Why must traveling waves have the same amplitude to form a standing wave?












11












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I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?






waves







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  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    19 hours ago
















11












$begingroup$


I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?






waves







share|cite|improve this question









$endgroup$












  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    19 hours ago














11












11








11


3



$begingroup$


I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?






waves







share|cite|improve this question









$endgroup$




I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?






waves




waves






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Julia KimJulia Kim

826




826












  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    19 hours ago


















  • $begingroup$
    Related question and discussions here
    $endgroup$
    – Aaron Stevens
    19 hours ago
















$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago




$begingroup$
Related question and discussions here
$endgroup$
– Aaron Stevens
19 hours ago










2 Answers
2






active

oldest

votes


















26












$begingroup$

If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.



In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.



enter image description here



If the travelling waves are of unequal amplitude then there is a net transfer of energy.

If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.



In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
enter image description here



If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.



The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
    $endgroup$
    – Aaron Stevens
    19 hours ago












  • $begingroup$
    The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
    $endgroup$
    – Carl Witthoft
    12 hours ago






  • 2




    $begingroup$
    @CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
    $endgroup$
    – Farcher
    12 hours ago










  • $begingroup$
    @Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
    $endgroup$
    – Carl Witthoft
    10 hours ago










  • $begingroup$
    All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
    $endgroup$
    – tpg2114
    5 hours ago





















14












$begingroup$

You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1 for the physicist's old trick of deducing general results from what happens in between special cases.
    $endgroup$
    – J.G.
    13 hours ago








  • 2




    $begingroup$
    Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
    $endgroup$
    – David Richerby
    11 hours ago






  • 1




    $begingroup$
    @DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
    $endgroup$
    – Yakk
    5 hours ago












  • $begingroup$
    @Yakk That seems to fill the gap -- thanks.
    $endgroup$
    – David Richerby
    5 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









26












$begingroup$

If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.



In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.



enter image description here



If the travelling waves are of unequal amplitude then there is a net transfer of energy.

If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.



In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
enter image description here



If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.



The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
    $endgroup$
    – Aaron Stevens
    19 hours ago












  • $begingroup$
    The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
    $endgroup$
    – Carl Witthoft
    12 hours ago






  • 2




    $begingroup$
    @CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
    $endgroup$
    – Farcher
    12 hours ago










  • $begingroup$
    @Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
    $endgroup$
    – Carl Witthoft
    10 hours ago










  • $begingroup$
    All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
    $endgroup$
    – tpg2114
    5 hours ago


















26












$begingroup$

If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.



In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.



enter image description here



If the travelling waves are of unequal amplitude then there is a net transfer of energy.

If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.



In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
enter image description here



If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.



The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
    $endgroup$
    – Aaron Stevens
    19 hours ago












  • $begingroup$
    The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
    $endgroup$
    – Carl Witthoft
    12 hours ago






  • 2




    $begingroup$
    @CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
    $endgroup$
    – Farcher
    12 hours ago










  • $begingroup$
    @Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
    $endgroup$
    – Carl Witthoft
    10 hours ago










  • $begingroup$
    All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
    $endgroup$
    – tpg2114
    5 hours ago
















26












26








26





$begingroup$

If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.



In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.



enter image description here



If the travelling waves are of unequal amplitude then there is a net transfer of energy.

If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.



In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
enter image description here



If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.



The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.



enter image description here






share|cite|improve this answer











$endgroup$



If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.



In this animation taken from Acoustics and Vibration Animations the amplitude of the reflected wave is the same as that of the incident wave.



enter image description here



If the travelling waves are of unequal amplitude then there is a net transfer of energy.

If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.



In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
enter image description here



If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.



The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









FarcherFarcher

51k339107




51k339107








  • 1




    $begingroup$
    S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
    $endgroup$
    – Aaron Stevens
    19 hours ago












  • $begingroup$
    The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
    $endgroup$
    – Carl Witthoft
    12 hours ago






  • 2




    $begingroup$
    @CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
    $endgroup$
    – Farcher
    12 hours ago










  • $begingroup$
    @Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
    $endgroup$
    – Carl Witthoft
    10 hours ago










  • $begingroup$
    All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
    $endgroup$
    – tpg2114
    5 hours ago
















  • 1




    $begingroup$
    S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
    $endgroup$
    – Aaron Stevens
    19 hours ago












  • $begingroup$
    The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
    $endgroup$
    – Carl Witthoft
    12 hours ago






  • 2




    $begingroup$
    @CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
    $endgroup$
    – Farcher
    12 hours ago










  • $begingroup$
    @Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
    $endgroup$
    – Carl Witthoft
    10 hours ago










  • $begingroup$
    All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
    $endgroup$
    – tpg2114
    5 hours ago










