[math-fun] Physically intuitive?
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3] Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3] This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg
No, forces are applied, power is an intrinsic property, so you might rather say "Given two movers with equal average power, one that goes twice as long also goes eight times as far"; however, this is also incorrect. The integral equation P = Int F*v*dt, says that power depends on resistive force. This is somewhat confusing when thinking of, say, a sprinter or a marathon runner because it isn't clear what the resistive force is. It is more complicated than you might guess: http://www.georgeron.com/2017/09/kipchoge-running-power-marathon.html If you want a high school problem to work out, here is one I wrote earlier this month: https://0x0.st/i14t.pdf --Brad On Mon, Apr 20, 2020 at 12:06 PM Bill Gosper <billgosper@gmail.com> wrote:
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3]
Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3]
This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Way too unspecified. Brad, you seem to be considering "steady-state" motion at a constant velocity, where the power counteracts some resistive force. And your integral is wrong. P = F v. The time integral of this is E, energy. Check the units. Bill, you would seem to I be considering accelerating an object, where the power is used to the increase the kinetic energy if the object. It's all going to be dependent on the initial velocity before the power/force is applied. Let me take a stab at an example elementary physics problem: You have a block of mass m on a frictionless surface, initially at rest at x = 0. Starting at t = 0, you are going to apply a force to the block in the positive x direction in such a way that the rate of change of the kinetic energy of the block is a constant, ie, dK/dt = P, where P is the power "being applied"/required/however you want to term it in words. Now: a) What is the kinetic energy of the block as a function of time? b) What is the velocity of the block as a function of time? c) What is the position of the block as a function of time? d) How does the position of the block at a time 2t compare with the position at time t? a) The block starts at rest so K(t=0)=0, and K is increasing at a constant rate dK/dt=P, hence: K = P t ((for t>=0)) b) The kinetic energy of the moving block is K=(1/2) m v^2, so v = Sqrt[2K/m] = Sqrt[2 P t / m] c) The block starts at x=0 so x(t=0)=0, and v=dx/dt, hence x = Int v dt = Int Sqrt[2 P t / m] dt = ( Sqrt[2 P / m] (2/3) t^(3/2) ) d) x(2t)/x(t) = ... = 2^(3/2), so the block goes Sqrt[8] times as far. On Mon, Apr 20, 2020, 10:28 Brad Klee <bradklee@gmail.com> wrote:
No, forces are applied, power is an intrinsic property, so you might rather say "Given two movers with equal average power, one that goes twice as long also goes eight times as far"; however, this is also incorrect. The integral equation P = Int F*v*dt, says that power depends on resistive force. This is somewhat confusing when thinking of, say, a sprinter or a marathon runner because it isn't clear what the resistive force is. It is more complicated than you might guess:
http://www.georgeron.com/2017/09/kipchoge-running-power-marathon.html
If you want a high school problem to work out, here is one I wrote earlier this month:
--Brad
On Mon, Apr 20, 2020 at 12:06 PM Bill Gosper <billgosper@gmail.com> wrote:
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3]
Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3]
This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Arg, wrong again! Yes that is the work integral, so needs another 1/delta t. Thanks for the correction. However, the mistaken formula doesn't change my argument, which is that P = m*(delta x)^2 / (delta t)^3 is not the correct formula because actually, P is proportional to a Force F, which usually isn't proportional to (delta x) / (delta t)^2. The math works out nice in this block problem, but we don't really need another block problem. It isn't physically realistic anyways. What is a "frictionless surface"? And what about drag? Is the block in a vacuum? Why does the block matter? Etc. In the problem about runners, the resistive force combines a percentage of body weight and drag force. Drag is significant especially for sprinters like Usain Bolt, who are running very fast. The drag proportionality is v^2. We need runners b/c physical fitness helps to improve bodily health, and to improve chances of surviving viral sickness! An even better runner problem is to look closely at what happens during the acceleration phase of a 100m dash. Not only are forces imbalanced, the resistive force increases as a function of time and /or velocity. Yet Usain Bolt can still exceed 2 kilowatts of human body power. https://www.popsci.com/science/article/2013-07/physics-record-breaking-run/ Amazingly, some of the beetles reflect mostly green, left-circular polarized light. If that is not enough they are stronger on average than the strongest humans, most likely more powerful, and easy to find in Arkansas (if you are willing to break open a dung pat): https://www.livescience.com/8145-super-bug-world-strongest-insect-revealed.h... https://www.inaturalist.org/observations/41877497 I don't know if anyone has directly measured the beetle power (I haven't read about it), but it should be done! Too bad about the march purge. The labs are shut down right now anyways... Cheers, Brad --Brad On Mon, Apr 20, 2020 at 1:21 PM William R Somsky <wrsomsky@gmail.com> wrote:
Way too unspecified.
