<p>A simple way to estimate the implications is to simplify the problem. If the orbits of Earth, Sun and Moon were circular and in the same plane, and the Earth had no tilt, the only remaining tidal constituents would be $M_2$ and $S_2$ (also the overtides and combination tides: $M_4$, $MS_4$...). The rest of the tidal constituents can be expressed as linear combinations of their rates of change based on the remaining forcings: <a href="https://en.wikipedia.org/wiki/Orbital_eccentricity" rel="nofollow noreferrer">orbital eccentricity</a> and <a href="https://en.wikipedia.org/wiki/Orbital_inclination" rel="nofollow noreferrer">inclination</a> and <a href="https://en.wikipedia.org/wiki/Axial_tilt" rel="nofollow noreferrer">axial tilt</a>.  </p>

<p>The modulation of the amplitude $A_M$ of the main constituent wave, $\eta_M = A_M cos (\omega_M t)$, by another constituent of amplitude $A_c$ is: $\eta_t= (A_M+A_ccos(\omega_c t))cos(\omega_Mt)$. Expressing it as a linear combinations of the rates of change gives $\eta_t= A_Mcos(\omega_Mt)+{A_c\over2}cos(\omega_M-\omega_c)t+{A_c\over2}cos(\omega_M+\omega_c)t$, such that the original amplitude remains unchanged, while two extra components with slightly different frequencies appear.</p>

<p>In the case of the effect of the axial tilt, it modulates the main frequencies based on the solar parameters (the inclination of the Moon orbit is not affected by the axial tilt). The period of the solar declination is a sidereal year (365.2564 days). In the case of the modulation of the $M_2$, the resulting periodicities are 1 cycle/solar day &plusmn; 1 cycle/sidereal year. The two resulting constituents are $P_1$ and $K_1$. $P_1$ is known as the solar declinational diurnal constituent. The case of $K_1$ is slightly more complex as the inclination of the Moon also results in a contribution at this frequency. The rest of the tidal constituents are also modified accordingly, but their amplitudes are much smaller.</p>

<p><strong>In short, the removal of the axial tilt will result in the removal of the $P_1$ tidal constituent (and some other minor ones) and an amplitude decrease in the $K_1$ constituent.</strong></p>

<p><a href="https://i.sstatic.net/nzx2Z.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nzx2Z.gif" alt="K1 from AVISO"></a></p>

<p>$K_1$ amplitude and phase from <a href="http://www.aviso.altimetry.fr" rel="nofollow noreferrer">AVISO</a>. The maximum $K_1$ amplitudes of around 0.4m are found south of Alaska and west of the Kamchatka Peninsula in the Sea of Okhotsk. The <a href="https://www.mikepoweredbydhi.com/-/media/shared%20content/mike%20by%20dhi/flyers%20and%20pdf/product-documentation/mike-21-toolbox-global-tide-model.pdf?la=en" rel="nofollow noreferrer">amplitude of the $P_1$ constituent</a> is about half of $K_1$ with a maximum of around 0.2m in the Sea of Okhotsk. Overall, the effect of the tilt on tides is pretty minor relative to the rest of the effects.</p>

CC BY-SA 3.0
2016-08-29T15:55:34.060Z
2016-08-29T15:55:34.060Z
arkaia
<p>I cannot answer your hypothetical question, but the tilt does probably affect the tides and there have been some studies on the topic. Hopefully they can inspire to further ideas and speculations. It appears that the main outcome from the report should be for you to understand how the moon, the Sun and the Earth control the tide and how the <a href="https://en.wikipedia.org/wiki/Tidal_force" rel="nofollow noreferrer">tidal force</a> works. </p>

<p><a href="https://i.sstatic.net/NiwZD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NiwZD.png" alt="Laskar 1986 Creative common Wikimedia"></a></p>

<p><em>Change in axial tilt over time from Laskar (1986)</em></p>

<p>We cannot remove the tilt completely, but we can look at the constant axial tilt change in the <a href="https://en.wikipedia.org/wiki/Milankovitch_cycles" rel="nofollow noreferrer">Milankovitch cycles</a>. Some believe that the impact of this 2.4° change is rather large. <a href="http://clivebest.com/blog/?p=7300" rel="nofollow noreferrer">Clive Best blog</a> gives the main arguments and suggests some good references. See also <a href="https://wattsupwiththat.com/2014/01/14/do-super-tides-kick-start-interglacials/" rel="nofollow noreferrer">this discussion</a> and <a href="http://www.pnas.org/content/97/8/3814.abstract" rel="nofollow noreferrer">this paper</a> on tidal change and glaciation.</p>

