<p>Any Planetarium software that simulates precession will tell you this. If you know what to ask it.</p>
<p>Delta cancri has a Right ascension of 8hr 45min (or about 131 degrees). It has an ecliptic longitude of about 129 degrees.  These are angles measured from the vernal equinox parallel to equator and the ecliptic respectively.  I'm not sure how you get 84 degrees, that value is not correct. Note that Delta cancri will still be in close to the same position relative to distant stars 5000 years ago, or 1000 years in the future. It has a slow proper motion.</p>
<p>But if you mean the ecliptic longitude, relative to a moving equinox then planetarium software (eg Stellarium) tells me it was +32 in 5000BC and will be 143 degrees in 3000 AD.</p>
<p>You can then interpolate between those values. Linear interpolation would be reasonable, as the rate of precession has changed only a little over that time period. Planetarium software could be used to get a more precise value.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Year</th>
<th>Elip. Long.</th>
</tr>
</thead>
<tbody>
<tr>
<td>-5000</td>
<td>32.44</td>
</tr>
<tr>
<td>-4900</td>
<td>33.79411111</td>
</tr>
<tr>
<td>-4800</td>
<td>35.14883333</td>
</tr>
<tr>
<td>-4700</td>
<td>36.50416667</td>
</tr>
<tr>
<td>-4600</td>
<td>37.86011111</td>
</tr>
<tr>
<td>-4500</td>
<td>39.21666667</td>
</tr>
<tr>
<td>-4400</td>
<td>40.57383333</td>
</tr>
<tr>
<td>-4300</td>
<td>41.93161111</td>
</tr>
<tr>
<td>-4200</td>
<td>43.29</td>
</tr>
<tr>
<td>-4100</td>
<td>44.649</td>
</tr>
<tr>
<td>-4000</td>
<td>46.00861111</td>
</tr>
<tr>
<td>-3900</td>
<td>47.36883333</td>
</tr>
<tr>
<td>-3800</td>
<td>48.72966667</td>
</tr>
<tr>
<td>-3700</td>
<td>50.09111111</td>
</tr>
<tr>
<td>-3600</td>
<td>51.45316667</td>
</tr>
<tr>
<td>-3500</td>
<td>52.81583333</td>
</tr>
<tr>
<td>-3400</td>
<td>54.17911111</td>
</tr>
<tr>
<td>-3300</td>
<td>55.543</td>
</tr>
<tr>
<td>-3200</td>
<td>56.9075</td>
</tr>
<tr>
<td>-3100</td>
<td>58.27261111</td>
</tr>
<tr>
<td>-3000</td>
<td>59.63833333</td>
</tr>
<tr>
<td>-2900</td>
<td>61.00466667</td>
</tr>
<tr>
<td>-2800</td>
<td>62.37161111</td>
</tr>
<tr>
<td>-2700</td>
<td>63.73916667</td>
</tr>
<tr>
<td>-2600</td>
<td>65.10733333</td>
</tr>
<tr>
<td>-2500</td>
<td>66.47611111</td>
</tr>
<tr>
<td>-2400</td>
<td>67.8455</td>
</tr>
<tr>
<td>-2300</td>
<td>69.2155</td>
</tr>
<tr>
<td>-2200</td>
<td>70.58611111</td>
</tr>
<tr>
<td>-2100</td>
<td>71.95733333</td>
</tr>
<tr>
<td>-2000</td>
<td>73.32916667</td>
</tr>
<tr>
<td>-1900</td>
<td>74.70161111</td>
</tr>
<tr>
<td>-1800</td>
<td>76.07466667</td>
</tr>
<tr>
<td>-1700</td>
<td>77.44833333</td>
</tr>
<tr>
