Issue when reusing Column headers

[swopnil123]

One more thing I noticed. If we use ditto to access the first column from previously defined table, then then summation in the second column will go wrong.

Can you see this?

Originally posted by @swopnil123 in https://github.com/ronkok/Hurmet/issues/34#issuecomment-2429513390

[swopnil123]

¢` format = "h4" `

#### Vertical Distribution of Seismic Force

From earlier analysis

::: indented
¢` S_DS = 1.0 `          ¢` I = 1.0 `

¢` V_base = 400kips `, base shear        ¢` T = 0.75 sec `, fundamental period
of the structure
:::

From ASCE-16 section 12.8.3, Equivalent Lateral Procedure

::: indented
¢` k = {   1 if T ≤ 0.5;   2 if T ≥ 2.5;   1 + (T - 0.5) // 2 otherwise } = ? `

¢` C_vs = (w × h^k) / (Σ (w × h^k)) `, vertical distribution coefficient, Eq.
12.8-12

¢` F = C_vs V_base `, local force, Eq. 12.8-11
:::

¶

: dist: Vertical Distribution of Seismic Force
+--------+----------+---------+----------+-----------+--------------+--------+
| Floor  | Weight\  | Height\ | w × h^k^ | C~vs~     | Force\       | Shear\ |
|        | kips     | ft      |          |           | kips         | kips   |
+========+:========:+:=======:+:========:+:=========:+:============:+:======:+
| roof   | 950      | 70      | =B1×C1^k | =D1/D_end | ==E1× V_base | =F1    |
+--------+----------+---------+----------+-----------+--------------+--------+
| fifth  | 1,250    | 56      | "        | "         | "            | =G1+F2 |
+--------+----------+---------+----------+-----------+--------------+--------+
| fourth | "        | 42      | "        | "         | "            | "      |
+--------+----------+---------+----------+-----------+--------------+--------+
| third  | "        | 28      | "        | "         | "            | "      |
+--------+----------+---------+----------+-----------+--------------+--------+
| second | "        | 14      | "        | "         | "            | "      |
+--------+----------+---------+----------+-----------+--------------+--------+
| total  | =sum(up) | 0       | =sum(up) |           | =sum(up)     | =F_end |
+--------+----------+---------+----------+-----------+--------------+--------+
{#dist ."four-rules spreadsheet" colWidths="NaN NaN NaN NaN NaN NaN NaN"}

Next, let’s find the design diaphragm forces.

::: indented
¢` F_pxTrial = (∑F_(i=1)^n) / (∑W_(i=1)^n) W_i `, Eqn 12.10-1

¢` F_pxMin = 0.2 S_DS I w_px `. This usually controls.

¢` F_pxMax = 0.4 S_DS I w_px `
:::

: fram: Diaphragm Forces
+--------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| Floor  | Weight\  | ΣW\    | F\       | ΣF\      | F~pxTrial~\ | F~pxMin~\      | F~pxMax~\ | F~px~\              |
|        | kips     | kips   | kips     | kips     | kips        | kips           | kips      | kips                |
+:======:+:========:+:======:+:========:+:========:+:===========:+:==============:+:=========:+:===================:+
| roof   | 950      | =B1    | =dist.F1 | =dist.G1 | =E1/C1×B1   | =0.2×S_DS×I×B1 | =2×G1     | =min(max(F1,G1),H1) |
+--------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| fifth  | 1,250    | =C1+B2 | "        | "        | "           | "              | "         | "                   |
+--------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| fourth | "        | "      | "        | "        | "           | "              | "         | "                   |
+--------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| third  | "        | "      | "        | "        | "           | "              | "         | "                   |
+--------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| second | "        | "      | "        | "        | "           | "              | "         | "                   |
+--------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| total  | =sum(up) | =B_end | "        | "        |             |                |           |                     |
+--------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
{#fram ."four-rules spreadsheet" colWidths="69 71 NaN NaN NaN 83 64 NaN NaN"}

Same example discussed earlier.

[ronkok]

I have not been able to reproduce the error you describe. I copied the text from your comment above, then invoked the Paste from Markdown command in a blank document. The result has the correct value at the bottom of the second column.