<p>However, it's difficult to specify how much of the impact that is directly related to the tilt change, as sea level change and glaciation also affects and interacts with the tides pattern. </p>

<p><a href="https://i.sstatic.net/EiZ9e.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/EiZ9e.png" alt="Figure from Wilmes and Green 2014"></a></p>

<p><em>This model from Wilmes and Green (2014) suggests much higher tides at high latitudes during LGM, 21.000 years ago. The model is based on changes in sea-level and glaciation.</em></p>

<p>From the astronomical data, it however does appear that the impact (tractional acceleration) of the Moon was much higher at large latitudes when the obliquity of the tilt is large (see figure) and this resulted in higher tides at latitudes where glaciation forms. </p>

<p>With less or no tilt, tides at large latitudes should be even smaller and maybe contribute to less breakup of ice sheets. </p>

CC BY-SA 3.0
2016-08-29T09:33:22.677Z
2016-08-29T09:33:22.677Z
user2821
<p>I would like to know what would be the impacts/effects on the tides if earth were to have no tilt (perpendicular to plane of ecliptic)? </p>

CC BY-SA 3.0
2016-08-28T14:57:28.717Z
What's the actual problem that you are trying to solve?
410 gone
CC BY-SA 3.0
2016-08-28T23:12:56.913Z
What other assumptions are you considering (i.e. does the Earth still have a moon whose orbit is tilted about 5° from the ecliptic, that that moon is about the same size; or do we assume that there is no moon, or that, since the Earth was never knocked on its side, that the moon orbits in the ecliptic)?
BillDOe
CC BY-SA 3.0
2016-08-29T02:13:16.880Z
The question is too general to be answerable. In the mid-oceans the tidal bulge is very small. It is greatly amplified close to land, and is highly sensitive to the continental slope and land configuration. To answer in full would require running a global program on a super-computer. There's nothing wrong with running 'what-if' models, but there has to be a point. 'Zero tilt' will never happen, so is there a point to this?
Gordon Stanger
CC BY-SA 3.0
2016-08-29T06:51:18.970Z
in my science class, we have to write a report and one of the sections is focussed on if the earth had a differnt tilt, such as no tilt, 90 degree tilt etc; we then have to do research on this. I chose no tilt, and could not find any information on this, so that is why I asked. In this case, i am assuming that the earth still as a moon, and that moon does orbit in the ecliptic. I do agree though, it is kind of pointless, as it will never happen.                                                                                                                                      -Kait-
Kait
CC BY-SA 3.0
2016-08-29T16:00:50.433Z
I think the question has some clear answers as can be seen below. "What if" questions are extremely useful as they can help simplify very complex problems. In this case, it helps understand the way the different tidal constituents are originated and their relative effects.
arkaia
<ocean><earth-rotation><tides><axial-obliquity>
What would be the impact on tides if the earth had no tilt?
Kait
# What would be the impact on tides if the earth had no tilt?

(Asked by: username_0 on 2016-08-28)

I would like to know what would be the impacts/effects on the tides if earth were to have no tilt (perpendicular to plane of ecliptic)?

**Tags:** ocean, earth-rotation, tides, axial-obliquity

> **Comments:**
> **username_1**: What's the actual problem that you are trying to solve?  
> **username_2**: What other assumptions are you considering (i.e. does the Earth still have a moon whose orbit is tilted about 5° from the ecliptic, that that moon is about the same size; or do we assume that there is no moon, or that, since the Earth was never knocked on its side, that the moon orbits in the ecliptic)?  
> **username_3**: The question is too general to be answerable. In the mid-oceans the tidal bulge is very small. It is greatly amplified close to land, and is highly sensitive to the continental slope and land configuration. To answer in full would require running a global program on a super-computer. There's nothing wrong with running 'what-if' models, but there has to be a point. 'Zero tilt' will never happen, so is there a point to this?  
> **username_0**: in my science class, we have to write a report and one of the sections is focussed on if the earth had a differnt tilt, such as no tilt, 90 degree tilt etc; we then have to do research on this. I chose no tilt, and could not find any information on this, so that is why I asked. In this case, i am assuming that the earth still as a moon, and that moon does orbit in the ecliptic. I do agree though, it is kind of pointless, as it will never happen.                                                                                                                                      -username_0-  
> **username_4**: I think the question has some clear answers as can be seen below. "What if" questions are extremely useful as they can help simplify very complex problems. In this case, it helps understand the way the different tidal constituents are originated and their relative effects.  