<td>-1600</td>
<td>78.82261111</td>
</tr>
<tr>
<td>-1500</td>
<td>80.1975</td>
</tr>
<tr>
<td>-1400</td>
<td>81.573</td>
</tr>
<tr>
<td>-1300</td>
<td>82.94911111</td>
</tr>
<tr>
<td>-1200</td>
<td>84.32583333</td>
</tr>
<tr>
<td>-1100</td>
<td>85.70316667</td>
</tr>
<tr>
<td>-1000</td>
<td>87.08111111</td>
</tr>
<tr>
<td>-900</td>
<td>88.45966667</td>
</tr>
<tr>
<td>-800</td>
<td>89.83883333</td>
</tr>
<tr>
<td>-700</td>
<td>91.21861111</td>
</tr>
<tr>
<td>-600</td>
<td>92.599</td>
</tr>
<tr>
<td>-500</td>
<td>93.98</td>
</tr>
<tr>
<td>-400</td>
<td>95.36161111</td>
</tr>
<tr>
<td>-300</td>
<td>96.74383333</td>
</tr>
<tr>
<td>-200</td>
<td>98.12666667</td>
</tr>
<tr>
<td>-100</td>
<td>99.51011111</td>
</tr>
<tr>
<td>0 = 1 BC</td>
<td>100.8941667</td>
</tr>
<tr>
<td>100</td>
<td>102.2788333</td>
</tr>
<tr>
<td>200</td>
<td>103.6641111</td>
</tr>
<tr>
<td>300</td>
<td>105.05</td>
</tr>
<tr>
<td>400</td>
<td>106.4365</td>
</tr>
<tr>
<td>500</td>
<td>107.8236111</td>
</tr>
<tr>
<td>600</td>
<td>109.2113333</td>
</tr>
<tr>
<td>700</td>
<td>110.5996667</td>
</tr>
<tr>
<td>800</td>
<td>111.9886111</td>
</tr>
<tr>
<td>900</td>
<td>113.3781667</td>
</tr>
<tr>
<td>1000</td>
<td>114.7683333</td>
</tr>
<tr>
<td>1100</td>
<td>116.1591111</td>
</tr>
<tr>
<td>1200</td>
<td>117.5505</td>
</tr>
<tr>
<td>1300</td>
<td>118.9425</td>
</tr>
<tr>
<td>1400</td>
<td>120.3351111</td>
</tr>
<tr>
<td>1500</td>
<td>121.7283333</td>
</tr>
<tr>
<td>1600</td>
<td>123.1221667</td>
</tr>
<tr>
<td>1700</td>
<td>124.5166111</td>
</tr>
<tr>
<td>1800</td>
<td>125.9116667</td>
</tr>
<tr>
<td>1900</td>
<td>127.3073333</td>
</tr>
<tr>
<td>2000</td>
<td>128.7036111</td>
</tr>
<tr>
<td>2100</td>
<td>130.1005</td>
</tr>
<tr>
<td>2200</td>
<td>131.498</td>
</tr>
<tr>
<td>2300</td>
<td>132.8961111</td>
</tr>
<tr>
<td>2400</td>
<td>134.2948333</td>
</tr>
<tr>
<td>2500</td>
<td>135.6941667</td>
</tr>
<tr>
<td>2600</td>
<td>137.0941111</td>
</tr>
<tr>
<td>2700</td>
<td>138.4946667</td>
</tr>
<tr>
<td>2800</td>
<td>139.8958333</td>
</tr>
<tr>
<td>2900</td>
<td>141.2976111</td>
</tr>
<tr>
<td>3000</td>
<td>142.7</td>
</tr>
</tbody>
</table>
</div>
CC BY-SA 4.0
2023-11-18T13:39:44.110Z
It has an ecliptic longitude of 128 deg? what exactly does that mean?
codingEnthu
CC BY-SA 4.0
2023-11-18T13:40:35.453Z
When I say ecliptic longitude I'm talking about the actual live distance from vernal equinox, that as of now for Delta Cancri is around 83 to 84
codingEnthu
CC BY-SA 4.0
2023-11-18T13:43:43.563Z
> But if you mean the ecliptic longitude, relative to a moving equinox then planetarium software (eg Stellarium) tells me it was +32 in 5000BC and will be 143 degrees in 3000 AD.