Please try the same (copy and paste) steps. If you get the same results, we can explore how your other document arrived at an incorrect result.

[swopnil123]

Please use this

¢` format = "h4" `

#### Vertical Distribution of Seismic Force

From earlier analysis

::: indented
¢` S_DS = 1.0 `          ¢` I = 1.0 `

¢` V_base = 400kips `, base shear        ¢` T = 0.75 sec `, fundamental period
of the structure
:::

From ASCE-16 section 12.8.3, Equivalent Lateral Procedure

::: indented
¢` k = {   1 if T ≤ 0.5;   2 if T ≥ 2.5;   1 + (T - 0.5) // 2 otherwise } = ? `

¢` C_vs = (w × h^k) / (Σ (w × h^k)) `, vertical distribution coefficient, Eq.
12.8-12

¢` F = C_vs V_base `, local force, Eq. 12.8-11
:::

¶

: dist: Vertical Distribution of Seismic Force
+--------+----------+---------+----------+-----------+--------------+--------+
| Floor  | Weight\  | Height\ | w × h^k^ | C~vs~     | Force\       | Shear\ |
|        | kips     | ft      |          |           | kips         | kips   |
+========+:========:+:=======:+:========:+:=========:+:============:+:======:+
| roof   | 950      | 70      | =B1×C1^k | =D1/D_end | ==E1× V_base | =F1    |
+--------+----------+---------+----------+-----------+--------------+--------+
| fifth  | 1,250    | 56      | "        | "         | "            | =G1+F2 |
+--------+----------+---------+----------+-----------+--------------+--------+
| fourth | "        | 42      | "        | "         | "            | "      |
+--------+----------+---------+----------+-----------+--------------+--------+
| third  | "        | 28      | "        | "         | "            | "      |
+--------+----------+---------+----------+-----------+--------------+--------+
| second | "        | 14      | "        | "         | "            | "      |
+--------+----------+---------+----------+-----------+--------------+--------+
| total  | =sum(up) | 0       | =sum(up) |           | =sum(up)     | =F_end |
+--------+----------+---------+----------+-----------+--------------+--------+
{#dist ."four-rules spreadsheet" colWidths="NaN NaN NaN NaN NaN NaN NaN"}

Next, let’s find the design diaphragm forces.

::: indented
¢` F_pxTrial = (∑F_(i=1)^n) / (∑W_(i=1)^n) W_i `, Eqn 12.10-1

¢` F_pxMin = 0.2 S_DS I w_px `. This usually controls.

¢` F_pxMax = 0.4 S_DS I w_px `
:::

: fram: Diaphragm Forces
+----------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| Floor    | Weight\  | ΣW\    | F\       | ΣF\      | F~pxTrial~\ | F~pxMin~\      | F~pxMax~\ | F~px~\              |
|          | kips     | kips   | kips     | kips     | kips        | kips           | kips      | kips                |
+:========:+:========:+:======:+:========:+:========:+:===========:+:==============:+:=========:+:===================:+
| =dist.A1 | =dist.B1 | =B1    | =dist.F1 | =dist.G1 | =E1/C1×B1   | =0.2×S_DS×I×B1 | =2×G1     | =min(max(F1,G1),H1) |
+----------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| “        | “        | =C1+B2 | "        | "        | "           | "              | "         | "                   |
+----------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| “        | "        | "      | "        | "        | "           | "              | "         | "                   |
+----------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| “        | "        | "      | "        | "        | "           | "              | "         | "                   |
+----------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| “        | "        | "      | "        | "        | "           | "              | "         | "                   |
+----------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
| total    | =sum(up) | =B_end | "        | "        |             |                |           |                     |
+----------+----------+--------+----------+----------+-------------+----------------+-----------+---------------------+
{#fram ."four-rules spreadsheet" colWidths="69 71 NaN NaN NaN 83 64 NaN NaN"}

[ronkok]

Yes, now I can reproduce the error. I'll take a look at it. By the way, thank you for writing your example in Markdown. That makes it much easier for me to reproduce what you are seeing.

[ronkok]

The bug is fixed, now. Thank you for bringing this to my attention.