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## 2 answers

### Answer by username_4 on 2016-08-29. Score: 5

A simple way to estimate the implications is to simplify the problem. If the orbits of Earth, Sun and Moon were circular and in the same plane, and the Earth had no tilt, the only remaining tidal constituents would be $M\_2$ and $S\_2$ (also the overtides and combination tides: $M\_4$, $MS\_4$...). The rest of the tidal constituents can be expressed as linear combinations of their rates of change based on the remaining forcings: [orbital eccentricity](https://en.wikipedia.org/wiki/Orbital_eccentricity) and [inclination](https://en.wikipedia.org/wiki/Orbital_inclination) and [axial tilt](https://en.wikipedia.org/wiki/Axial_tilt).

The modulation of the amplitude $A\_M$ of the main constituent wave, $\eta\_M = A\_M cos (\omega\_M t)$, by another constituent of amplitude $A\_c$ is: $\eta\_t= (A\_M+A\_ccos(\omega\_c t))cos(\omega\_Mt)$. Expressing it as a linear combinations of the rates of change gives $\eta\_t= A\_Mcos(\omega\_Mt)+{A\_c\over2}cos(\omega\_M-\omega\_c)t+{A\_c\over2}cos(\omega\_M+\omega\_c)t$, such that the original amplitude remains unchanged, while two extra components with slightly different frequencies appear.

In the case of the effect of the axial tilt, it modulates the main frequencies based on the solar parameters (the inclination of the Moon orbit is not affected by the axial tilt). The period of the solar declination is a sidereal year (365.2564 days). In the case of the modulation of the $M\_2$, the resulting periodicities are 1 cycle/solar day ± 1 cycle/sidereal year. The two resulting constituents are $P\_1$ and $K\_1$. $P\_1$ is known as the solar declinational diurnal constituent. The case of $K\_1$ is slightly more complex as the inclination of the Moon also results in a contribution at this frequency. The rest of the tidal constituents are also modified accordingly, but their amplitudes are much smaller.

**In short, the removal of the axial tilt will result in the removal of the $P\_1$ tidal constituent (and some other minor ones) and an amplitude decrease in the $K\_1$ constituent.**

[![K1 from AVISO](https://i.sstatic.net/nzx2Z.gif)](https://i.sstatic.net/nzx2Z.gif)

$K\_1$ amplitude and phase from [AVISO](http://www.aviso.altimetry.fr). The maximum $K\_1$ amplitudes of around 0.4m are found south of Alaska and west of the Kamchatka Peninsula in the Sea of Okhotsk. The [amplitude of the $P\_1$ constituent](https://www.mikepoweredbydhi.com/-/media/shared%20content/mike%20by%20dhi/flyers%20and%20pdf/product-documentation/mike-21-toolbox-global-tide-model.pdf?la=en) is about half of $K\_1$ with a maximum of around 0.2m in the Sea of Okhotsk. Overall, the effect of the tilt on tides is pretty minor relative to the rest of the effects.


---

### Answer by username_5 on 2016-08-29. Score: 2

I cannot answer your hypothetical question, but the tilt does probably affect the tides and there have been some studies on the topic. Hopefully they can inspire to further ideas and speculations. It appears that the main outcome from the report should be for you to understand how the moon, the Sun and the Earth control the tide and how the [tidal force](https://en.wikipedia.org/wiki/Tidal_force) works.

[![Laskar 1986 Creative common Wikimedia](https://i.sstatic.net/NiwZD.png)](https://i.sstatic.net/NiwZD.png)

*Change in axial tilt over time from Laskar (1986)*

We cannot remove the tilt completely, but we can look at the constant axial tilt change in the [Milankovitch cycles](https://en.wikipedia.org/wiki/Milankovitch_cycles). Some believe that the impact of this 2.4° change is rather large. [Clive Best blog](http://clivebest.com/blog/?p=7300) gives the main arguments and suggests some good references. See also [this discussion](https://wattsupwiththat.com/2014/01/14/do-super-tides-kick-start-interglacials/) and [this paper](http://www.pnas.org/content/97/8/3814.abstract) on tidal change and glaciation.

However, it's difficult to specify how much of the impact that is directly related to the tilt change, as sea level change and glaciation also affects and interacts with the tides pattern.

[![Figure from Wilmes and Green 2014](https://i.sstatic.net/EiZ9e.png)](https://i.sstatic.net/EiZ9e.png)

*This model from Wilmes and Green (2014) suggests much higher tides at high latitudes during LGM, 21.000 years ago. The model is based on changes in sea-level and glaciation.*

From the astronomical data, it however does appear that the impact (tractional acceleration) of the Moon was much higher at large latitudes when the obliquity of the tilt is large (see figure) and this resulted in higher tides at latitudes where glaciation forms.

With less or no tilt, tides at large latitudes should be even smaller and maybe contribute to less breakup of ice sheets.


---