Brad, you seem to be considering "steady-state" motion at a constant velocity, where the power counteracts some resistive force. And your integral is wrong. P = F v. The time integral of this is E, energy. Check the units.
Bill, you would seem to I be considering accelerating an object, where the power is used to the increase the kinetic energy if the object. It's all going to be dependent on the initial velocity before the power/force is applied.
Let me take a stab at an example elementary physics problem:
You have a block of mass m on a frictionless surface, initially at rest at x = 0. Starting at t = 0, you are going to apply a force to the block in the positive x direction in such a way that the rate of change of the kinetic energy of the block is a constant, ie, dK/dt = P, where P is the power "being applied"/required/however you want to term it in words.
Now: a) What is the kinetic energy of the block as a function of time? b) What is the velocity of the block as a function of time? c) What is the position of the block as a function of time? d) How does the position of the block at a time 2t compare with the position at time t?
a) The block starts at rest so K(t=0)=0, and K is increasing at a constant rate dK/dt=P, hence: K = P t ((for t>=0))
b) The kinetic energy of the moving block is K=(1/2) m v^2, so v = Sqrt[2K/m] = Sqrt[2 P t / m]
c) The block starts at x=0 so x(t=0)=0, and v=dx/dt, hence x = Int v dt = Int Sqrt[2 P t / m] dt = ( Sqrt[2 P / m] (2/3) t^(3/2) )
d) x(2t)/x(t) = ... = 2^(3/2), so the block goes Sqrt[8] times as far.
On Mon, Apr 20, 2020, 10:28 Brad Klee <bradklee@gmail.com> wrote:
No, forces are applied, power is an intrinsic property, so you might rather say "Given two movers with equal average power, one that goes twice as long also goes eight times as far"; however, this is also incorrect. The integral equation P = Int F*v*dt, says that power depends on resistive force. This is somewhat confusing when thinking of, say, a sprinter or a marathon runner because it isn't clear what the resistive force is. It is more complicated than you might guess:
http://www.georgeron.com/2017/09/kipchoge-running-power-marathon.html
If you want a high school problem to work out, here is one I wrote earlier this month:
--Brad
On Mon, Apr 20, 2020 at 12:06 PM Bill Gosper <billgosper@gmail.com> wrote:
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3]
Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3]
This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
One of the most powerful animals is the mantis shrimp (neither a mantis, nor a shrimp, but …): https://www.youtube.com/watch?v=E0Li1k5hGBE <https://www.youtube.com/watch?v=E0Li1k5hGBE>
On Apr 20, 2020, at 6:46 PM, Brad Klee <bradklee@gmail.com> wrote:
Arg, wrong again! Yes that is the work integral, so needs another 1/delta t. Thanks for the correction.
However, the mistaken formula doesn't change my argument, which is that P = m*(delta x)^2 / (delta t)^3 is not the correct formula because actually, P is proportional to a Force F, which usually isn't proportional to (delta x) / (delta t)^2.
The math works out nice in this block problem, but we don't really need another block problem. It isn't physically realistic anyways. What is a "frictionless surface"? And what about drag? Is the block in a vacuum? Why does the block matter? Etc.
In the problem about runners, the resistive force combines a percentage of body weight and drag force. Drag is significant especially for sprinters like Usain Bolt, who are running very fast. The drag proportionality is v^2.
We need runners b/c physical fitness helps to improve bodily health, and to improve chances of surviving viral sickness!
