This forluma is used then you push an object under water and let it go. Like if you throw a rock underwater. Submarines have an engine constantly pushing them forward so you can calculate KE normaly.
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M3X_2.0
yeah but are you sure that you'd need sub rel speeds to do that?
Ugarik
Apperantly you do. The result is so insainly high is because the speed grows expanentialy with the distance. If it crosses 1 meter in 0.2 seconds (same average speed), the result would be:
(e^(1*1.645)-1)/(1.645*0.2) = 12.7 m/s
(e^(1*1.645)-1)/(1.645*0.2) = 12.7 m/s
Ugarik
Honestly I think this formula makes sence. But I should probably ask DontTalkDT if everything is correct here
M3X_2.0
Ok, I contacted him
DontTalkDT
As I say in my status I currently have absolutely no time to invest in this wiki. So given that I don't even know how exactly the formula was derived I can't really judge it properly right now.
My common sense tells me the result is probably flawed. Like, at those speeds the water would probably literally vaporize in the path of the sphere. Mythbusters once fired bullets underwater and those already did better than this sub-relativistic projectile does according to the formula. Although they did find some inverse relationship between energy in distance it's still quite extreme.
One probable flaw of the drag coefficient. I doubt the general value applies under those extreme conditions.
Maybe compare your results with formula 65 in this document for a start. Although even that formula isn't made for relativistic speeds, I think.
My common sense tells me the result is probably flawed. Like, at those speeds the water would probably literally vaporize in the path of the sphere. Mythbusters once fired bullets underwater and those already did better than this sub-relativistic projectile does according to the formula. Although they did find some inverse relationship between energy in distance it's still quite extreme.
One probable flaw of the drag coefficient. I doubt the general value applies under those extreme conditions.
Maybe compare your results with formula 65 in this document for a start. Although even that formula isn't made for relativistic speeds, I think.
Ugarik
Okay, I devided the formula of the velocity after it crosses x meters underwater too, using the same method. It's V*e^(-bx)
Let's say a steel ball (8080 kg/m^3 density) with the diameter of 10 cm enters water at 1000 m/s, at right angle. What would be it's velocity after it crosses 5 meters underwater.
Using the formula 65 from the doc.
1000*exp(-3*1000*0.866*5/(2*8080*0.1)) = 0.32 m/s
Now using my formula we need to find b value first
Cross section area: pi*0.05^2 = 0.007854 m^2
Mass: 4/3*pi*0.05^2 = 0.0005236, 0.0005236*8080 = 4.2307 kg
b = (1000*0.007854*0.866/4.2307) = 1.6077 m^-1
V = 1000*e^(-5*1.6077) = 0.32 m/s
Wow, I honestly didn't expect it
Let's say a steel ball (8080 kg/m^3 density) with the diameter of 10 cm enters water at 1000 m/s, at right angle. What would be it's velocity after it crosses 5 meters underwater.
Using the formula 65 from the doc.
1000*exp(-3*1000*0.866*5/(2*8080*0.1)) = 0.32 m/s
Now using my formula we need to find b value first
Cross section area: pi*0.05^2 = 0.007854 m^2
Mass: 4/3*pi*0.05^2 = 0.0005236, 0.0005236*8080 = 4.2307 kg
b = (1000*0.007854*0.866/4.2307) = 1.6077 m^-1
V = 1000*e^(-5*1.6077) = 0.32 m/s
Wow, I honestly didn't expect it
DontTalkDT
From a look I would guess your result might be able to be transformed into formula 65 then if one sets in the formulas for mass and crossectional area as parameters (and maybe rounds Pi or something?). That would be good confirmation.
If that's the case the formula is probably (Again: I haven't checked your derivation) fine for cases of meteor speeds at least (up to 72 km/s), but for much higher stuff like relativistic results I'm still iffy about it. Like, if I set into your result correctly, 15cm ball going 13 meters over 2 second would already be FTL which obviously is a problem. As said, at a certain point I feel like the normal drag of water doesn't apply anymore since the projectile would just make the water stop existing in its usual form.
And this probably shouldn't be used for speed calcs, but I guess that might have been clear.
If that's the case the formula is probably (Again: I haven't checked your derivation) fine for cases of meteor speeds at least (up to 72 km/s), but for much higher stuff like relativistic results I'm still iffy about it. Like, if I set into your result correctly, 15cm ball going 13 meters over 2 second would already be FTL which obviously is a problem. As said, at a certain point I feel like the normal drag of water doesn't apply anymore since the projectile would just make the water stop existing in its usual form.
And this probably shouldn't be used for speed calcs, but I guess that might have been clear.