Yes where to find the decade wise data for this? or atleast century wise data, as this is not linear we can't calculate from a simple formula
codingEnthu
CC BY-SA 4.0
2023-11-18T13:46:45.653Z
As of now Delta Cancri is 129 degrees 3 minutes and 19.8 seconds.  relative to a moving equinox. measured parallel to the ecliptic.  Not 84 degrees You can generate this for any date with planetarium software (but it does increase linearly for delta cancri)
James K
CC BY-SA 4.0
2023-11-18T13:52:18.787Z
Are you getting this info from stellarium?
codingEnthu
CC BY-SA 4.0
2023-11-18T13:53:06.693Z
Yes................
James K
CC BY-SA 4.0
2023-11-18T14:08:37.517Z
Oh my bad, sorry instead of adding the difference to 106, I subtracted it. 

it is 106 + 23 = 129, I mistakenly did 106 - 23 = 83. Got it.

But my main question is how to get decade or century wise data of this? Because we can't do linear interpolation. It is non linear.
codingEnthu
CC BY-SA 4.0
2023-11-18T19:40:31.960Z
Why do you say "not linear"?  Precession is very stable, and hasn't changed much in millions of years. The ecliptic plane is likewise very stable.  Unless you are doing some super accurate positioning (which I doubt) linear interpolation would be good.  The eclptic longitude increases by 5028.8 arc seconds per century, though that value is itself increasing by 2.2 arcsecond per century, per century. With that you can get a very accurate position
James K
CC BY-SA 4.0
2023-11-19T01:31:07.617Z
[Wikipedia](https://en.wikipedia.org/wiki/Axial_precession#Values) gives this approximation for precession: `5028.796195*T + 1.1054348*T*T`, where `T` is Julian centuries since J2000.0, which is consistent with the numbers in James's last comment.
PM 2Ring
CC BY-SA 4.0
2023-11-19T18:16:59.160Z
@PM2Ring thanks for the approximation, it gives an error of 10 years for 1500 years approx. It is it possible to get the approximate equation for 0 AD? or 500 BC?
codingEnthu
CC BY-SA 4.0
2023-11-19T18:25:19.600Z
How do you know there is an error?
James K
CC BY-SA 4.0
2023-11-20T08:39:31.247Z
@JamesK Thanks a lot for the table, it really helps me 
Just a pedantic question in the table you have have given J2000 value to be  128.7036111, but when I calculated from Stellarium I got 128.72425 (128 Deg 43 Min 27.3 Sec)
codingEnthu
CC BY-SA 4.0
2023-11-20T08:41:23.637Z
@JamesK I used Stellarium to see when Delta Cancri was 106 Deg, it was in 368 Nov 11. But when I used (128.72425 - 106)*3600 = 5028.796195*T + 1.1054348*T*T equation to see when the equation predicts it to be 106 Deg, then the value of T is 16.210008904634684. Which is 1621 years from J2000, which is 379 AD, so there is an error of around 10 to 11 years.
codingEnthu
CC BY-SA 4.0
2023-11-20T18:36:09.970Z
Yes, that's about right.  The stellarium model is probably a little more complex. It might be documented somewhere, but it is just a model. You'd expect that sort of divergence of he models over 2000 years
James K
CC BY-SA 4.0
2023-11-18T13:18:42.043Z
2023-11-19T01:15:28.220Z
2023-11-19T01:15:28.220Z
James K
<p>I need historical data of tropical longitude of Delta Cancri.</p>
<p>Where can I get it. As of now it is approximately 128 Deg from Vernal Equinox.</p>
<p>Where can I get it's historical data, say from 5000 BC to 3000 AD?</p>

<star><history><ephemeris><equinox>
Where can I get historical data of tropical longitude of Delta Cancri
codingEnthu
# Where can I get historical data of tropical longitude of Delta Cancri

(Asked by: username_0 on 2023-11-18)

I need historical data of tropical longitude of Delta Cancri.

Where can I get it. As of now it is approximately 128 Deg from Vernal Equinox.

Where can I get it's historical data, say from 5000 BC to 3000 AD?

**Tags:** star, history, ephemeris, equinox


---

## 1 answers

### Answer by username_1 on 2023-11-18. Score: 2 | Accepted answer

Any Planetarium software that simulates precession will tell you this. If you know what to ask it.

Delta cancri has a Right ascension of 8hr 45min (or about 131 degrees). It has an ecliptic longitude of about 129 degrees. These are angles measured from the vernal equinox parallel to equator and the ecliptic respectively. I'm not sure how you get 84 degrees, that value is not correct. Note that Delta cancri will still be in close to the same position relative to distant stars 5000 years ago, or 1000 years in the future. It has a slow proper motion.

But if you mean the ecliptic longitude, relative to a moving equinox then planetarium software (eg Stellarium) tells me it was +32 in 5000BC and will be 143 degrees in 3000 AD.

You can then interpolate between those values. Linear interpolation would be reasonable, as the rate of precession has changed only a little over that time period. Planetarium software could be used to get a more precise value.