An even better runner problem is to look closely at what happens during the acceleration phase of a 100m dash. Not only are forces imbalanced, the resistive force increases as a function of time and /or velocity. Yet Usain Bolt can still exceed 2 kilowatts of human body power.
https://www.popsci.com/science/article/2013-07/physics-record-breaking-run/
Amazingly, some of the beetles reflect mostly green, left-circular polarized light. If that is not enough they are stronger on average than the strongest humans, most likely more powerful, and easy to find in Arkansas (if you are willing to break open a dung pat):
https://www.livescience.com/8145-super-bug-world-strongest-insect-revealed.h... https://www.inaturalist.org/observations/41877497
I don't know if anyone has directly measured the beetle power (I haven't read about it), but it should be done! Too bad about the march purge. The labs are shut down right now anyways...
Cheers,
Brad
--Brad
On Mon, Apr 20, 2020 at 1:21 PM William R Somsky <wrsomsky@gmail.com> wrote:
Way too unspecified.
Brad, you seem to be considering "steady-state" motion at a constant velocity, where the power counteracts some resistive force. And your integral is wrong. P = F v. The time integral of this is E, energy. Check the units.
Bill, you would seem to I be considering accelerating an object, where the power is used to the increase the kinetic energy if the object. It's all going to be dependent on the initial velocity before the power/force is applied.
Let me take a stab at an example elementary physics problem:
You have a block of mass m on a frictionless surface, initially at rest at x = 0. Starting at t = 0, you are going to apply a force to the block in the positive x direction in such a way that the rate of change of the kinetic energy of the block is a constant, ie, dK/dt = P, where P is the power "being applied"/required/however you want to term it in words.
Now: a) What is the kinetic energy of the block as a function of time? b) What is the velocity of the block as a function of time? c) What is the position of the block as a function of time? d) How does the position of the block at a time 2t compare with the position at time t?
a) The block starts at rest so K(t=0)=0, and K is increasing at a constant rate dK/dt=P, hence: K = P t ((for t>=0))
b) The kinetic energy of the moving block is K=(1/2) m v^2, so v = Sqrt[2K/m] = Sqrt[2 P t / m]
c) The block starts at x=0 so x(t=0)=0, and v=dx/dt, hence x = Int v dt = Int Sqrt[2 P t / m] dt = ( Sqrt[2 P / m] (2/3) t^(3/2) )
d) x(2t)/x(t) = ... = 2^(3/2), so the block goes Sqrt[8] times as far.
On Mon, Apr 20, 2020, 10:28 Brad Klee <bradklee@gmail.com> wrote:
No, forces are applied, power is an intrinsic property, so you might rather say "Given two movers with equal average power, one that goes twice as long also goes eight times as far"; however, this is also incorrect. The integral equation P = Int F*v*dt, says that power depends on resistive force. This is somewhat confusing when thinking of, say, a sprinter or a marathon runner because it isn't clear what the resistive force is. It is more complicated than you might guess:
http://www.georgeron.com/2017/09/kipchoge-running-power-marathon.html
If you want a high school problem to work out, here is one I wrote earlier this month:
--Brad
On Mon, Apr 20, 2020 at 12:06 PM Bill Gosper <billgosper@gmail.com> wrote:
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3]
Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3]
This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
One of the better colloquium talks I saw was about spring-latch mechanisms in animals, and mentioned the mantis shrimp. The talk was along the lines of: https://link.springer.com/article/10.1007/BF00225821 https://www.researchgate.net/profile/Wulfila_Gronenberg/publication/22698761... With latch mechanisms, large mechanical energy can be generated and stored at low power and then released at high power as needed. The same day I found three species of Onthophagus, I also found an Alaus oculatus (the largest Elateridae I have seen in situ) near a rather large gallery of Bess beetles. While I was handling it, it snapped at me, and the power of it surprised me completely! I should have thought to try and record it in slow motion, but it's too late now. After a night in the freezer... Cheers, Brad On Mon, Apr 20, 2020 at 7:45 PM Tom Knight <tk@csail.mit.edu> wrote:
One of the most powerful animals is the mantis shrimp (neither a mantis, nor a shrimp, but …): https://www.youtube.com/watch?v=E0Li1k5hGBE < https://www.youtube.com/watch?v=E0Li1k5hGBE>
On Apr 20, 2020, at 6:46 PM, Brad Klee <bradklee@gmail.com> wrote:
Arg, wrong again! Yes that is the work integral, so needs another 1/delta t. Thanks for the correction.