Ugarik
Okay, I'll show the derivation:
1. Write down Newton's second law equation:
2. Replace F with negative drag force formula (negative because it opposes the motion), and replace a with time derivative of velocity:
3. Since everything except v^2 is constant, replace it with b
4. Swap v^2 and dt and integrate both sides:
5. Solve for t (you'll see why in the last step)
6. Solve for v
7. Integrate to get the distance formula
8. Since initial speed v_0 is our goal, and time t and distance D are given, solve for v_0. That's the final formula.
9. For further proof, replace t with the formula from step 5
10. Simplify and solve for v. That's formula gives the same result as №65 from that file.
11. To prove this even further, replace b with the formula in step 3.
12. Define mass as the density of the material times the volume. And input the formula for cross-sectional area and volume from the diameter (assuming a solid sphere):
13. Simplify to get the formula 65.
1. Write down Newton's second law equation:
2. Replace F with negative drag force formula (negative because it opposes the motion), and replace a with time derivative of velocity:
3. Since everything except v^2 is constant, replace it with b
4. Swap v^2 and dt and integrate both sides:
5. Solve for t (you'll see why in the last step)
6. Solve for v
7. Integrate to get the distance formula
8. Since initial speed v_0 is our goal, and time t and distance D are given, solve for v_0. That's the final formula.
9. For further proof, replace t with the formula from step 5
10. Simplify and solve for v. That's formula gives the same result as №65 from that file.
11. To prove this even further, replace b with the formula in step 3.
12. Define mass as the density of the material times the volume. And input the formula for cross-sectional area and volume from the diameter (assuming a solid sphere):
13. Simplify to get the formula 65.
M3X_2.0
so, what happened?
Ugarik
This forluma seems to be correct, however there should abviously be some limitations. For example the initial speed should not be highers than the average one x100.
So I think you can use it as long as the final speed is not insainly high.
So I think you can use it as long as the final speed is not insainly high.
M3X_2.0
Ok, the formula is V*e^(-bx), right?
Ugarik
No, you don't have the timeframe in this formula. It's the formula. You need to use (e^(Db)-1)/(bt)
Nerd1435
Drag force is directly proportional to the cross-sectional surface area and square of it's instantaneous velocity.
F = -k*A*v^2
a = -k*A*v^2/m
dv/dt = -k*A*v^2/m
dv/v^2 = -k*A*dt/m
Integrate it from v->v1 to v2, t-> 0 to 2
1/v1 - 1/v2 = -2*k*A/m
If you are aware of it's speed at the end, you can get its speed at the start.
F = -k*A*v^2
a = -k*A*v^2/m
dv/dt = -k*A*v^2/m
dv/v^2 = -k*A*dt/m
Integrate it from v->v1 to v2, t-> 0 to 2
1/v1 - 1/v2 = -2*k*A/m
If you are aware of it's speed at the end, you can get its speed at the start.
Nerd1435
If you put a =v*dv/dx instead of dv/dt, then you can get an equation in terms of v and x 
v*dv/dx = -k*A*v^2/m
dv/v = -k*A*dx/m
Integrate v from v1 to v2, x from 0 to X (Distance)
Ln(v2/v1) = -k*A*X/m
V2 = v1*e^(-k*A*X/m)
You have 2 equations, 2 variables. (You already know X, k, A, etc...)
Solve them and you get v1, v2
v*dv/dx = -k*A*v^2/m
dv/v = -k*A*dx/m
Integrate v from v1 to v2, x from 0 to X (Distance)
Ln(v2/v1) = -k*A*X/m
V2 = v1*e^(-k*A*X/m)
You have 2 equations, 2 variables. (You already know X, k, A, etc...)
Solve them and you get v1, v2
Nerd1435
(1/v1)*(1-e^(k*A*X/m)) = -2*k*A/m
V1 = m*[e^(k*A*X/m) - 1]/(2*k*A)
This can be said to be the final formula.
Just note that the "2" was actually time and I've already substituted the value before.
V1 = m*[e^(k*A*X/m) - 1]/(2*k*A)
This can be said to be the final formula.
Just note that the "2" was actually time and I've already substituted the value before.
Nerd1435
Sh*T! Its proportional to v^2, I'll edit it.
Edit: Corrected it and seems like it's redundant the formula I obtained seems to have been posted already. F.
Edit: Corrected it and seems like it's redundant the formula I obtained seems to have been posted already. F.
Nerd1435
Tf turned out tk be 3,718,529.0074456 m/s 
To project rubber ball of radius 10cm by 10m in 2sec
Sus
F
F
Sorry for spamming. It looked interesting so I commented.
To project rubber ball of radius 10cm by 10m in 2sec
Sus
F
F
Sorry for spamming. It looked interesting so I commented.
Ugarik
You got the same formula as mine. You just have a different definition of k
I got this using google, you sure that "e" is really necessary? You literally got Mach 12349.63 of an object crossing 10m in 2 seconds underwater, but submarines aren't sub rel