1




1




$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago






$begingroup$
S.McGrew would say the second case is also a standing wave based on comments/discussions here. You have made the argument much better than I was trying to.
$endgroup$
– Aaron Stevens
19 hours ago














$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago




$begingroup$
The second case still has a standing wave, plus a DC component. Shortly thereafter you will discover nonlinear crystals and Second-Harmonic Generation.
$endgroup$
– Carl Witthoft
12 hours ago




2




2




$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago




$begingroup$
@CarlWitthoft Is it really a dc (fixed value) component? Is it not a standing wave modulated by a travelling wave?
$endgroup$
– Farcher
12 hours ago












$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago




$begingroup$
@Farcher you may well be right. Over in optics-land, the frequencies are somewhat higher and we look at the situation a bit differently.
$endgroup$
– Carl Witthoft
10 hours ago












$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago






$begingroup$
All of the images from the Acoustics and Vibration Animations page have a clear statement at the top that they cannot be used in other websites... I know they are handy, but they shouldn't be put into answers without contacting the author and getting explicit permission.
$endgroup$
– tpg2114
5 hours ago













14












$begingroup$

You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1 for the physicist's old trick of deducing general results from what happens in between special cases.
    $endgroup$
    – J.G.
    13 hours ago








  • 2




    $begingroup$
    Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
    $endgroup$
    – David Richerby
    11 hours ago






  • 1




    $begingroup$
    @DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
    $endgroup$
    – Yakk
    5 hours ago












  • $begingroup$
    @Yakk That seems to fill the gap -- thanks.
    $endgroup$
    – David Richerby
    5 hours ago
















14












$begingroup$

You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1 for the physicist's old trick of deducing general results from what happens in between special cases.
    $endgroup$
    – J.G.
    13 hours ago








  • 2




    $begingroup$
    Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
    $endgroup$
    – David Richerby
    11 hours ago






  • 1




    $begingroup$
    @DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
    $endgroup$
    – Yakk
    5 hours ago












  • $begingroup$
    @Yakk That seems to fill the gap -- thanks.
    $endgroup$
    – David Richerby
    5 hours ago














14












14








14





$begingroup$

You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.






share|cite|improve this answer









$endgroup$



You can think of this in a different way. Suppose you have a standing wave made from two oppositely-directed waves with equal amplitudes. What if amplitude of one of them reduces a bit? You think the sum should still be a standing wave. OK, but what if you continue reducing the amplitude down to $0$? In this case you'll be left only with a single traveling wave, which is not a standing wave anymore. At which point should the standing wave switch to traveling one? The answer is simple: it stops being pure standing wave at the moment when the amplitudes of the two oppositely-traveling components become different. Instead, you have a superposition of a traveling wave and a standing one.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 19 hours ago









RuslanRuslan

9,69343173




9,69343173








  • 1




    $begingroup$
    +1 for the physicist's old trick of deducing general results from what happens in between special cases.
    $endgroup$
    – J.G.
    13 hours ago








  • 2




    $begingroup$
    Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
    $endgroup$
    – David Richerby
    11 hours ago






  • 1




    $begingroup$
    @DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
    $endgroup$
    – Yakk
    5 hours ago












  • $begingroup$
    @Yakk That seems to fill the gap -- thanks.
    $endgroup$
    – David Richerby
    5 hours ago














  • 1




    $begingroup$
    +1 for the physicist's old trick of deducing general results from what happens in between special cases.
    $endgroup$
    – J.G.
    13 hours ago








  • 2




    $begingroup$
    Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
    $endgroup$
    – David Richerby
    11 hours ago






  • 1




    $begingroup$
    @DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
    $endgroup$
    – Yakk
    5 hours ago












  • $begingroup$
    @Yakk That seems to fill the gap -- thanks.
    $endgroup$
    – David Richerby
    5 hours ago








1




1




$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago






$begingroup$
+1 for the physicist's old trick of deducing general results from what happens in between special cases.
$endgroup$
– J.G.
13 hours ago






2




2




$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago




$begingroup$
Hmm. I think this needs a bit more argument. You say that, obviously, the change-over is when the amplitudes become different. But what if the asker comes back and says that, obviously, the change-over is when the other amplitude becomes zero? Why does it have to be your endpoint, rather than the other one? I suppose your comment about it being a superposition kind of addresses this but it feels like you need a little more.
$endgroup$
– David Richerby
11 hours ago




1




1




$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago






$begingroup$
@DavidRicherby There is no continuous process that converts a standing wave to one moving at a fast speed travelling wave over an arbitrary short interval; there is a continuous process that converts a standing wave to a very slow travelling wave over an arbitrary short interval. Basically, "slowly becomes moving" is continuous, while "instantly stops/starts" is not.
$endgroup$
– Yakk
5 hours ago














$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago




$begingroup$
@Yakk That seems to fill the gap -- thanks.
$endgroup$
– David Richerby
5 hours ago


















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