. Any object that is falling from a certain height will eventually reach terminal velocity and doesn't accelerate anymore. Now I know that Toothless will eventually flap his wings but let's just say that he dived from a very high attitude. At a certain point, he won't gain anymore speed due to him reaching terminal velocity. So, the formula's statement will not be correct if Toothless already reached terminal velocity. (There's just one flaw in my statement, I haven't figure out how long will Toothless take for him to reach terminal velocity. So he might hit the ground first before ever reaching terminal velocity, thus, making my point invalid.)

Also, just asking, while Toothless is flying, shouldn't you count in the force that gravity puts on him?
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Zer0x
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Yes, you're right. The formula doesn't work when v0 > a (because the ln from a negative value isn't defined), but I think we don't need to pay much attention to that case.

He will never reach terminal velocity (a), because t would have to become infinite.

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Also, just asking, while Toothless is flying, shouldn't you count in the force that gravity puts on him?
It is in TheCube42's calculation, but not directly. That was the meaning of the whole thing, to calculate the lift that counter the weight.
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mr.k.c.w.
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Zer0x
08 Jun 2014, 12:19
Yes, you're right. The formula doesn't work when v0 > a (because the ln from a negative value isn't defined), but I think we don't need to pay much attention to that case.

He will never reach terminal velocity (a), because t would have to become infinite.
You're right about that, but what I think is that as long as we can define the distance were the acceleration started, you can't say that t would be infinite. Though we need to pin-point the velocity of one particular moment and the time to reach that velocity...
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HideousZippleback
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I'm sad because I don't understand any of the math and equation stuff going on here...

And what about Cloudjumper? Could he really physically fly in real life with four wings?
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mr.k.c.w.
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HideousZippleback
08 Jun 2014, 21:45
I'm sad because I don't understand any of the math and equation stuff going on here...
Actually, I'm still a little foggy about the formulas but when it comes to basic physics concepts , I've learned a bit from watching a good few seasons of Mythbusters. You should too, it's a good show.
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Seriously, I, [INSERT USERNAME HERE] have nothing to say here. Continue on with your viewing of not-so regularly scheduled posts.


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HideousZippleback
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mr.k.c.w.
08 Jun 2014, 23:15
HideousZippleback
08 Jun 2014, 21:45
I'm sad because I don't understand any of the math and equation stuff going on here...
Actually, I'm still a little foggy about the formulas but when it comes to basic physics concepts , I've learned a bit from watching a good few seasons of Mythbusters. You should too, it's a good show.
I LOVE Mythbusters! I just don't know where to watch it online... >.>
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