However, the mistaken formula doesn't change my argument, which is that P = m*(delta x)^2 / (delta t)^3 is not the correct formula because actually, P is proportional to a Force F, which usually isn't proportional to (delta x) / (delta t)^2.
The math works out nice in this block problem, but we don't really need another block problem. It isn't physically realistic anyways. What is a "frictionless surface"? And what about drag? Is the block in a vacuum? Why does the block matter? Etc.
In the problem about runners, the resistive force combines a percentage of body weight and drag force. Drag is significant especially for sprinters like Usain Bolt, who are running very fast. The drag proportionality is v^2.
We need runners b/c physical fitness helps to improve bodily health, and to improve chances of surviving viral sickness!
An even better runner problem is to look closely at what happens during the acceleration phase of a 100m dash. Not only are forces imbalanced, the resistive force increases as a function of time and /or velocity. Yet Usain Bolt can still exceed 2 kilowatts of human body power.
https://www.popsci.com/science/article/2013-07/physics-record-breaking-run/
Amazingly, some of the beetles reflect mostly green, left-circular polarized light. If that is not enough they are stronger on average than the strongest humans, most likely more powerful, and easy to find in Arkansas (if you are willing to break open a dung pat):
https://www.livescience.com/8145-super-bug-world-strongest-insect-revealed.h...
https://www.inaturalist.org/observations/41877497
I don't know if anyone has directly measured the beetle power (I haven't read about it), but it should be done! Too bad about the march purge. The labs are shut down right now anyways...
Cheers,
Brad
--Brad
On Mon, Apr 20, 2020 at 1:21 PM William R Somsky <wrsomsky@gmail.com> wrote:
Way too unspecified.
Brad, you seem to be considering "steady-state" motion at a constant velocity, where the power counteracts some resistive force. And your integral is wrong. P = F v. The time integral of this is E, energy. Check the units.
Bill, you would seem to I be considering accelerating an object, where the power is used to the increase the kinetic energy if the object. It's all going to be dependent on the initial velocity before the power/force is applied.
Let me take a stab at an example elementary physics problem:
You have a block of mass m on a frictionless surface, initially at rest at x = 0. Starting at t = 0, you are going to apply a force to the block in the positive x direction in such a way that the rate of change of the kinetic energy of the block is a constant, ie, dK/dt = P, where P is the power "being applied"/required/however you want to term it in words.
Now: a) What is the kinetic energy of the block as a function of time? b) What is the velocity of the block as a function of time? c) What is the position of the block as a function of time? d) How does the position of the block at a time 2t compare with the position at time t?
a) The block starts at rest so K(t=0)=0, and K is increasing at a constant rate dK/dt=P, hence: K = P t ((for t>=0))
b) The kinetic energy of the moving block is K=(1/2) m v^2, so v = Sqrt[2K/m] = Sqrt[2 P t / m]
c) The block starts at x=0 so x(t=0)=0, and v=dx/dt, hence x = Int v dt = Int Sqrt[2 P t / m] dt = ( Sqrt[2 P / m] (2/3) t^(3/2) )
d) x(2t)/x(t) = ... = 2^(3/2), so the block goes Sqrt[8] times as far.
On Mon, Apr 20, 2020, 10:28 Brad Klee <bradklee@gmail.com> wrote:
No, forces are applied, power is an intrinsic property, so you might rather say "Given two movers with equal average power, one that goes twice as long also goes eight times as far"; however, this is also incorrect. The integral equation P = Int F*v*dt, says that power depends on resistive force. This is somewhat confusing when thinking of, say, a sprinter or a marathon runner because it isn't clear what the resistive force is. It is more complicated than you might guess:
http://www.georgeron.com/2017/09/kipchoge-running-power-marathon.html
If you want a high school problem to work out, here is one I wrote earlier this month:
--Brad
On Mon, Apr 20, 2020 at 12:06 PM Bill Gosper <billgosper@gmail.com> wrote:
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3]
Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3]
This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Power = Force*Speed = Force*Distance/Time = Work/Time It doesn't matter whether the force is producing acceleration or just overcoming friction or gravity. Brent On 4/20/2020 3:46 PM, Brad Klee wrote:
Arg, wrong again! Yes that is the work integral, so needs another 1/delta t. Thanks for the correction.