| Year | Elip. Long. |
| --- | --- |
| -5000 | 32.44 |
| -4900 | 33.79411111 |
| -4800 | 35.14883333 |
| -4700 | 36.50416667 |
| -4600 | 37.86011111 |
| -4500 | 39.21666667 |
| -4400 | 40.57383333 |
| -4300 | 41.93161111 |
| -4200 | 43.29 |
| -4100 | 44.649 |
| -4000 | 46.00861111 |
| -3900 | 47.36883333 |
| -3800 | 48.72966667 |
| -3700 | 50.09111111 |
| -3600 | 51.45316667 |
| -3500 | 52.81583333 |
| -3400 | 54.17911111 |
| -3300 | 55.543 |
| -3200 | 56.9075 |
| -3100 | 58.27261111 |
| -3000 | 59.63833333 |
| -2900 | 61.00466667 |
| -2800 | 62.37161111 |
| -2700 | 63.73916667 |
| -2600 | 65.10733333 |
| -2500 | 66.47611111 |
| -2400 | 67.8455 |
| -2300 | 69.2155 |
| -2200 | 70.58611111 |
| -2100 | 71.95733333 |
| -2000 | 73.32916667 |
| -1900 | 74.70161111 |
| -1800 | 76.07466667 |
| -1700 | 77.44833333 |
| -1600 | 78.82261111 |
| -1500 | 80.1975 |
| -1400 | 81.573 |
| -1300 | 82.94911111 |
| -1200 | 84.32583333 |
| -1100 | 85.70316667 |
| -1000 | 87.08111111 |
| -900 | 88.45966667 |
| -800 | 89.83883333 |
| -700 | 91.21861111 |
| -600 | 92.599 |
| -500 | 93.98 |
| -400 | 95.36161111 |
| -300 | 96.74383333 |
| -200 | 98.12666667 |
| -100 | 99.51011111 |
| 0 = 1 BC | 100.8941667 |
| 100 | 102.2788333 |
| 200 | 103.6641111 |
| 300 | 105.05 |
| 400 | 106.4365 |
| 500 | 107.8236111 |
| 600 | 109.2113333 |
| 700 | 110.5996667 |
| 800 | 111.9886111 |
| 900 | 113.3781667 |
| 1000 | 114.7683333 |
| 1100 | 116.1591111 |
| 1200 | 117.5505 |
| 1300 | 118.9425 |
| 1400 | 120.3351111 |
| 1500 | 121.7283333 |
| 1600 | 123.1221667 |
| 1700 | 124.5166111 |
| 1800 | 125.9116667 |
| 1900 | 127.3073333 |
| 2000 | 128.7036111 |
| 2100 | 130.1005 |
| 2200 | 131.498 |
| 2300 | 132.8961111 |
| 2400 | 134.2948333 |
| 2500 | 135.6941667 |
| 2600 | 137.0941111 |
| 2700 | 138.4946667 |
| 2800 | 139.8958333 |
| 2900 | 141.2976111 |
| 3000 | 142.7 |

> **Comments:**
> **username_0**: It has an ecliptic longitude of 128 deg? what exactly does that mean?  
> **username_0**: When I say ecliptic longitude I'm talking about the actual live distance from vernal equinox, that as of now for Delta Cancri is around 83 to 84  
> **username_0**: > But if you mean the ecliptic longitude, relative to a moving equinox then planetarium software (eg Stellarium) tells me it was +32 in 5000BC and will be 143 degrees in 3000 AD.  Yes where to find the decade wise data for this? or atleast century wise data, as this is not linear we can't calculate from a simple formula  
> **username_1**: As of now Delta Cancri is 129 degrees 3 minutes and 19.8 seconds.  relative to a moving equinox. measured parallel to the ecliptic.  Not 84 degrees You can generate this for any date with planetarium software (but it does increase linearly for delta cancri)  
> **username_0**: Are you getting this info from stellarium?  
> **username_1**: Yes................  
> **username_0**: Oh my bad, sorry instead of adding the difference to 106, I subtracted it.   it is 106 + 23 = 129, I mistakenly did 106 - 23 = 83. Got it.  But my main question is how to get decade or century wise data of this? Because we can't do linear interpolation. It is non linear.  
> **username_1**: Why do you say "not linear"?  Precession is very stable, and hasn't changed much in millions of years. The ecliptic plane is likewise very stable.  Unless you are doing some super accurate positioning (which I doubt) linear interpolation would be good.  The eclptic longitude increases by 5028.8 arc seconds per century, though that value is itself increasing by 2.2 arcsecond per century, per century. With that you can get a very accurate position  
> **username_2**: [Wikipedia](https://en.wikipedia.org/wiki/Axial_precession#Values) gives this approximation for precession: `5028.796195*T + 1.1054348*T*T`, where `T` is Julian centuries since J2000.0, which is consistent with the numbers in James's last comment.  
> **username_0**: @PM2Ring thanks for the approximation, it gives an error of 10 years for 1500 years approx. It is it possible to get the approximate equation for 0 AD? or 500 BC?  
> **username_1**: How do you know there is an error?  
> **username_0**: @JamesK Thanks a lot for the table, it really helps me  Just a pedantic question in the table you have have given J2000 value to be  128.7036111, but when I calculated from Stellarium I got 128.72425 (128 Deg 43 Min 27.3 Sec)  
> **username_0**: @JamesK I used Stellarium to see when Delta Cancri was 106 Deg, it was in 368 Nov 11. But when I used (128.72425 - 106)*3600 = 5028.796195*T + 1.1054348*T*T equation to see when the equation predicts it to be 106 Deg, then the value of T is 16.210008904634684. Which is 1621 years from J2000, which is 379 AD, so there is an error of around 10 to 11 years.  
> **username_1**: Yes, that's about right.  The stellarium model is probably a little more complex. It might be documented somewhere, but it is just a model. You'd expect that sort of divergence of he models over 2000 years  

---