However, the mistaken formula doesn't change my argument, which is that P = m*(delta x)^2 / (delta t)^3 is not the correct formula because actually, P is proportional to a Force F, which usually isn't proportional to (delta x) / (delta t)^2.
The math works out nice in this block problem, but we don't really need another block problem. It isn't physically realistic anyways. What is a "frictionless surface"? And what about drag? Is the block in a vacuum? Why does the block matter? Etc.
In the problem about runners, the resistive force combines a percentage of body weight and drag force. Drag is significant especially for sprinters like Usain Bolt, who are running very fast. The drag proportionality is v^2.
We need runners b/c physical fitness helps to improve bodily health, and to improve chances of surviving viral sickness!
An even better runner problem is to look closely at what happens during the acceleration phase of a 100m dash. Not only are forces imbalanced, the resistive force increases as a function of time and /or velocity. Yet Usain Bolt can still exceed 2 kilowatts of human body power.
https://www.popsci.com/science/article/2013-07/physics-record-breaking-run/
Amazingly, some of the beetles reflect mostly green, left-circular polarized light. If that is not enough they are stronger on average than the strongest humans, most likely more powerful, and easy to find in Arkansas (if you are willing to break open a dung pat):
https://www.livescience.com/8145-super-bug-world-strongest-insect-revealed.h... https://www.inaturalist.org/observations/41877497
I don't know if anyone has directly measured the beetle power (I haven't read about it), but it should be done! Too bad about the march purge. The labs are shut down right now anyways...
Cheers,
Brad
--Brad
On Mon, Apr 20, 2020 at 1:21 PM William R Somsky <wrsomsky@gmail.com> wrote:
Way too unspecified.
Brad, you seem to be considering "steady-state" motion at a constant velocity, where the power counteracts some resistive force. And your integral is wrong. P = F v. The time integral of this is E, energy. Check the units.
Bill, you would seem to I be considering accelerating an object, where the power is used to the increase the kinetic energy if the object. It's all going to be dependent on the initial velocity before the power/force is applied.
Let me take a stab at an example elementary physics problem:
You have a block of mass m on a frictionless surface, initially at rest at x = 0. Starting at t = 0, you are going to apply a force to the block in the positive x direction in such a way that the rate of change of the kinetic energy of the block is a constant, ie, dK/dt = P, where P is the power "being applied"/required/however you want to term it in words.
Now: a) What is the kinetic energy of the block as a function of time? b) What is the velocity of the block as a function of time? c) What is the position of the block as a function of time? d) How does the position of the block at a time 2t compare with the position at time t?
a) The block starts at rest so K(t=0)=0, and K is increasing at a constant rate dK/dt=P, hence: K = P t ((for t>=0))
b) The kinetic energy of the moving block is K=(1/2) m v^2, so v = Sqrt[2K/m] = Sqrt[2 P t / m]
c) The block starts at x=0 so x(t=0)=0, and v=dx/dt, hence x = Int v dt = Int Sqrt[2 P t / m] dt = ( Sqrt[2 P / m] (2/3) t^(3/2) )
d) x(2t)/x(t) = ... = 2^(3/2), so the block goes Sqrt[8] times as far.
On Mon, Apr 20, 2020, 10:28 Brad Klee <bradklee@gmail.com> wrote:
No, forces are applied, power is an intrinsic property, so you might rather say "Given two movers with equal average power, one that goes twice as long also goes eight times as far"; however, this is also incorrect. The integral equation P = Int F*v*dt, says that power depends on resistive force. This is somewhat confusing when thinking of, say, a sprinter or a marathon runner because it isn't clear what the resistive force is. It is more complicated than you might guess:
http://www.georgeron.com/2017/09/kipchoge-running-power-marathon.html
If you want a high school problem to work out, here is one I wrote earlier this month:
--Brad
On Mon, Apr 20, 2020 at 12:06 PM Bill Gosper <billgosper@gmail.com> wrote:
In[395]:= UnitConvert[Quantity[42, "Watts"], "Kilograms" "Meters"^2/"Seconds"^3]
Out[395]= Quantity[42, ("Kilograms" ("Meters")^2)/("Seconds")^3]
This seems to say: If you apply a given amount of power (to a freely moving object) for twice as long, you go eight times as far. —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Either I don’t know, or I don’t care. Since the thread title isn’t asking for standard formulae, I am erring toward the side of real-world examples. Here’s another one, FIREFLY POWER. A firefly lights it’s lantern to 0.03 lumens, and a scientist (so he claims) estimates the generating power as 0.045mW. Is the estimate believable? Why or why not? —Brad
On Apr 20, 2020, at 9:55 PM, Brent Meeker via math-fun <math-fun@mailman.xmission.com> wrote:
Power = Force*Speed = Force*Distance/Time = Work/Time
It doesn't matter whether the force is producing acceleration or just overcoming friction or gravity.
Brent
Well, that’s the power delivered by the light, since 1 lux delivers approximately 1/683 watts for light at the optimal human sensitivity peak. Fireflies are pretty close to that peak, I think. So he is probably calculating .03 lumens/683 as the power, which is roughly right. Now, I don’t know what you are really asking. The firefly will use a lot more power to create that light, which is probably what you want to know. How far up the chemical pathway do you want to go?
On Apr 23, 2020, at 12:01 AM, Brad Klee <bradklee@gmail.com> wrote:
Either I don’t know, or I don’t care. Since the thread title isn’t asking for standard formulae, I am erring toward the side of real-world examples. Here’s another one, FIREFLY POWER.
A firefly lights it’s lantern to 0.03 lumens, and a scientist (so he claims) estimates the generating power as 0.045mW.
Is the estimate believable? Why or why not?
—Brad
On Apr 20, 2020, at 9:55 PM, Brent Meeker via math-fun <math-fun@mailman.xmission.com> wrote:
Power = Force*Speed = Force*Distance/Time = Work/Time
It doesn't matter whether the force is producing acceleration or just overcoming friction or gravity.
Brent
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According to [1], the conversion factor is about 715 at 540 nm. This is very near the mercury reference line at 546 nm, which is on center of the firefly emission, see also [2]. The conversion factor 683 is also reasonable, and for both, 0.045 mW is an overestimate, by efficiency factors either 93.2% or 97.6%. The question is not really about power, rather about efficiency. LEDs are very comparable in terms of quantum mechanical process, but typical efficiency statistics are much lower, 75-100 lm / Watt. The target is to get a comparable statistic for fireflies. The chemical pathway probably should be traced back to inhibition of mitochondrial oxygen consumption, whether or not this is accomplished using nitric oxide [3]. I don't know, but could hope that the fireflies would be considerably better than 10% efficient. Of course, the inverse process of photosynthesis is more important. Something similar happens... photon capture is extremely efficient, but energy conversion and dispersal involves more losses. --Brad [1] https://www.wolframalpha.com/input/?i=1+lumen+in+photons+per+second [2] https://rupress.org/jgp/article/48/1/95/31452/The-Spectral-Distribution-of-F... [3] https://academic.oup.com/jinsectscience/article/14/1/56/2386239 On Thu, Apr 23, 2020 at 7:43 AM Tom Knight <tk@csail.mit.edu> wrote:
Well, that’s the power delivered by the light, since 1 lux delivers approximately 1/683 watts for light at the optimal human sensitivity peak. Fireflies are pretty close to that peak, I think. So he is probably calculating .03 lumens/683 as the power, which is roughly right. Now, I don’t know what you are really asking. The firefly will use a lot more power to create that light, which is probably what you want to know. How far up the chemical pathway do you want to go?
On Apr 23, 2020, at 12:01 AM, Brad Klee <bradklee@gmail.com> wrote:
Either I don’t know, or I don’t care. Since the thread title isn’t asking for standard formulae, I am erring toward the side of real-world examples. Here’s another one, FIREFLY POWER.
A firefly lights it’s lantern to 0.03 lumens, and a scientist (so he claims) estimates the generating power as 0.045mW.
Is the estimate believable? Why or why not?
—Brad
On Apr 20, 2020, at 9:55 PM, Brent Meeker via math-fun < math-fun@mailman.xmission.com> wrote:
Power = Force*Speed = Force*Distance/Time = Work/Time
It doesn't matter whether the force is producing acceleration or just overcoming friction or gravity.
Brent
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participants (5)
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Bill Gosper -
Brad Klee -
Brent Meeker -
Tom Knight -
William R Somsky